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Stream: learning: reading & references

Topic: reading through Baez's topos theory blog posts

David Egolf (Mar 26 2024 at 17:08):

I'm going to try an experiment, where I work through John Baez's topos theory blog posts while "thinking out loud" in this topic. (For discussion of this idea, see #meta: meta > "learning out loud"? ).

A few disclaimers: I'm not sure how far I'll get! Also, my discussion here is unlikely to be self-contained; if you want to follow along, you may need to reference the blog posts. And finally, I am almost certainly going to make a lot of mistakes!

Please feel free to join in, or just to point out when I'm confused about something! I can't guarantee I'll have the energy to give you a full response if you do so, but please know that I will still appreciate whatever you choose to post.

David Egolf (Mar 26 2024 at 17:29):

I'm going to take a puzzle/exercise-based approach. I find it helps me focus my thoughts to have a particular thing I'm trying to figure out. (Sometimes I'll even jump straight to an exercise before reading a section! Then that exercise helps motivate my reading.)

The first exercise I want to contemplate is this:

In many of these examples something nice happens. First, suppose we have $s \in F U$ and an open cover of $U$ by open sets $U_i$. Then we can restrict $s$ to $U_i$ getting something we can call $s|_{U_i}$. We can then further restrict this to $U_i \cap U_j$. And by the definition of presheaf, we have

$(s|_{U_i})|_{U_i \cap U_j} = (s|_{U_j})|_{U_i \cap U_j}$

In other words, if we take a guy in $F U$ and restrict it to a bunch of open sets covering $U$, the resulting guys agree on the overlaps $U_i \cap U_j$. Check that this follows from the definition of functor and some other facts!

Morgan Rogers (he/him) (Mar 26 2024 at 17:35):

The exercise is "Check that...", right? Do you have an idea of where to start? :wink:

David Egolf (Mar 26 2024 at 17:40):

Morgan Rogers (he/him) said:

The exercise is "Check that...", right? Do you have an idea of where to start? :wink:

Yes, that's right! And I think I do have an idea of where to start! It'll take me a minute to type it out, though.

David Egolf (Mar 26 2024 at 17:45):

We have a functor $F:\mathcal{O}(X)^{\mathrm{op}} \to \mathsf{Set}$. Here, $\mathcal{O}(X)^{\mathrm{op}}$ is the category of open sets of a topological space $X$, where we have a unique morphism from $U$ to $V$ exactly if $V \subseteq U$. I think of a morphism from $U$ to $V$ as saying "$U$ contains $V$".

The first thing I want to note is that $\mathcal{O}(X)^{\mathrm{op}}$ is a poset. Consequently, all diagrams commute in $\mathcal{O}(X)^{\mathrm{op}}$! In particular, this diagram commutes for any $U_i, U_j \subseteq U$:
diagram

David Egolf (Mar 26 2024 at 17:50):

Next, I'll use the fact that functors send commutative diagrams to commutative diagrams. That means that this diagram in $\mathsf{Set}$ also commutes:
diagram

by $r_{U \to V}$ I mean the "restriction function" that restricts things from $U$ to $V$, for $V \subseteq U$. This is the image under $F$ of the unique morphism from $U$ to $V$ in $\mathcal{O}(X)^{\mathrm{op}}$.

David Egolf (Mar 26 2024 at 17:53):

Now, let's pick some $s \in FU$. Since this diagram commutes, we have that $r_{U_i \to U_i \cap U_j} \circ r_{U \to U_i}(s) = r_{U_j \to U_i \cap U_j} \circ r_{U \to U_j}(s)$. I believe this is just different notation for the thing we wanted to prove!

Peva Blanchard (Mar 26 2024 at 18:01):

(side question: how do you make the diagrams (pictures) so fast?)

David Egolf (Mar 26 2024 at 18:12):

Peva Blanchard said:

(side question: how do you make the diagrams (pictures) so fast?)

I use this amazing website to draw the diagrams: https://q.uiver.app/ .
Then all I have to do is take screenshots, and paste them into my draft!

David Egolf (Mar 27 2024 at 17:33):

Alright, we next have our first official "puzzle"!

Puzzle. Let $X = \mathbb{R}$ and for each open set $U \subseteq \mathbb{R}$ take $F U$ to be the set of continuous real-valued functions on $U$. Show that with the usual concept of restriction of functions, $F$ is a presheaf and in fact a sheaf.

I'll start by seeking to show that $F$ is a presheaf.

David Egolf (Mar 27 2024 at 17:39):

To show that $F$ is a presheaf, we need to show that it is a functor $F: \mathcal{O}(\mathbb{R})^{\mathrm{op}} \to \mathsf{Set}$. Now, for each open $U \subseteq \R$, we have that $FU$ is the set of continuous-real valued functions on $U$. To talk about "continuous" functions means that there needs to be some topology on $U$. I think it's reasonable to assume that $U$ is equipped with the subspace topology it inherits from $\mathbb{R}$.

I'm a bit intrigued by the fact that we then have $FU = \mathsf{Top}(U, \mathbb{R})$, where $U$ is equipped with the subspace topology. This leads me to consider the functor $\mathsf{Top}(-,\mathbb{R}): \mathsf{Top}^{\mathrm{op}} \to \mathsf{Set}$.

David Egolf (Mar 27 2024 at 17:43):

I'd now like dream up some functor $G: \mathcal{O}(\mathbb{R})^{\mathrm{op}} \to \mathsf{Top}^{\mathrm{op}}$, so that we can express $F$ as $F = \mathsf{Top}(-, \mathbb{R}) \circ G$.

The first idea that comes to mind is to let $G$ act as follows:

• it sends each open subset of $\mathbb{R}$ to that same subset, viewed as a topological space equipped with the subspace topology it inherits from $\mathbb{R}$
• it sends each morphism $U \to V$ to the morphism from $U$ to $V$ corresponding to the inclusion function from $V$ to $U$

David Egolf (Mar 27 2024 at 17:44):

It remains to show that $G$ is a really a functor, and that $F = \mathsf{Top}(-, \mathbb{R}) \circ G$.

Peva Blanchard (Mar 27 2024 at 18:17):

I would have gone for a more direct proof that $F$ is functor.

David Egolf (Mar 27 2024 at 18:22):

To show that $G$ is a functor, I think the only tricky thing to check (unless I'm missing something) is as follows: We want to show that if $U$ and $V$, with $V \subseteq U$, are open subsets of $\mathbb{R}$ equipped with the subspace topology inherited from $\mathbb{R}$, then the inclusion function $i:V \to U$ is continuous.

To show this, I want to use this property of the subspace topology: "Let $Y$ be as subspace of $X$ and let $i:Y \to X$ be the inclusion map. Then for any topological space $Z$ a map $f: Z \to Y$ is continuous if and only if the composite map $i \circ f$ is continuous".

In our case, we have the inclusion map $i_{V \to U}: V \to U$ and the two inclusions to $\mathbb{R}$, namely: $i_V: V \to \mathbb{R}$ and $i_U: U \to \mathbb{R}$. Since $i_U \circ i_{V \to U} = i_V$ and $i_V$ is continuous, we conclude that $i_{V \to U}: V \to U$ is also continuous.

David Egolf (Mar 27 2024 at 18:24):

@Peva Blanchard I was strongly considering aiming for a more direct proof! I'll be interested to take a look at the "spoiler" in a bit!

David Egolf (Mar 27 2024 at 18:49):

It remains to show that $F = \mathsf{Top}(-, \mathbb{R}) \circ G$. To show this, it suffices to show the following:

1. $\mathsf{Top}(-, \mathbb{R}) \circ G$ sends each open subset $U$ of $\mathbb{R}$ to the set of real-valued continuous function on $U$
2. $\mathsf{Top}(-, \mathbb{R}) \circ G$ sends each morphism in $\mathcal{O}(\mathbb{R})^{\mathrm{op}}$ to a function that restricts functions

Let's consider $r:U \to V$ in $\mathcal{O}(\mathbb{R})^{\mathrm{op}}$. So, $V \subseteq U$. First, we note that $\mathsf{Top}(-, \mathbb{R}) \circ G(U)$ first equips $U$ with the subspace topology and then gives the set of all real-valued continuous functions on $U$.

Next, we consider $\mathsf{Top}(-, \mathbb{R}) \circ G(r)$. $G(r)$ is the morphism $i^{\mathrm{op}}$ from $U$ to $V$ corresponding to the (continuous) inclusion function $i:V \to U$. Then, $\mathsf{Top}(-, \mathbb{R})(i^{\mathrm{op}}) = i^*: \mathsf{Top}(U, \mathbb{R}) \to \mathsf{Top}(V, \mathbb{R})$. Here, $i^*$ acts by precomposition, so it sends a continuous function $f:U \to \mathbb{R}$ to the function $f \circ i:V \to \mathbb{R}$. We note that $\mathsf{Top}(-, \mathbb{R}) \circ G(r) = i^*$ is indeed acting to restrict functions, as desired.

David Egolf (Mar 27 2024 at 18:52):

Peva Blanchard said:

I would have gone for a more direct proof that $F$ is functor.

spoiler

Regarding the definition of $F$. On object: $F$ sends an open subset $U$ to the set $FU$ of continuous real-valued functions. On arrow: $F$ maps the containment $U \supseteq V$ to the function $FU \rightarrow FV$ that sends a continuous function $f$ on $U$ to its restriction on $V$. At this point, we could dive into the details of why the restriction of a continuous function on a subset of its domain is indeed a continuous function, given the relevant induced topologies on their domains. But that would be an exercise in topology, so ... I'll invoke the "reductio ad obvious-um" ...

The fact that $F$ preserve identities and composition amounts to:
* restricting to the entire domain is the same as not restricting anything at all
* when $U \supseteq V \supseteq W$, restricting a function $f : U \rightarrow \mathbb{R}$ on $V$ and then restricting on $W$ is the same as restricting $f$ directly on $W$.

I think the crux of the exercise is in proving that $F$ also satisfies the sheaf condition.

I've taken a look at this now! Thanks for chiming in! I went for the less direct approach in part because it felt like it made it easier for me to realize that there are topology things going on. (The "reduction ad obvious-um" is great :laughing:, but I wanted to try and work out the details this time!)

I think there's going to be a bit more topology involved in showing that $F$ also satisfies the sheaf condition... :upside_down:

Peva Blanchard (Mar 27 2024 at 19:09):

:smile: I see, I was not sure which level of details you wanted to work at.

Btw, I find this format very nice. I read John's blog post series on topos not that long ago, but I never took the time to do the puzzles in detail. If you don't mind I'll probably join you on some puzzles, using the "spoiler" feature.

David Egolf (Mar 27 2024 at 19:18):

Peva Blanchard said:

:smile: I see, I was not sure which level of details you wanted to work at.

Btw, I find this format very nice. I read John's blog post series on topos not that long ago, but I never took the time to do the puzzles in detail. If you don't mind I'll probably join you on some puzzles, using the "spoiler" feature.

That sounds awesome!

Julius Hamilton (Mar 28 2024 at 11:53):

(deleted)

Julius Hamilton (Mar 28 2024 at 11:55):

David Egolf said:

I'm going to take a puzzle/exercise-based approach. I find it helps me focus my thoughts to have a particular thing I'm trying to figure out. (Sometimes I'll even jump straight to an exercise before reading a section! Then that exercise helps motivate my reading.)

The first exercise I want to contemplate is this:

In many of these examples something nice happens. First, suppose we have $s \in F U$ and an open cover of $U$ by open sets $U_i$. Then we can restrict $s$ to $U_i$ getting something we can call $s|_{U_i}$. We can then further restrict this to $U_i \cap U_j$. And by the definition of presheaf, we have

$(s|_{U_i})|_{U_i \cap U_j} = (s|_{U_j})|_{U_i \cap U_j}$

In other words, if we take a guy in $F U$ and restrict it to a bunch of open sets covering $U$, the resulting guys agree on the overlaps $U_i \cap U_j$. Check that this follows from the definition of functor and some other facts!

I’ll aim to attempt this exercise today.

David Egolf (Mar 28 2024 at 16:44):

Julius Hamilton said:

I’ll aim to attempt this exercise today.

Sounds great! By the way, I think it's very much in the spirit of this topic to "think out loud" a bit on these exercises. So, if you feel like it, please feel free to share some thoughts on the exercise - whether you're stuck on it or whether you have completed it!

David Egolf (Mar 28 2024 at 16:45):

Here's the puzzle I was working on above:

Puzzle. Let $X = \mathbb{R}$ and for each open set $U \subseteq \mathbb{R}$ take $F U$ to be the set of continuous real-valued functions on $U$. Show that with the usual concept of restriction of functions, $F$ is a presheaf and in fact a sheaf.

We saw above that $F$ is a presheaf. It remains to show that $F$ is a sheaf.

David Egolf (Mar 28 2024 at 16:56):

$F$ is a sheaf if we can do the following:

• First, pick some $U \in \mathcal{O}(\mathbb{R})$, and a bunch of $U_i$ also in $\mathcal{O}(\mathbb{R})$ such that $U = \cup_i U_i$
• Next, for each $i$, pick some $s_i \in FU_i$ such that for all $i$ and $j$ if $U_i \cap U_j$ is nonempty, then ${s_i}|_{U_i \cap U_j} = {s_j}|_{U_i \cap U_j}$
• Then, given all these choices above, we always need to be able to "glue together" these $s_i$ to get a $s \in FU$ such that $s|_{U_i} = s_i$ for each $i$. Further, this $s$ needs to be unique.

David Egolf (Mar 28 2024 at 17:03):

To give an analogy to imaging, we might think of each $s_i$ as a picture of some region in space. Then we would like to be able to "stitch together" a bunch of pictures that agree on their overlaps to get one picture of a larger area. Depending on what conditions we place on the pictures in question, this may or may not be possible! For example if we care about "plausibility" in some sense, note that we can't always stitch together plausible images of small areas to make a plausible image of some larger area.

David Egolf (Mar 28 2024 at 17:12):

Let us assume that we have selected some $U \in \mathcal{O}(\mathbb{R})$, a bunch of $U_i \in \mathcal{O}(\mathbb{R})$ so that $U = \cup_i U_i$, and for each $i$ an $s_i \in FU_i$ so that these selected continuous-real-valued functions "agree on overlaps" in the sense mentioned above. We wish to show that these $s_i$ can be glued together to form a real-valued continuous function $s:U \to \mathbb{R}$ that restricts to $s_i$ on $U_i$, and further that this resulting function is unique.

Now, a function $s:U \to \mathbb{R}$ is completely determined by the value that it takes at each point in $U$. Since $\cup_i U_i = U$, an arbitrary point $x$ of $U$ is in some $U_i$. And since $s|_{U_i} = s_i$ we have $s(x) = s_i(x)$. So, the value of $s$ is determined at each point once we pick $s_i$ for each $i$. So, if $s$ exists, it is certainly unique.

David Egolf (Mar 28 2024 at 17:15):

Let's define $s:U \to \mathbb{R}$ as follows. For $x \in U$, if $x \in U_i$, let $s(x) = s_i(x)$. This definition gives us a function, because if $x \in U_i$ and $x \in U_j$, then $s_i(x) = s_j(x)$. It remains to show that $s$ is continuous.

David Egolf (Mar 28 2024 at 17:21):

At this point, I recall the "gluing lemma" from topology (the quote below is from "Introduction to Topological Manifolds" by Lee):

Let $X$ and $Y$ be topological spaces, and let $\{A_i\}$ be either an arbitrary open cover of $X$ or a finite closed cover of $X$. Suppose that we are given continuous maps $f_i: A_i \to Y$ that agree on overlaps: ${f_i}|_{A_i \cap A_j} = {f_j}_{A_i \cap A_j}$. Then there exists a unique continuous map $f: X \to Y$ whose restriction to each $A_i$ is equal to $f_i$ .

David Egolf (Mar 28 2024 at 17:23):

This lemma assures us that $s$ is indeed continuous! I think that concludes this puzzle. (If I was working this out offline, I'd consider trying to prove the relevant part of this "gluing lemma". But to keep this topic a bit more focused, I'll not do that here).

Reid Barton (Mar 28 2024 at 17:28):

Here's another (possibly tricky) puzzle for you. When you proved that $F$ was a presheaf, you introduced another functor, $G$. Does this proof that $F$ is a sheaf have some formulation that involves $G$?

Reid Barton (Mar 28 2024 at 17:29):

(To make things a bit simpler, I suggest getting rid of the "op"s in the definition of $G$.)

John Baez (Mar 28 2024 at 18:16):

I would like to see a proof of the gluing lemma! To my mind this is the most interesting part of the whole puzzle. It's also not hard to prove. In fact, I never knew anyone had stated it formally as a "lemma" in some book.

The reason it's important is this: to show that $\mathbb{R}$-valued continuous functions form a sheaf, it turns out that the crucial step - the only step where something about continuity matters - is the step where you show that continuity is a "local" property.

That is, you can check if a function is continuous by running around your space, looking at each point, and asking "is the function continuous here?" Your function is continuous iff the answer is "yes" at each point.

Later I give an example of a property that's not local: namely, for an $\mathbb{R}$-valued function to be bounded is not local. So there's no sheaf of bounded $\mathbb{R}$-valued functions.

So, by outsourcing this "gluing lemma" to some textbook, I think you're missing out on an insight that this puzzle was designed to deliver.

Peva Blanchard (Mar 28 2024 at 18:16):

David Egolf said:

This lemma assures us that $s$ is indeed continuous! I think that concludes this puzzle. (If I was working this out offline, I'd consider trying to prove the relevant part of this "gluing lemma". But to keep this topic a bit more focused, I'll not do that here).

This gluing business is really what clicked for me when trying to understand sheaves. Continuity is a "local" property, so it goes well with gluing. I'll try to prove the gluing lemma.

David Egolf (Mar 28 2024 at 19:06):

Thanks for pointing that out, @John Baez ! (And @Reid Barton , thanks for suggesting another puzzle - I'll potentially take a look at it in a bit! It sounds intriguing.)

I was already a bit "spoiled" on the proof of the gluing lemma, when I looked it up in my book earlier. Lee uses what he calls the "Local Criterion for Continuity" to prove the lemma in the case where the $\{A_i\}$ form an open cover of $X$. Here is a statement of the "Local Criterion for Continuity" from Lee's book:

A map $f:X \to Y$ between topological spaces is continuous if and only if each point of $X$ has a neighborhood on which (the restriction of ) $f$ is continuous.

If we accept this, then we can use this to prove that our $s \in FU$ is continuous (we recall that $s:U \to \mathbb{R}$). Let us pick an arbitrary point $x$ from $U \subseteq \mathbb{R}$. We want to show that there is a neighborhood of $x$ on which the restriction of $s$ is continuous. Since $U = \cup_i U_i$, we know that $x \in U_i$ for some $i$. The restriction of $s$ to $U_i$ is $s_i$, which we know is continuous. By the "Local Criterion for Continuity", we conclude that $s: U \to \mathbb{R}$ is continuous.

David Egolf (Mar 28 2024 at 19:08):

It might be good to take this a step further and prove the "Local Criterion for Continuity" described above. I'll probably give this a try in a bit!

John Baez (Mar 28 2024 at 22:23):

Yes, that "local criterion for continuity" is the key step. It's actually a wonderful fact: while continuity is defined in a somewhat "global" way for a function $f: X \to Y$, there turns out to be a concept of a function being "continuous at a point", such that $f$ is continuous iff it is continuous at each point $x \in X$.

If you know how to prove this in your sleep, then of course there's no need to prove it here! But otherwise, it's worth thinking about.

Julius Hamilton (Mar 29 2024 at 13:03):

David Egolf said:

Julius Hamilton said:

I’ll aim to attempt this exercise today.

Sounds great! By the way, I think it's very much in the spirit of this topic to "think out loud" a bit on these exercises. So, if you feel like it, please feel free to share some thoughts on the exercise - whether you're stuck on it or whether you have completed it!

That’s exactly what I want to do. I get self conscious that my amateur sloppiness feels like fluff. But I will. One sec.

Julius Hamilton (Mar 29 2024 at 13:20):

David Egolf said:

I'm going to take a puzzle/exercise-based approach. I find it helps me focus my thoughts to have a particular thing I'm trying to figure out. (Sometimes I'll even jump straight to an exercise before reading a section! Then that exercise helps motivate my reading.)

The first exercise I want to contemplate is this:

In many of these examples something nice happens. First, suppose we have $s \in F U$ and an open cover of $U$ by open sets $U_i$. Then we can restrict $s$ to $U_i$ getting something we can call $s|_{U_i}$. We can then further restrict this to $U_i \cap U_j$. And by the definition of presheaf, we have

$(s|_{U_i})|_{U_i \cap U_j} = (s|_{U_j})|_{U_i \cap U_j}$

In other words, if we take a guy in $F U$ and restrict it to a bunch of open sets covering $U$, the resulting guys agree on the overlaps $U_i \cap U_j$. Check that this follows from the definition of functor and some other facts!

I live under time constraints (like all of us) so if it seems like I could easily get the answers to these questions just by reading more, I’m just trying to make it clear that I was encouraged by David egolf to “think out loud” and it is helpful for me to be able to learn on-the-go like this. Thanks.

I took a real analysis class as an undergraduate but did not take a lot of other standard math classes. I never really studied topology.

The definition seems simple (I already looked it up). I want to make my thinking super rigorous which is why I am trying to formulate everything in Coq lately. That’s been fun but means I also need to learn more about Coq itself.

(This is meant to be on-topic, I’m saying I prefer to learn by expressing math in Coq).

Set is a fundamental keyword in Coq - I think a Type. I know there is the meta-type Sort as well. I am not clear in some regards how Coq treats types and sets differently. For example, I don’t know if Types have zero implication of their size, ie how many terms are assumed to exist in any given type.

I need to express “some specific element / term of type Set, but I don’t know which one (becusss it’s meant to be arbitrary)”. I think this is the Parameter keyword but not sure yet.

Parameter X : Set.

A topology is arguably of Coq Sort “Record”. I understand Record to be yet another of these mathematical ways of describing “collections” of things. A Record tries to have no commitment to any theories of mathematics, like Sets do. A record is a Type, but it can contain “multiple things”. So, a topology is:

A Record T
Consisting of a set X
And a set O of subsets of X
Such that the 3 topology conditions hold. They are:
Empty set and X are in S.
Arbitrary unions.
Finite intersections.
(I never took the time to think about why unions can be arbitrary but intersections have to be finite).

Actually, Topology should not be a record, it should use keyword Definition, since a Record is better for a single instance of an object? Not sure.

Record T : Type :=
| Parameter X : Set
| Parameter S : Set :=
(assume we just import a “power set function” until I have time to define one myself)
| (I haven’t learn how to state constraints on a type yet, but I need to state here the topology requirements. I guess it’s of type Prop, since they are Boolean. Some like Definition has_empty : S -> Prop := (assume we import a definition of empty set and set membership).

And so on.

There’s my thinking aloud before I head off to work. I wanted to work up to questions I had about presheafs. I was stuck on thinking about inclusion mappings.

Julius Hamilton (Mar 29 2024 at 13:21):

Baez’s post mentions that we need to reverse the direction of the arrows (O(X)^{op}) and I was trying to fully understand.

Julius Hamilton (Mar 29 2024 at 13:24):

I have actually been spending a lot of time thinking about what the real nature of being “functional” is. I know the common definition. But I want to see clearly why the properties of categories come from mappings being functional (sometimes).

An inclusion mapping is functional. It maps an element of a subset of S to the same element in the set S. How do you define that mathematically?

Julius Hamilton (Mar 29 2024 at 13:26):

I guess we can express “mapped to itself” with an expression like f: S1 -> S such that S1 \subset S and f(x) = x. This is allowed by the axiom of extensionality? Though they are elements in different sets, there is already an equality relation defined on the elements of the two respective sets.

Julius Hamilton (Mar 29 2024 at 13:26):

How can we reverse an inclusion mapping? It wouldn’t be functional. An inclusion mapping is injective but not surjective.

Julius Hamilton (Mar 29 2024 at 13:26):

Thanks!

David Egolf (Mar 29 2024 at 17:08):

Thanks for joining in @Julius Hamilton ! I don't know enough about Coq to understand your questions relating to it (hopefully someone else here does, though!).

I can talk a bit about how I understand $\mathcal{O}(X)$ and $\mathcal{O}(X)^{\mathrm{op}}$ though, in case that is helpful to you.

Julius Hamilton (Mar 29 2024 at 17:08):

Yeah please do

David Egolf (Mar 29 2024 at 17:14):

In my understanding, $\mathcal{O}(X)$ is a category that we can create when we're given a topology on the set $X$. To create this category, we need to say what its objects and morphisms are.

As you mentioned above, any topology on $X$ has a collection of subsets of $X$ - the "open sets". The open sets of our topology are the objects of $\mathcal{O}(X)$.

And given two open sets $U$ and $V$ in our topology on $X$, we can ask this question: Is $U$ a subset of $V$? If the answer to this question is "yes!", then we put a morphism from $U$ to $V$. Otherwise, we don't put a morphism from $U$ to $V$.

David Egolf (Mar 29 2024 at 17:17):

Then $\mathcal{O}(X)^{\mathrm{op}}$ is another category: it's just like $\mathcal{O}(X)$ except we turn all the arrows around. So now we put a morphism from $U$ to $V$ exactly if $V \subseteq U$.

David Egolf (Mar 29 2024 at 17:19):

I don't really think of these morphisms as functions. I think of them more like "yes!" answers to a yes/no question.

I'm not sure I was really addressing your point of confusion :sweat_smile:. Hopefully this is still somewhat helpful!

Julius Hamilton (Mar 29 2024 at 17:28):

Ok. Yeah that simplifies it dramatically.

Julius Hamilton (Mar 29 2024 at 17:28):

I think I was confused regarding how simple a morphism can be. I’ll think more about that.

Julius Hamilton (Mar 29 2024 at 17:39):

In many of these examples something nice happens. First, suppose we have $s \in F U$ and an open cover of $U$ by open sets $U_i$.

s is some arbitrary set.
U is an open set in O(X).
Does it matter if it turned out that s was a set in O(X)? Since O(X) is surely a sub-category of Set?

I’ll assume an open cover of U is a union of sets in O(X) such that U \subset of the union.

Then we can restrict $s$ to $U_i$ getting something we can call $s|_{U_i}$.

Baez’s post said we need to flip the morphisms precisely so we can “restrict” the functor. So this is saying, “only those elements of s such that there exists an element x in U_i for which F(x) \in s”. ?

We can’t do this without flipping the morphisms? I’d like to think about how. (Thinking out loud :wink:)

David Egolf (Mar 29 2024 at 17:53):

The notation $s \in FU$ means that $s$ is an element of the set $FU$. For example, depending on what $F$ does, $s$ might be a continuous real-valued function with domain of $U$.

John Baez (Mar 29 2024 at 17:56):

@Julius Hamilton - I think it's very helpful to focus on a specific example of a sheaf when trying to understand the definition of sheaf, and (unsurprisingly) I recommend the example David is talking about, where $FU$ is the set of continuous functions $f: U \to \mathbb{R}$.

Or, if "continuous" is distracting to you, think about the sheaf where $FU$ is the set of all functions $f: U \to \mathbb{R}$.

Or, if $\mathbb{R}$ is distracting to you, replace it with some other set.

If you read all the sheaf axioms keeping some example like this in mind, they should make more sense.

John Baez (Mar 29 2024 at 18:05):

(The reason I picked a sheaf of continuous functions is because topos theory originated as a generalization of topology, as the name suggests - so ideas from topology can help explain what people are doing in topos theory. There are also other ways to get into topos theory, but my course notes - and the book they're based on - start with topology. Luckily you only need to know a small amount of topology.)

David Egolf (Mar 29 2024 at 18:53):

In the spirit of trying to better understand why we can detect continuity of a function in a local way, I'll now to try to prove this:

A map $f:X \to Y$ between topological spaces is continuous if and only if each point of $X$ has a neighborhood on which (the restriction of ) $f$ is continuous.

I got stuck along the way. To get unstuck, I referenced Lee's book on topological manifolds. So, what I wrote below below closely follows the proof in Lee's book.

If $f$ is continuous, then for any point $x \in X$, we have that $X$ is a neighborhood of $x$ on which the restriction of $f$ (which is just $f$) is continuous.

For the other direction, we assume that each point of $X$ has a neighborhood (an open set) on which the restriction of $f$ is continuous. We wish to show that $f: X \to Y$ must then be continuous. To show that $f$ is continuous, we consider an arbitrary open subset $V$ of $Y$. We wish to show that the preimage of $V$ under $f$ is open in $X$.

I'll call this preimage $f^*(V)$. Now, a set $A$ is open exactly if every point $x$ in $A$ is contained in an open subset $U_x \subseteq A$. Thinking of openness in this way seems likely helpful, as it involves a condition that can be checked at each point - and we are trying to understand why continuity involves a condition that can be checked at each point.

Now, we know that any $x \in f^*(V)$ has some neighborhood $U_x$ such that $f|_{U_x}: U_x \to Y$ is continuous. Since $f|_{U_x}$ is continuous and $V$ is open, $(f|_{U_x})^*(V)$ is also open in the subspace topology on $U_x$. Thus, it is the intersection of some open set $A$ (in the topology on $X$) with $U_x$. Since $U_x$ is open and $A$ is open, $(f|_{U_x})^*(V)$ is also open (in the topology on $X$).

We are hoping that $(f|_{U_x})^*(V)$ is an open subset of $f^*(V)$ that contains $x$. This would provide the neighborhood of "breathing room" about $x$ in $f^*(V)$ that we need for $f^*(V)$ to be open.

By definition, $(f|_{U_x})^*(V)$ is the subset of $U_x$ that maps into $V$ under $f$. So, its elements are exactly those that are: (1) in $U_x$ and (2) map to $V$ under $f$. Thus, $(f|_{U_x})^*(V) = U_x \cap f^*(V)$. Note that $x \in U_x \cap f^*(V)$. We also note that $(f|_{U_x})^*(V) = U_x \cap f^*(V)$ is a subset of $f^*(V)$.

We conclude that an arbitrary point of $f^*(V)$ has a neighborhood contained in $f^*(V)$. Therefore $f$ is continuous.

Julius Hamilton (Mar 29 2024 at 19:02):

John Baez said:

Julius Hamilton - I think it's very helpful to focus on a specific example of a sheaf when trying to understand the definition of sheaf, and (unsurprisingly) I recommend the example David is talking about, where $FU$ is the set of continuous functions $f: U \to \mathbb{R}$.

Or, if "continuous" is distracting to you, think about the sheaf where $FU$ is the set of all functions $f: U \to \mathbb{R}$.

Or, if $\mathbb{R}$ is distracting to you, replace it with some other set.

If you read all the sheaf axioms keeping some example like this in mind, they should make more sense.

A presheaf $F$ is a functor from $O(X)^{op}$ to ${\mathbf Set}$. (Just restating definitions to exercise myself).

Baez says that we can see why we would want to take the opposite category of $O(X)$ if we think of one possible presheaf $F$ sending each $U \in Ob(O(X))$ to that set in $\{\mathbf Set\}$ that contains all real-valued functions over $U$.

Let’s consider some examples.

If $X$ is the real numbers, let’s consider a common topology over the reals. (Which is?)

Topologies are a way to express geometric concepts. Why are they so fundamentally defined in terms of “open sets”?

Perhaps it has to do with continuity and limits. Maybe it allows us to define the epsilon-delta condition without recourse to a distance metric?

Julius Hamilton (Mar 29 2024 at 19:06):

David Egolf said:

The first thing I want to note is that $\mathcal{O}(X)^{\mathrm{op}}$ is a poset. Consequently, all diagrams commute in $\mathcal{O}(X)^{\mathrm{op}}$!

Why?

Eric M Downes (Mar 29 2024 at 19:29):

Julius Hamilton said:

David Egolf said:

The first thing I want to note is that $\mathcal{O}(X)^{\mathrm{op}}$ is a poset. Consequently, all diagrams commute in $\mathcal{O}(X)^{\mathrm{op}}$!

Why?

1. Can you distinguish between the "generator arrows" of a poset, and the morphisms only in the closure? The latter are "required to be there" to satisfy the associativity of composition, given the generators; hint: three objects, with two morphisms $f,g;~dom.g=cod.f$ and $dom.f\neq cod.g$ apply transitivity of partial order; what have you got?)
2. Given two distinct objects $X,Y\in obj.\mathcal{O}(X)$, how many generator arrows are you allowed to draw? That is, how big is the homset $hom(X,Y)$?
3. Does moving to the opposite category change any of this?
4. So, if you draw a diagram with only morphisms within $\mathcal{O}(X)$ (no functors to other cats allowed!), how do (1) and (2) constrain said diagrams?

I'm sure there are more sophisticated ways of thinking about this, but that is how I approach it.

David Egolf (Mar 29 2024 at 19:58):

Julius Hamilton said:

David Egolf said:

The first thing I want to note is that $\mathcal{O}(X)^{\mathrm{op}}$ is a poset. Consequently, all diagrams commute in $\mathcal{O}(X)^{\mathrm{op}}$!

Why?

Here's how I think of it ("it" being "all diagrams commute in a poset"):

When we say (part of) a diagram "commutes", we mean that two different sequences of morphisms compose to the same morphism. If a category is a poset, it has at most one morphism from $A$ to $B$, for any objects $A$ and $B$. Therefore, if I have two different sequences of morphisms from $A$ to $B$, when I compose the morphisms in each sequence there's only one possible morphism for me to get as a result!

Consequently, these two sequences of morphisms must compose to the same morphism. And hence the corresponding (part of a) diagram must commute.

Julius Hamilton (Mar 29 2024 at 22:00):

David Egolf said:

Julius Hamilton said:

David Egolf said:

The first thing I want to note is that $\mathcal{O}(X)^{\mathrm{op}}$ is a poset. Consequently, all diagrams commute in $\mathcal{O}(X)^{\mathrm{op}}$!

Why?

Here's how I think of it ("it" being "all diagrams commute in a poset"):

When we say (part of) a diagram "commutes", we mean that two different sequences of morphisms compose to the same morphism. If a category is a poset, it has at most one morphism from $A$ to $B$, for any objects $A$ and $B$. Therefore, if I have two different sequences of morphisms from $A$ to $B$, when I compose the morphisms in each sequence there's only one possible morphism for me to get as a result!

Consequently, these two sequences of morphisms must compose to the same morphism. And hence the corresponding (part of a) diagram must commute.

That is such a beautifully simple explanation. You have a knack for clear simple understanding.

Julius Hamilton (Mar 29 2024 at 22:19):

Eric M Downes said:

1. Can you distinguish between the "generator arrows" of a poset, and the morphisms only in the closure?

I’d never thought of that before. I’m curious to know what those elements might be called in an abstract algebra setting. I’ve been thinking about how a one object category is a monoid, and if there is a corresponding abstract algebraic structure for a “multi-object category”. The thing is, not all the arrows (elements of the “structure”) compose with one another. All algebraic structures I know of are defined by closure and by “totality”.

I think your point is that all the arrows in a thin category are generators. Now I have to think about what how categories can have non-generator arrows.
I think the most basic way of expressing the compositional requirement of the arrows in a category is, “if they can compose, they do.” “If they are composable, then compose.”

I’ll come to the other points you made in a bit.

Julius Hamilton (Mar 30 2024 at 00:31):

David Egolf said:

Now, let's pick some $s \in FU$. Since this diagram commutes, we have that $r_{U_i \to U_i \cap U_j} \circ r_{U \to U_i}(s) = r_{U_j \to U_i \cap U_j} \circ r_{U \to U_j}(s)$. I believe this is just different notation for the thing we wanted to prove!

I believe I follow your argument piece by piece but want to digest it more.

Every diagram in a poset commutes.
Functors preserve commutative diagrams (why)?
In $O(X)^{op}$, the morphisms essentially say “contains”.
If we think of $F$ as mapping a set to a function defined on that set, we can think of the “contains” morphism in $O(X)^{op}$ as corresponding, under $F$, to a restriction of some function on some subset of its domain.
Baez basically just asks us to show that restricting some $s \in FU$ (which can be thought of as a function) to an open subset $U_i \in U$, and restrict it to some other subset $U_j \in U$, you can further restrict both of those restricted functions to $U_i \cap U_j$, and they are the same. Which David showed.

Questions:

1. What is an example of a function which does not have this property?
2. What if we stop thinking of $F$ as mapping a domain to a function over that domain? It is harder for me to regain the intuition of what the arrows in ${\mathbf {Set} }$ “mean” or “correspond to”.

Julius Hamilton (Mar 30 2024 at 00:40):

David Egolf said:

Alright, we next have our first official "puzzle"!

Puzzle. Let $X = \mathbb{R}$ and for each open set $U \subseteq \mathbb{R}$ take $F U$ to be the set of continuous real-valued functions on $U$. Show that with the usual concept of restriction of functions, $F$ is a presheaf and in fact a sheaf.

I'll start by seeking to show that $F$ is a presheaf.

In order to show it is a presheaf, I think we have to show $X$ has a natural topology and forms a thin category, and then that $F$ fulfills the functor axioms (if we reverse the direction of the arrows). I’ve been trying to tell myself “a functor is a morphism in the category of categories” as a single idea to remind myself of the definition. I think the important thing is if two arrows $f, g$ in $C$ compose, then arrows $Ff, Fg$ must compose in such a way that $Fg \circ Ff = F(g \circ f)$. Which basically says, “when you map the morphisms over, you can take the composition before, or you can take it after”.

Julius Hamilton (Mar 30 2024 at 00:42):

I’ll follow the rest of David’s proof a little later.

David Egolf (Mar 30 2024 at 02:35):

I don't have energy right now to respond in detail to your comments, @Julius Hamilton. But I did notice that above you asked "why?", regarding the fact that functors send commutative diagrams to commutative diagrams. You might find it a helpful exercise to pick a particular (simple) commutative diagram, and then aim to show that applying a functor $F$ to that diagram yields a commutative diagram.

Julius Hamilton (Mar 30 2024 at 02:35):

All good brotha rest up. Yes I will think about that.

David Tanzer (Mar 30 2024 at 04:47):

To prove it, need to first formalize what it means for a diagram to be commutative.

David Tanzer (Mar 30 2024 at 06:16):

As an aside, in an alternative definition, one could take preserving diagram commutativity as a defining characteristic of a functor; then recover $T(f \circ g) = T(f) \circ T(g)$ by applying this to a simple diagram. Not as efficient as a technical definition, but seems conceptually useful.

David Egolf (Mar 30 2024 at 16:14):

Reid Barton said:
Before moving on to the next puzzle, I want think about this for a bit:

Here's another (possibly tricky) puzzle for you. When you proved that $F$ was a presheaf, you introduced another functor, $G$. Does this proof that $F$ is a sheaf have some formulation that involves $G$?

The words "possibly tricky" strike some fear into my heart, but this sounds fun to think about - so I'll see what I can figure out...

David Egolf (Mar 30 2024 at 16:22):

First, let's recall the context:

• $F: {\mathcal{O}(\mathbb{R})}^{\mathrm{op}} \to \mathsf{Top}$ sends each open subset of $\mathbb{R}$ to the set of real-valued continuous functions on that subset. And $F$ sends a morphism $:A \to B$ in ${\mathcal{O}(\mathbb{R})}^{\mathrm{op}}$ to the restriction function from $\mathsf{Top}(A, \mathbb{R})$ to $\mathsf{Top}(B, \mathbb{R})$.
• $G: {\mathcal{O}(\mathbb{R})}^{\mathrm{op}} \to \mathsf{Top}^\mathrm{op}$ sends each open subset of $\mathbb{R}$ to the same subset of $\mathbb{R}$, equipped with the subspace topology. And $G$ sends a morphism $:A \to B$ to the morphism from $A$ to $B$ in $\mathsf{Top}^{\mathrm{op}}$ corresponding to the inclusion function from $B$ to $A$.
• We saw above that $F = \mathsf{Top}(-, \mathbb{R}) \circ G$ is a sheaf.

David Egolf (Mar 30 2024 at 21:53):

Imagine we have two open sets $U_1$ and $U_2$ (in the standard topology on $\mathbb{R}$) with $U = U_1 \cup U_2$. For $\mathsf{Top}(-, \mathbb{R}) \circ G$ to be a sheaf, I think we need to be able to construct a unique real-valued continuous function $s:GU \to \mathbb{R}$ from a pair of real-valued continuous functions $s_1:GU_1 \to \mathbb{R}$ and $s_2:GU_2 \to \mathbb{R}$, provided that $s_1$ and $s_2$ agree on $GU_1 \cap GU_2$. [Or should they agree on $G(U_1 \cap U_2)$? Still figuring this out...]

We also want to be able to go the other way: given the $s:GU \to \mathbb{R}$ constructed from $s_1$ and $s_2$ above, we want to be able to recover $s_1$ and $s_2$ from $s$ by appropriately restricting $s$.

When we have two batches of information that we want to be equivalent, that makes me start to think of limits or colimits!

David Egolf (Mar 30 2024 at 22:05):

I'm pretty sure what I just wrote above isn't quite right. Or, at least, I'm quite confused about it.

But here's a picture illustrating the very rough idea I have in mind, that I'm still working to clearly spell out:
picture

This diagram is in $\mathsf{Top}^{\mathrm{op}}$.

David Egolf (Mar 30 2024 at 22:07):

Very roughly, I'm starting to wonder if $G$ should do something like "preserving pullbacks", if $\mathsf{Top}(-, \mathbb{R}) \circ G$ is to be a sheaf. But it will take me some thinking to express this idea more clearly!

John Baez (Mar 30 2024 at 22:16):

I think you're on the right track. I think all this can be simplified a bit.

John Baez (Mar 30 2024 at 22:24):

I go into this in a bit more detail in Part 3 of my course. Just two short paragraphs. But you seem to be enjoying discovering this stuff on your own, which is really better.

John Baez (Mar 30 2024 at 22:27):

If you try to develop a subject on your own, the way you're doing, it can become much easier to understand what people are doing when you read the 'official' treatment.

John Baez (Mar 30 2024 at 22:36):

By the way, another tiny point:

David Egolf said:

Here is a statement of the "Local Criterion for Continuity" from Lee's book:

A map $f:X \to Y$ between topological spaces is continuous if and only if each point of $X$ has a neighborhood on which (the restriction of ) $f$ is continuous.

I'm losing track of who said what where, but I think I saw someone derive this "local criterion for continuity" from something more basic, which might be called the "local criterion for openness:

A subset $S$ of a topological space is open $\iff$ each point $x \in S$ is contained in some open set $O_x$ contained in $S$.

This is amusingly easy to prove. For the $\implies$ direction just take $O_x = S$. For the $\Leftarrow$ direction just note $S = \bigcup_{x \in S} O_x$ and use the fact that a union of open sets is open.

John Baez (Mar 30 2024 at 22:38):

So we can say openness is a 'local' condition: to see if a set is open, you can run around checking some condition about all its points, and the set is open iff this condition holds for all its points.

And this implies that continuity is also a local condition: to see if a function is continuous, you can run around checking some condition at all points of its domain, and the function is continuous iff this condition holds for all those points.

David Egolf (Mar 30 2024 at 22:40):

Yes, I made implicit use of this "local criterion for openness" above! Before doing so, I had never noticed this connection between continuity being "locally detectable" and openness being "locally detectable"! Cool stuff! :smile:

John Baez (Mar 30 2024 at 22:40):

Yes!

And there's a "for all" in both of these "local criteria". I haven't thought about it hard, but I bet this is connected to the fact that the sheaf condition can be stated in terms of limits. (A "for all" is a limit, and the pullback you were looking at is also a limit.)

John Baez (Mar 30 2024 at 22:43):

I guess for now the main moral is that sheaves are all about "locally detectable" properties.

Peva Blanchard (Mar 30 2024 at 22:46):

I'm wondering then what makes a property "local".

It is as if it was something that is trivially parallelizable: I imagine a (possibly uncountable) set of agents that would check for each point $x$ whether the property holds "around" that point, and, most importantly, they don't need to communicate/synchronize. And the agents are indistinguishable (each of them runs the same check procedure).

John Baez (Mar 30 2024 at 23:00):

That sounds right! It's a nice thought. I can try to make it slightly more precise. There's an agent for each point. Each agent runs the same check procedure, which can be checked in an arbitrarily small neighborhood of their point. Then at the end we report the answer "true" if and only if they all get the answer "true".

John Baez (Mar 30 2024 at 23:02):

Important properties of functions $f : \mathbb{R} \to \mathbb{R}$ like continuity, differentiability, smoothness, analyticity, upper and lower continuity, and measurability all work like this.

John Baez (Mar 30 2024 at 23:02):

Later in my posts I talk about the idea of a 'germ', which is connected to this stuff.

Peva Blanchard (Mar 30 2024 at 23:22):

Oh yes, I remember now the formal definition of 'germ', but it's the first time that I get an inverse "tree-like" mental picture of it. Here's what I have in mind.

Initially, we have one agent that checks whether the property $P$ holds over the open set $X$. The agent can spawn (possibly uncountably many) new agents, that are clones of their genitor, each of them being responsible for an open subset of $X$. The parent agent reports true iff all its children reports true. And the process continues like this.

This is a very poor algorithm: depending on the topological space, this could take a transfinite number of steps and a transfinite number of agents.

What I find amusing is that the opposite category of open subsets of $X$ somehow describes this big clone-spawning branching process. The points of $X$ are exactly the (infinitely long) branches.

Peva Blanchard (Mar 30 2024 at 23:32):

ps: To be more precise, I should force agents to merge when they work on the same open subset.

David Egolf (Mar 31 2024 at 00:10):

The idea of "local agents that can work in parallel" reminds me a whole lot of an ultrasound reconstruction technique I know of, where the reconstruction at each point can be computed in isolation of the reconstruction at different points. But this is not quite analogous because the agents in this case would report a number, not just a "yes!" or "no!" regarding whether some property holds.

David Egolf (Mar 31 2024 at 00:11):

One could also consider checking a reconstructed image point by point, and at each point asking if the reconstruction at that point is "plausible" (in some sense) given the observations. (This would probably involve assessing whether the observed data "relevant to this point" is similar enough relative to what we'd expect if our reconstruction as this point reflects the true object).

However, just because a reconstructed image agrees (in some sense) with the observed data at each point does not imply that the entire reconstructed image agrees with the observed data. So, in this example, "reconstruction plausibility" is not "locally detectable".

David Egolf (Mar 31 2024 at 00:16):

This has been a fruitful bonus question to think about!

However, starting next week, I'm hoping to move on to the next puzzle in the blog post, which is this:

Let $X = \mathbb{R}$ and for each open set $U \subseteq \mathbb{R}$ take $F U$ to be the set of bounded continuous real-valued functions on $U$. Show that with the usual concept of restriction of functions, $F$ is a separated presheaf but not a sheaf.

Peva Blanchard (Mar 31 2024 at 00:44):

David Egolf said:

The idea of "local agents that can work in parallel" reminds me a whole lot of an ultrasound reconstruction technique I know of, where the reconstruction at each point can be computed in isolation of the reconstruction at different points. But this is not quite analogous because the agents in this case would report a number, not just a "yes!" or "no!" regarding whether some property holds.

I'll cheat a bit (because I remember later posts from John's series). I think these Boolean agents actually are the truth values of the topos of sheaves over $X$. When the root agent reports "yes!", then the property holds everywhere. When one of its children does, then the property holds on the associated subset. I think we can think of this "collection of boolean agents over $X$" as a specific sheaf.

Now regarding agents that would report numbers. I think we should remember that the parent agent aggregates the results of its children. In the "yes/no" case, the aggregation is simply a big conjunction. If, oth, "point"-children report numbers, then the parent can aggregate just by making a tuple out of them, indexed by their location. In other words, each agent really reports real-valued function, and the parent aggregation process amounts to gluing the functions reported by its children, i.e., taking a categorical limit.

I will stop there, as otherwise, it's just going to be another burrito tutorial.

Anyway, a lot of things clicked today, so thank you :)

John Baez (Mar 31 2024 at 01:44):

Good stuff, Peva! I'm too tired to think hard about what you said, so I'll just report the slogan "topos theory is the study of local truth".

Madeleine Birchfield (Mar 31 2024 at 15:16):

Do presheaf categories still have subobject classifiers if Set does not have a subobject classifier because one is a constructive predicativist?

Morgan Rogers (he/him) (Mar 31 2024 at 15:23):

Did you already see the nLab page [[predicative topos]]?

Eric M Downes (Mar 31 2024 at 18:13):

Julius Hamilton said:

Eric M Downes said:

1. Can you distinguish between the "generator arrows" of a poset, and the morphisms only in the closure?

I’d never thought of that before. I’m curious to know what those elements might be called in an abstract algebra setting. I’ve been thinking about how a one object category is a monoid, and if there is a corresponding abstract algebraic structure for a “multi-object category”.

"Generators" is most common for such elements in most contexts; famously the symmetric group $S_n$ can be generated by just two maps $m\mapsto m+1\pmod{n}$ and $(12)$. What is the operation under which a permutation group specifically can be said to be closed?

A family of subsets $\mathcal{F}(X)$ can generate a topology $\mathrm{cl}_{\cup,\cap}(\mathcal{F})$. The topology is the closure under arbitrary unions and finite intersections. Is a topology a category?

There is such a thing as a delooping; simplest context is a finite commutative monoid, in which every element is an object. Draw an arrow $x\to y$ just when $\exists z;~y=zx$ (Green's relations). Take your favorite finite commutative monoid (without inverses if you want to deal with fewer arrows), how few arrows can you specify, such that asserting closure under composition of arrows fills in the rest of the cayley table?

The above arrow drawing requires associativity of elements. "Magmas" are the non-associative binops. For a familiar structure that is closed in an elemental sense but not closed in another very meaningful sense, consider the rock-paper-scissors magma
$\begin{array}{c|ccc}& r&p&s\\ \hline r & r & p & r\\ p & p & p & s\\ s & r & s & s\end{array}$
This binary operator is not associative. You can rephrase the associativity condition as a kind of closure (or lack-thereof) under a certain familiar operation. What is it? How many elements must the closed structure have?

Eric M Downes (Mar 31 2024 at 18:36):

Julius Hamilton said:

I think your point is that all the arrows in a thin category are generators. Now I have to think about what how categories can have non-generator arrows.

No, you can have non-generator arrows in a thin category. Consider a poset $\big(\{x,y,z\},\leq\big)$ and there are two "generator" arrows
$x\leq y, y\leq z$
what third arrow must also be present?

Eric M Downes (Mar 31 2024 at 18:44):

(And, having answered that question, and recalling there is at most one arrow between any two objects in a thin category, you should understand why all diagrams* in a thin category commute.)

• -- sometimes we draw diagrams like pullbacks etc. with certain key arrows having a caveat "when this arrow exists", so the "all diagrams" here is modulo the existential quantifier.

David Egolf (Apr 01 2024 at 16:29):

Here's the next puzzle I want to work through:

Let $X = \mathbb{R}$ and for each open set $U \subseteq \mathbb{R}$ take $F U$ to be the set of bounded continuous real-valued functions on $U$. Show that with the usual concept of restriction of functions, $F$ is a separated presheaf but not a sheaf.

We start by showing $F$ is a functor $F: {\mathcal{O}(\mathbb{R})}^{\mathrm{op}} \to \mathsf{Set}$, which means it is a presheaf. On objects, $F$ sends an open set $U \subseteq \mathbb{R}$ to the set $FU$ of bounded continuous real-valued functions on $U$. Note: to determine if a function from $U$ it continuous, we need to put a topology on $U$. To talk about continuity, we equip $U$ with the subspace topology it inherits from $\mathbb{R}$.

On morphisms, $F$ sends a morphism $r:A \to B$ to the corresponding restriction function, which sends a bounded continuous real-valued function $f:A \to \mathbb{R}$ to $f \circ i:B \to \mathbb{R}$, where $i: B \to A$ is the inclusion map. We saw earlier that inclusion maps like $i$ are continuous, and therefore the restriction of a continuous function is continuous. Further, restricting a bounded function yields a bounded function.

For each object $U$ in ${\mathcal{O}(\mathbb{R})}^{\mathrm{op}}$, $F$ sends the identity morphism $1_U: U \to U$ to the identity function on $FU$. This is because restricting a function to its own domain leaves the function unchanged.

If we have the situation $r \circ r' = r''$ in ${\mathcal{O}(\mathbb{R})}^{\mathrm{op}}$, then we have $F(r) \circ F(r') = F(r'')$. That is because restricting a function to some domain in two steps yields the same result as restricting it to that domain all at once.

We conclude that $F$ is a functor $F: {\mathcal{O}(\mathbb{R})}^{\mathrm{op}} \to \mathsf{Set}$ and hence a presheaf.

David Egolf (Apr 01 2024 at 16:34):

Next, we show that $F$ is a separated presheaf but not a sheaf. If $F$ was a sheaf, we'd always be able to do the following:

• pick a bunch of $s_i \in FU_i$ from various $U_i$, such that these bounded real-valued continuous functions agree pairwise on the overlap of their domains
• glue these $s_i$ together to form an $s \in FU$ where: (1) $U = \cup_i U_i$ and (2) $s|_{U_i} = s_i$ for all $i$

If this $s$ always exists and is unique, then $F$ is a sheaf. If $s$ doesn't always exist, but is unique when it exists, then $F$ is a separated presheaf.

David Egolf (Apr 01 2024 at 16:40):

In this puzzle, if $s$ exists it is unique. For any $x \in U$, since $U = \cup_i U_i$, we have that $x \in U_i$ for some $i$. Then, since $s|_{U_i} = s_i$, we have that $s(x) = s_i(x)$. So, the value of $s$ at every point is fixed (if it exists) once we pick all our $s_i$.

But $s$ doesn't always exist! That's because if you glue together an infinite number of bounded real-valued continuous functions that agree on overlaps, you don't always get a bounded function! Intuitively, if you run around and check that each little bit of a function is locally bounded, you can't conclude that the whole thing is bounded.

We conclude that $F$ is a separated presheaf and not a sheaf.

David Egolf (Apr 01 2024 at 16:56):

There is quite a bit of discussion before the next puzzle! So, I'll try to introduce the next puzzle a little.

To my understanding, part of the goal of the next puzzle is to work towards a notion of morphism between categories of sheaves. And since each category of sheaves is an "elementary topos", this is relevant for thinking about morphisms between elementary topoi.

David Egolf (Apr 01 2024 at 17:03):

And why do we care about morphisms between topoi? Here are a couple possible reasons:

• The blog notes that if we think of each topos as an "alternate universe" for doing mathematics, then we'd like to be able to translate math between different alternate universes.
• (A rough intuitive guess): In the case where our topoi are categories of sheaves, then maybe (intuitively) each one lets us talk about "locally detectable properties" on potentially different topological spaces. A morphism between them might involve relating locally detectable properties on different spaces. This rough idea catches my interest because it makes me think of medical imaging.

David Egolf (Apr 01 2024 at 17:06):

At any rate, here is the next puzzle:

Show that $f_\ast F$ is a presheaf. That is, explain how we can restrict an element of $(f_\ast F)(V)$ to any open set contained in $V$, and check that we get a presheaf this way.

Here, $f_\ast F$ is defined as: $(f_\ast F)(V) = F(f^{-1} V)$ for each open subset $V$ of a topological space $Y$. Note that $f: X \to Y$ is a continuous function and $F$ is a presheaf on $X$. We also have $f^{-1} V = \{x \in X :\; f(x) \in V \} \subseteq X$. Note that $f^{-1} V$ is open because $V$ is open and $f$ and continuous.

Roughly, our goal here is to make a presheaf on $Y$ given a continuous function $f:X \to Y$ and a presheaf $F$ on $X$.

Peva Blanchard (Apr 01 2024 at 21:59):

David Egolf said:

In this puzzle, if $s$ exists it is unique. For any $x \in U$, since $U = \cup_i U_i$, we have that $x \in U_i$ for some $i$. Then, since $s|_{U_i} = s_i$, we have that $s(x) = s_i(x)$. So, the value of $s$ at every point is fixed (if it exists) once we pick all our $s_i$.

But $s$ doesn't always exist! That's because if you glue together an infinite number of bounded real-valued continuous functions that agree on overlaps, you don't always get a bounded function! Intuitively, if you run around and check that each little bit of a function is locally bounded, you can't conclude that the whole thing is bounded.

We conclude that $F$ is a separated presheaf and not a sheaf.

I wanted to provide a concrete counter-example.

Consider the topological space $X = (0,1]$, the half-open unit interval, with the induced topology from $\mathbb{R}$. For all $n \in \mathbb{N}$, let $U_n = \left(\frac{1}{1 + n}, 1\right]$ and $f_n(x) = \frac{1}{x}$ on $U_n$. The $U_n$ cover $X$, $X = \bigcup_n U_n$, and each $f_n$ is bounded. Moreover, for any $n \leq m$, $U_n \cap U_m = U_n$ and $f_n$ and $f_m$ obviously match on $U_n$. If the presheaf $\mathcal{F}$ of bounded functions were a sheaf, there would exist a bounded function $f$ defined on $X$ such that $f_{|U_n} = f_n$. In particular, for all $n$

$$sup~f \geq sup~f_n = n + 1$$

whence a contradiction, i.e., $\mathcal{F}$ is not a sheaf.

Peva Blanchard (Apr 01 2024 at 22:11):

There are variants of this idea: presheaf of Lipschitz functions (take $\sqrt{x}$ or $ln~x$ in the example above) (edit: $\frac{1}{x}$ works as well), presheaf of functions with bounded derivatives of order $k$ (just integrate $k$ times the previous examples).

David Egolf (Apr 01 2024 at 22:22):

Thanks for sharing some specific examples! I hadn't thought of those!

The counterexample I had in mind looks like this in picture form:
counterexample picture

The idea is to consider the identity function $f: \mathbb{R} \to \mathbb{R}$ that sends $x$ to $x$. Then, we can get each $s_i$ by restricting $f$ to, say, $U_i=(i, i+2)$. Each $s_i$ is then bounded, and the collection of $s_i$ agrees on overlaps, but when we try to glue together all the $s_i$, our resulting function isn't bounded anymore.

Peva Blanchard (Apr 01 2024 at 22:29):

oh yes, nice! The common pattern to get a "non-sheafy presheaf" is to start from a invalid candidate defined globally such that the restrictions of this candidate to specific open subsets satisfy a condition.

Each invalid candidate has a sort of singularity at some point (e.g., 0 in my example, and $+\infty$ in yours), and then we just restrict to open subsets that avoid "just enough" this point.

Peva Blanchard (Apr 01 2024 at 22:50):

Mmh, these examples do not work any more if $X$ is compact. Compact means that from any covering we can extract a finite cover of $X$. If $X$ is compact, is the presheaf of bounded functions on $X$ a sheaf? It seems so.

David Egolf (Apr 01 2024 at 22:59):

If $X$ is compact, couldn't it still have a non-compact open subset $U$? And then we could maybe set up an unbounded function $f$ defined on $U$ to show that we don't have a sheaf. (Restrict $f$ to a bunch of $U_i$ where $\cup_i U_i = U$, and where $f|_{U_i}$ is bounded for each $i$. Then these glue together to $f$, which is unbounded. So then, we can't always glue together a bunch of compatible $FU_i$ to make an element of $FU$.)

Peva Blanchard (Apr 01 2024 at 23:05):

I'm not sure that a compact set can contain a non-compact open subset. (I'm thinking about the closed unit interval $[0,1]$). (edit: oh my brain ...)

David Egolf (Apr 01 2024 at 23:07):

My initial thought was that something like $(0.4,0.6)$ would be a non-compact open subset of the topological space $[0,1]$. But I am a bit shaky on compactness, so maybe I'm just confused. (I'd need to review this stuff!)

Peva Blanchard (Apr 01 2024 at 23:09):

Oh yes you're right!

Peva Blanchard (Apr 01 2024 at 23:11):

I focused on the total space, but yes we can reproduce the example on any open subset.

Peva Blanchard (Apr 01 2024 at 23:16):

Now, I'm looking for a topological space $X$ such that the presheaf of bounded functions is indeed a sheaf. From our discussion, it suffices that any open subset of $X$ be compact, right? I'm wondering what kind of space is that.

Peva Blanchard (Apr 01 2024 at 23:17):

One example: any set $X$ equipped with the trivial topology ($\emptyset$, $X$ are the only open sets). Topologically, such a space behaves like a space with one point.

Peva Blanchard (Apr 01 2024 at 23:41):

Another example: a finite set $X$, with any topology.

Peva Blanchard (Apr 01 2024 at 23:42):

In other words: any set $X$ with a topology admitting a finite number of open sets.

David Egolf (Apr 01 2024 at 23:45):

I wonder if there are any examples where $X$ has an infinite number of open sets.
(Interesting stuff! I need to take a break for today - I have to manage my energy carefully - but of course please feel free to keep posting here.)

Peva Blanchard (Apr 01 2024 at 23:50):

Sure! I'll probably continue under another topic, so as not to divert the purpose of yours.

John Baez (Apr 01 2024 at 23:56):

Digression: here's another example of a separated presheaf that's not a sheaf, which I just thought of. Take $\mathbb{N}$ with its usual topology, where all subsets are open, and let $F$ be the presheaf where $F(U)$ for any $U \subset \mathbb{N}$ is the set of computable partial functions $F: \mathbb{N} \to \mathbb{N}$ whose domain includes $U$.

John Baez (Apr 01 2024 at 23:58):

So, simply put, $F(U)$ consists of all partially defined functions $f$ from the natural numbers to the natural numbers such that you can write a computer program which halts and spits out $f(n)$ when $n \in U$.

John Baez (Apr 01 2024 at 23:58):

For any finite $U$, these are all the functions from $U$ to $\mathbb{N}$.

John Baez (Apr 02 2024 at 00:00):

But for $U = \mathbb{N}$ there are lots of functions from $U$ to $N$ that aren't computable.

Peva Blanchard (Apr 02 2024 at 00:37):

Mmh, it's not as easy to come up with an explicit example (i.e., a witness of the non-sheafiness of $F$).

Peva Blanchard (Apr 02 2024 at 00:43):

Here's my attempt.

Let $\mathbb{N} = \bigcup_{i \in \mathbb{N}} U_i$ with $U_i = \{i\}$.

Define $s_i : U_i \rightarrow \mathbb{N}$ as the function that outputs $1$ if the $i$-th Turing machine halts, and $0$ otherwise. All the $s_i$'s agree on the intersections (the $U_i$'s are disjoint).

If $F$ were a sheaf, then there would be a total computable function on $\mathbb{N}$ that would solve Turing's halting problem, whence a contradiction.

Peva Blanchard (Apr 02 2024 at 00:46):

It's a bit weird, I had to convince myself that $s_i$ does belong to $F(U_i)$ (still not 100% sure). It seems trivially true since $U_i$ is finite. The strangeness comes from the fact that I am invoking the halting problem's oracle to define $s_i$.

In a sense, the covering of $\mathbb{N}$ by the $U_i$ acts like an oracle.

David Egolf (Apr 02 2024 at 17:01):

John Baez said:

So, simply put, $F(U)$ consists of all partially defined functions $f$ from the natural numbers to the natural numbers such that you can write a computer program which halts and spits out $f(n)$ when $n \in U$.