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Stream: learning: questions

Topic: Towards an understanding of pullbacks

Jacob Zelko (Oct 28 2022 at 16:31):

[Beginner Questions]

Background

Hi everyone,

I have recently been on a slow moving quest to understand fiber products. The path I have taken so far to make sure I have a sufficient understanding is to make sure I study the steps of: images -> preimages -> fibers -> bundles -> fiber products -> pullbacks. I went back to naive set theory to make sure I understood images and then preimages and then to fibers.

Where I am having a bit of difficulty in my study is twofold at the moment: 1) understanding the significance of fibers and 2) understanding the definition of a fiber within the category $Set$.

Questions

For 1), after I had done my reviews of images and preimages, when I came across the naive set theory definition of a fiber, I was rather nonplussed (definition taken from Wikipedia):

[T]he fiber of the element $y$ in the set $Y$ under a map $f:X\to Y$ is the inverse image of the singleton $\{y\}$ under $f$.

It just seemed like a rather uninteresting result and was surprised when the wiki page went on further about its applications in things like "partitions". Could someone explain to me the significance of fibers by themselves and why they are special enough to warrant study? Or, and this is my inference reading more about them, is the attractiveness of fibers stemming from how they could be used to build up other mathematical constructs?

For 2), one could imagine my surprise when I read about the application of fibers within the Category of $Set$ on nlab being that it seems much more interesting to me personally:

The fiber of a morphism or bundle $f:E→B$ over a point of $B$ is the collection of elements of $E$ that are mapped by $f$ to this point.

It felt very different from the definition given in naive set theory! To me, it feels like there is a missing step here about how we enriched the naive set theory definition of a fiber to a $Set$ category theory definition and I am not sure how that happened. I struggle to see the analogy between morphisms and bundles to singletons and sets. The best I could think of is that the inverse image is the morphism and the singleton is the "point"? Could anyone explain to me how fibers operate within $Set$ and how they relate back to naive set theory? Or are the two definitions rather different things and the word "fiber" is overloaded in these contexts?

Thanks all and I am happy to provide any additional information or clarification!

Morgan Rogers (he/him) (Oct 28 2022 at 16:40):

Have you encountered the fact that one can identify elements of a set with functions from a singleton into that set? In the same way, a point of a space corresponds to a map from "the" one-point space.

Ralph Sarkis (Oct 29 2022 at 19:57):

Did you study equalizers before trying pullbacks? The order is not that important, but fibers can also be seen as equalizers, so that could help.

Jacob Zelko (Oct 29 2022 at 23:19):

Hi @Morgan Rogers (he/him) ! Thank you for your time and comment -- if I am understanding the fact you are referring to, then, yes, I believe I have. To be clear, I imagine the following scenario:

Suppose I have the sets: $A = \R$, $B = \R$

The function: $f : x \rightarrow x^{2}$

And the subset: $D \subseteq B$, where $D$ is a singleton and $D = \{9\}$

If I take the inverse image of $D$ under $f$, $f^{-1}(D)$, the result would be the subset $\{-3, 3\} \subseteq A$. Therefore, using this singleton, $D$, I have identified two elements that are located in $A$ that map to the singleton subset $D$ that I did not know about before.

Is that the fact you are referring to?

Jacob Zelko (Oct 29 2022 at 23:20):

@Ralph Sarkis - I have not yet. Thanks for the pointer and I shall switch to researching them some!

Ralph Sarkis (Oct 29 2022 at 23:31):

No that is not the fact Morgan was referring to. There is a correspondence between elements of a set $S$ and functions from the terminal set $\mathbf{1} = \{*\}$ to $S$. Here it is:

• Given an element $x \in S$, you define the function $f_x: \mathbf{1} \to S$ that sends the unique element $*\in \mathbf{1}$ to $x$.
• Given a function $f:\mathbf{1} \to S$, you define the element $x_f = f(*) \in S$.
• You should check that these two processes are inverses, i.e., $x_{f_x} = x$ and $f_{x_f} = f$. You can then conclude $S \cong \mathrm{Hom}(\mathbf{1},S)$.

Unimportant aside, note that a common notation for the set of functions from $X$ to $Y$ is $Y^X$. What we proved above is that $S \cong S^{\mathbf{1}}$, which looks pretty neat imo.

We can now give a categorical definition of an element by noting that the singleton $\mathbf{1}$ is special in $\mathbf{Set}$ because it is the terminal object (disregard the footnote mark):
image.png

Puzzle: Try to instantiate this definition in other categories you know (my examples are the categories of posets, groups, and categories, but use the ones you prefer). Does this definition of element correspond to an intuitive notion in these categories?

Jacob Zelko (Oct 31 2022 at 23:00):

@Ralph Sarkis and @Morgan Rogers (he/him) , I am in the process of digesting this information -- may be a bit delayed as I am working through what all you suggested I look at and also digging into your comment some more Ralph as it made me realize I had some gaps in my understanding. Which, I am very grateful about! So going back and fixing up my understanding here.

Jacob Zelko (Nov 25 2022 at 19:28):

Hey @Ralph Sarkis , I have been puzzling over this for a while and have a couple questions about your response. Some of them will be very simple so I apologize in advance:

1. What do you mean by terminal set?
2. What does $1 = \{*\}$ mean? I understood the $*$ to mean any element within the set $1$. Admittedly, the $1$ as a set looks rather odd to me -- what kind of set is this?
3. What is the notation $f_{x}$ denoting?

Finally, zooming out, I suppose what I am struggling to see right now exploring this correspondence you explained is how it relates back to the "specialness" of fibers. Where is the relationship back to fibers? Sorry if these are asking all the wrong questions. Trying to wrap my head around the gaps I found in my understandings. Thanks!

Ralph Sarkis (Nov 25 2022 at 19:44):

No need for apologies.

1. I was referring to the concept of [[terminal object]]. We commonly use the notation $\mathbf{1}$ to denote the terminal object of a category. That is an object such that for any other object $X$, there is a unique morphism $X \to \mathbf{1}$. The terminal set is the terminal object in the category of sets, it is the set containing only one element $\ast$, we call it a singleton and write it as $\{\ast\}$. That is why I write $\mathbf{1} = \{\ast\}$. (Note that any set with one element works, I just choose to pick the one whose unique element I denote by $\ast$.) I let you convince youself that for any set $X$, there is only one function $X \to \{\ast\}$.
2. Partially answered above. I agree that $1$ can look weird for a set, but I use $\mathbf{1}$ (a bold face 1). It will be less weird if you remember that $\mathbf{1}$ contains only one element.
3. Both $f_x$ and $x_f$ are names that I gave to the things I defined:

• $f_x$ is the function that sends $\ast$ to $x$. For any $x \in S$, I define such an $f_x$. For instance if $S = \N$, I have $f_2$ that sends $\ast$ to $2$ (and I let you find all the others).
• $x_f$ is the element $f(\ast)$ that belongs to $S$. For any $f: \{\ast\} \to S$, I define such an $x_f$.

Tell me if everything is clear so we can go back to fibers.

Jacob Zelko (Nov 25 2022 at 22:17):

Hey @Ralph Sarkis -- thanks for the patience in working with me on understanding this! Here are my responses:

Response on 1: OH -- yea, that was my bad. Just was totally confused on the notation. After looking at the nlab post you linked and thinking on things some, this clarification was very helpful. I did try to convince myself of what you are saying regarding the statement

for any set $X$, there is only one function $X\rightarrow\{∗\}$

But I got confused when I considered the following scenario:

Suppose I have randomly chosen the set $A = \{1, 2, 3\}$ and the terminal set $\textbf{1} = \{6\}$. Say if I have the functions $f_{1} : \sum_{i \in 1}^{|A|} a_{i} \rightarrow \textbf{1}$ and $f_{2} : \prod A \rightarrow \textbf{1}$. Would not function 1 and 2 be valid morphisms to the terminal object in this case? What am I missing in this argument? (P.S. I may have screwed up the notation in $f_{2}$ but I was trying to say the factorial of the elements in $A$).

Response on 2: Yup, that is now crystal clear. Somehow, I guess I hadn't come across that notation yet. :face_palm:

Response on 3: Thanks for the explanation on the syntax! Yes, this makes a lot more sense. I wasn't sure if this was a "special" notation I had missed but this makes a lot more sense now that you have explained it.

Thanks! Once I am clear on 1, I am ready to jump back to fibers. :smile:

David Egolf (Nov 25 2022 at 22:40):

Hi @Jacob Zelko !
To my understanding, your $f_1$ and $f_2$ aren't actually functions from $A$ to $\{6\}$. A function from $A$ needs to specify what it maps each element of $A$ to. If $f: A \to \{6\}$, you need to specify what $f(1), f(2)$ and $f(3)$ are. (Hopefully I'm understanding what you wrote correctly!)
Hope this helps at least somewhat!

Jacob Zelko (Nov 25 2022 at 22:46):

Yea, that's right @David Egolf -- sorry about that! You are entirely correct. I was trying to somehow convey via $f_{1}$, collapsing $A$ into a singleton being $\{6\}$ through a summation of elements in the set and similarly taking the product of each element via $f_{2}$ ( I realized that I did in fact screw up that notation so edited the above function to reflect a product of elements) in $A$ and collapsing that to a singleton being $\{6\}$.

Jacob Zelko (Nov 25 2022 at 22:52):

I was thinking I could have a sort of intermediate step with $f_{1}$ and $f_{2}$ that collapses each element into a singleton and then maps that final singleton to the terminal object. All that to say, is I agree what I was saying is incorrect but am still a bit confused on how to convince myself of the statement "for any set $X$, there is only one function $X\rightarrow\{∗\}$". I guess to put it simply, it seems to me there is just more than one way to get to the terminal object $\textbf{1} = \{6\}$.

David Egolf (Nov 25 2022 at 22:55):

No need to apologize or anything - it's all part of the learning process! :smile:
As best as I can tell, you are using the word "function" differently than the usual definition I know. If I understand correctly, you are thinking of a function as a process which acts on an object to produce a new object?

David Egolf (Nov 25 2022 at 22:58):

An intuition that is closer to the definition I know is - think of a function as a labelling of the different parts of an object. So a function from $A$ needs to label each element of $A$ - three "labels" need to be provided. The way Wikipedia puts it is "In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y."

Jacob Zelko (Nov 25 2022 at 23:00):

Thanks! And you are right. I subconsciously shifted my thinking of function to mean "map an object to an object" or "map each element to an element" to something more akin to "map the transformation applied to an object (or objects) to an object". I should've been thinking more the former -- as it stands, what I was saying makes no sense at all in a mathematics setting.

Jacob Zelko (Nov 25 2022 at 23:01):

Definitely agree with your idea of a function. I confused myself here in trying to disentangle my thoughts and ended up even more confused. :joy: But on the function definition part, I am set straight again in my thinking.

Reid Barton (Nov 25 2022 at 23:02):

Often a morphism of a general category is described as some generic, nebulous notion like a "process"; but a morphism of a specific category is a specific thing; in the case of Set, a function.

David Egolf (Nov 25 2022 at 23:03):

Awesome! You may find it helpful to now write out the data of a function from $A$ to $\{6\}$ (what each element gets mapped to). The process of doing this should hopefully help clarify why there's only one such function.

Jacob Zelko (Nov 25 2022 at 23:06):

For sure @David Egolf -- be back in a bit! I'll try sketching it out right now.

Ralph Sarkis (Nov 25 2022 at 23:16):

David Egolf said:

Awesome! You may find it helpful to now write out the data of a function from $A$ to $\{6\}$ (what each element gets mapped to). The process of doing this should hopefully help clarify why there's only one such function.

When I first learn about functions, I would draw potatoes for the sets, points inside the potatoes for elements, and arrows between elements of one potato to another for the assignment. There are great pictures like these in the function section here.

Now draw a potato with only one element, call it $\mathbf{1}$. It should be obvious that for any other potato $X$, if you have to assign to every element of $X$ one element of $\mathbf{1}$, there is only one possible choice.

Jacob Zelko (Nov 25 2022 at 23:38):

So, I would not have expected the solution to my confusion was in the shape of a potato but that analogy and pictures were fantastic @Ralph Sarkis ! So, behold! The potato! Notepad2_1.png

Jacob Zelko (Nov 25 2022 at 23:42):

I think I realized my confusion on point 1 was coming from a bit ago! What I was confusing myself over were the details on how each element of a set (in this case $A$) terminates at the terminal object $\textbf{1}$. When I went back to my notebook and put pen to paper, I was still a bit confused as I was getting caught up in thinking, "well, there are multiple ways to turn A(1) (i.e. the value $1$) into $\{6\}$" and same with $2$ and $3$. But then thinking about abstracting to a general level on thinking about each element within a "potato" maps to the terminal object, I realized, it does not matter exactly how each element of a set gets to terminate at $\textbf{1}$ as long as it gets there.

Jacob Zelko (Nov 25 2022 at 23:44):

Am I finally thinking correctly about this now? When I was taking your approach @David Egolf of enumerating the functions (i.e. $f(1) = \{6\}$, $f(2)$, $f(3)$), I realized I was getting caught up in exactly how the elements could get mapped to $\textbf{1}$ when, categorically speaking, that doesn't so much matter.

David Egolf (Nov 25 2022 at 23:46):

I think you have the right idea! Although, strictly speaking, it's $f(1) = 6$, not $f(1) = \{6\}$. (Functions maps elements to elements, not elements to sets). But yes - a function is determined by what each element is mapped to.

Ralph Sarkis (Nov 25 2022 at 23:47):

Just a notation mistake, you write $f(1) = \{6\}$ but you should write $f(1) = 6$. There is an important difference between the set $\{6\}$ and the number $6$. They are not even the same kind of thing. This is also why we use different fonts for the number $1$ and the set $\mathbf{1}$ which are different things.

Caring for this informal notion of "kind of thing" (a more technical term would be type) can be very useful to avoid getting confused. When I am having trouble parsing some text, I first try to figure out if it typechecks, that is if things that should be of the same kind are of the same kind. For instance, if I see someone writes about a morphism $f:A \rightarrow B$, I make sure that $A$ and $B$ are objects of the same category, otherwise, it does not make sense to talk about morphisms between them.

Jacob Zelko (Nov 25 2022 at 23:57):

Right, we should have elements mapping to elements -- like you were both saying, it doesn't make sense for an element to map to a set. It should map from element to element (or in this case $\R$ to $\R$).

Ralph Sarkis (Nov 26 2022 at 00:08):

Sorry if the following confuses you, but I think you might still be missing something.

Jacob Zelko said:

it doesn't make sense for an element to map to a set

That is not what I said. You are trying to define a function from $\{1,2,3\}$ to $\{6\}$, so to each element of the first set, you should assign an element of the second set. Now, if you were defining a function from $\{1,2,3\}$ to $\{\text{fromage},6,\{6\}\}$, you do as before, but now you realize that in the second set you don't have just numbers, the elements of the second set are the word $\text{fromage}$, the number $6$ and the set containing the number $6$. The second set contains another set as one of its elements. Therefore, I can define $f: \{1,2,3\} \to \{\text{fromage},6,\{6\}\}$, by saying for instance $f(1) = 6$, $f(2) = \{6\}$ and $f(3) = 6$. The function $f$ maps a number($2$), an element from its domain, to a set $\{6\}$ which is an element of its codomain.

Jacob Zelko (Nov 26 2022 at 00:15):

I actually do follow what you are saying -- I should've been more specific: "it doesn't make sense in this example for an element of a set of integers to map to terminal set with only one integer". Like you said, sets could be formed of different things (types). Ugh, need to be thinking more abstractly.

David Egolf (Nov 26 2022 at 00:18):

I guess it's confusing how sometimes elements of a set are also sets! I should probably have also been more careful above.

David Egolf (Nov 26 2022 at 00:22):

By the way, if you want more practice with these concepts, you might also find it interesting to think about how many functions there are to a set $A$ from a set with a single element.

Jacob Zelko (Nov 26 2022 at 00:27):

Ah nah nah, y'all good @David Egolf ! I should've been more careful too.

Jacob Zelko (Nov 26 2022 at 00:45):

Ok, so, let me see end off today with thinking through how this all relates back to fibers:

Knowing that we are working in the category of $\textbf{Set}$, we know that in this category, there exists a terminal object $\textbf{1} = \{*\}$ and if we have any set $X$ within $\textbf{Set}$ then there is a unique morphism that maps $f: X \rightarrow \textbf{1}$. So relating all the way back to what @Morgan Rogers (he/him) was saying a while ago about "one can identify elements of a set with functions from a singleton into that set," is the attractiveness of fibers the ability to say "how" the elements of a given set $X$ were mapped to (and I'll use the terminal object for now but it could be any singleton) $\textbf{1}$?

Jacob Zelko (Nov 26 2022 at 00:45):

Of course, there could be other attractive features of fibers that I am missing but I am just working my way back to what Morgan was saying some months ago.

Jacob Zelko (Nov 26 2022 at 00:48):

P.S. Also, thanks for everyone helping me get back on track here! This was legendary! Everyone's patience is amazing as I am trying to understand this all! Y'all are amazing!:100: :heart:

Ralph Sarkis (Nov 26 2022 at 01:10):

Because $\mathbf{1}$ is terminal, there is a unique function $X \to \mathbf{1}$ for any set $X$. However, Morgan's suggestion was to look at functions $\mathbf{1} \to X$ and realize that these correspond with elements of $X$.

We needed to realize this because in given a function $f:X \to Y$, fibers are defined relative to an element of $Y$. We define the fiber of $f$ at $y$ as $f^{-1}(y) = \{x \in X \mid f(x) = y\}$. So in order to categorify this concept, we first need to categorify the notion of an element of a set. In the category of sets, $Y$ is just an object and at first glance we cannot access its elements. But with Morgan's trick, instead of working with elements of $Y$ explicitly, we can work with functions $\mathbf{1} \to Y$ which we have access to (they are morphisms in the category).

Morgan Rogers (he/him) (Nov 26 2022 at 12:45):

Excellent explanation @Ralph Sarkis :grinning_face_with_smiling_eyes:

@Jacob Zelko Try drawing a potato diagram for a pullback square defining a fiber. In the lower corners you will have a pair of spaces $X$ and $Y$ (give them just a handful of elements to keep things manageable) and a diagram for a function $X \to Y$. In the upper-right corner you will have your terminal-potato, a singleton set, equipped with a map to the lower-right corner picking out an element. Now figure out what the fiber looks like from there. This concept is not immediately appealing in isolation; one of the most interesting things about it is that, as we vary the function/element on the right-hand side of the square, the fibers collectively produce a decomposition of $X$ corresponding to the decomposition of $Y$ into its constituent elements. It's a "$Y$-indexed decomposition of $X$".

You could also use potatoes to depict topological spaces; have you met those before?