Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: ELI5: amazing right adjoint

Jacques Carette (Oct 13 2021 at 12:36):

Can someone try to give me a very concrete example? I've read the nLab page, Lawvere's Toward the Description in a Smooth Topos of the Dynamically Possible Motions and Deformations of a Continuous Body and attempted a few others, all of which presented things in increasingly general ways, without giving concrete examples. (There were plenty of highly abstract examples.)

I don't mind working out details, but I need a concrete starting point. [Hint: anything start with 'topos' will fail the ELI5 criteria; if you use the word 'presheaf' you're probably thinking at too high a level still, and need to expand things out. Please.]

Morgan Rogers (he/him) (Oct 13 2021 at 16:21):

What is ELI5?

Morgan Rogers (he/him) (Oct 13 2021 at 16:24):

Ah, "explain it like I'm 5". I assumed it was some categorical axiom heh.

Reid Barton (Oct 13 2021 at 16:27):

I know the interval (1-cube) in the category of cartesian cubical sets is supposed to be an example, though I never worked it out for myself. (I didn't say "presheaf"...)

Reid Barton (Oct 13 2021 at 16:28):

I'm surprised it's not on the nLab page for amazing right adjoint already.

John Baez (Oct 13 2021 at 16:38):

Once someone figures out an example, I can stick it on the nLab. I think Lawvere's favorite examples involve "infinitesimal spaces", - maybe like the "walking tangent vector" space in synthetic differential geometry?

John Baez (Oct 13 2021 at 16:38):

I get the idea of that "walking tangent vector" or "infinitesimal arrow" pretty well, but I haven't thought about this "amazing right adjoint".

Reid Barton (Oct 13 2021 at 16:44):

So this fails the ELI5 test, but here goes. Say $C$ is a category with an object $I$ (for "interval") such that $C$ has all products $X \times I$ for $X \in C$. The exponential ${-}^{yI}$ on $\hat C = \mathrm{Set}^{C^{\mathrm{op}}}$ has left adjoint ${-} \times yI$ (here $y$ is the Yoneda embedding). But $yX \times yI \cong y(X \times I)$, so ${-} \times yI$ takes representables to representables and as a left adjoint, it must be the left Kan extension along the functor ${-} \times I$ on $C$. That means that ${-}^{yI}$ is the restriction along this functor and so it has a right adjoint, given by right Kan extension along the same functor.

Morgan Rogers (he/him) (Oct 13 2021 at 16:48):

It's a shame you ruled out presheaves, because they provide the easiest example in my opinion, thanks to Proposition 2.5 of the nLab page.

Let $\mathcal{C}$ be a small category, and consider the category of functors $\mathcal{E}:= [\mathcal{C},\mathrm{Set}]$; this is the category of "copresheaves on $\mathcal{C}$", but the essential thing is that the objects of $\mathcal{E}$ are functors and the morphisms are natural transformations. For any object $F$ in this category, we can consider the representable functor $\mathrm{Hom}_{\mathcal{E}}(F,-)$ to sets, which always has a left adjoint given by sending any set $A$ to an $A$-indexed coproduct of copies of $F$. On the other hand, this representable functor has a right adjoint if and only if it is a (retract of a) representable copresheaf the latter being a functor of the form $\mathrm{Hom}_{\mathcal{C}}(C,-)$ for some object $C$ of $\mathcal{C}$.

Morgan Rogers (he/him) (Oct 13 2021 at 16:51):

The objects whose representable functors admit right adjoints are called "tiny objects", so the above says that we can identify the representable copresheaves on a category up to retracts as the subcategory of tiny objects.

John Baez (Oct 13 2021 at 17:03):

I don't think he ruled out examples that happen to be presheaves. I think he just wanted a concrete example, not a general example of the sort you just gave. So please give an example of your example!

Morgan Rogers (he/him) (Oct 13 2021 at 17:13):

The category of directed graphs is the category of copresheaves on the double-arrow category $s,t: E \rightrightarrows V$. By the above, the tiny objects are the digraph consisting of a single vertex and no edges, and the graph consisting of two vertices and a single edge from one to the other. I think this is a very instructive example to work out the details of!

Morgan Rogers (he/him) (Oct 13 2021 at 17:15):

(Note that this is normally presented as the category of presheaves on the double arrow category, but the result is identical; for the purposes of understanding this concept, it doesn't matter.)

Jacques Carette (Oct 13 2021 at 17:41):

Aha, that last example is indeed something I can sink my teeth in to. Thank you.

And indeed, I want to work my way up to understanding why this has anything to do with "infinitesimals". But I would like to do it via seeing 'tiny' objects in a number of sufficiently different concrete settings first.

Jens Hemelaer (Oct 14 2021 at 08:20):

Morgan Rogers (he/him) said:

The category of directed graphs is the category of copresheaves on the double-arrow category $s,t: E \rightrightarrows V$. By the above, the tiny objects are the digraph consisting of a single vertex and no edges, and the graph consisting of two vertices and a single edge from one to the other. I think this is a very instructive example to work out the details of!

There is some problem with the terminology. On the nlab there are two things that are called tiny:

• externally tiny: objects $X$ in $\mathcal{E}$ such that $\mathrm{Hom}_\mathcal{E}(X,-)$ preserves colimits.
• internally tiny: objects $X$ in $\mathcal{E}$ such that $Y \mapsto Y^X$ preserves colimits.

I don't know whether the two graphs that you mention are also internally tiny (I'll try to compute the exponentials).

Jens Hemelaer (Oct 14 2021 at 08:41):

The graph $\mathbf{y}V$ with a single vertex and no edges is the terminal object, so then the functor $Y \mapsto Y^{\mathbf{y}V}$ is the identity functor (which does preserve colimits). So this graph should be internally tiny as well. (The graph with a single vertex and no edges is not the terminal object, so this argument is wrong.)

But if $\mathbf{y}E$ is the graph with two vertices and an edge connecting them, then I believe $(\mathbf{y}E)^{\mathbf{y}E}$ has 4 edges, while $(\mathbf{y}E \sqcup \mathbf{y}E)^{\mathbf{y}E}$ has 32 edges... so $Y \mapsto Y^{\mathbf{y}E}$ does not preserve coproducts in this case.

John Baez (Oct 14 2021 at 12:20):

Jens Hemelaer said:

But if $\mathbf{y}E$ is the graph with two vertices and an edge connecting them, then I believe $(\mathbf{y}E)^{\mathbf{y}E}$ has 4 edges...

I'm having trouble seeing 4 edges. What are these 4 edges? For starters, what are the vertices of $(\mathbf{y}E)^{\mathbf{y}E}$?

John Baez (Oct 14 2021 at 12:23):

A vertex of $(\mathbf{y}E)^{\mathbf{y}E}$ corresponds to a morphism $f: \mathbf{y}E \to \mathbf{y}E$, right? And a morphism from $\mathbf{y}E$ to itself must map the one edge of this graph to itself. So I'm getting just one vertex.

Spencer Breiner (Oct 14 2021 at 13:00):

Similarly, the set of edges corresponds to morphisms $yE\times yE\to yE$. The product has one edge $(0,0)\to(1,1)$ and two isolated points $(1,0)$ and $(0,1)$. The diagonal edge has to map to the only edge in $yE$, but the isolated points can go to either vertex, making 4 options.

Jens Hemelaer (Oct 14 2021 at 13:21):

Ok, I miscalculated with the vertices...
A vertex of $(\mathbf{y}E)^{\mathbf{y}E}$ is the same thing as a morphism $\mathbf{y}V \times \mathbf{y}E \to \mathbf{y}E$. Here, $\mathbf{y}V \times \mathbf{y}E$ is the graph with two vertices and no edges. So there are 4 morphisms $\mathbf{y}V \times \mathbf{y}E \to \mathbf{y}E$ which means that $(\mathbf{y}E)^{\mathbf{y}E}$ has four vertices.

Jens Hemelaer (Oct 14 2021 at 13:28):

I made a similar mistake when I said that the functor $Y \mapsto Y^{\mathbf{y}V}$ is the identity functor.
You can compute that the functor $Y \mapsto Y^{\mathbf{y}V}$ sends a graph to the complete directed graph on its vertices.
This functor does not preserve colimits (not even coproducts), so $\mathbf{y}V$ is not an internally tiny object.

Jens Hemelaer (Oct 14 2021 at 18:33):

Here are the directed graphs $(\mathbf{y}E)^{\mathbf{y}E}$ and $(\mathbf{y}E \sqcup \mathbf{y}E)^{\mathbf{y}E}$ (hopefully no miscalculations). The loops are not shown in the pictures, the first should have a loop at vertex "st", and the second should have a loop at vertices 1 and 11. exponential1.png exponential2.png

Jens Hemelaer (Oct 14 2021 at 18:38):

In general, exponentials in a presheaf category are hard to compute... so it is difficult to come up with an easy and interesting example where taking exponentials has a right adjoint (the amazing right adjoint).

John Baez (Oct 14 2021 at 18:53):

I bet you're used to reflexive graphs, Jens. For reflexive graphs I believe $Y \mapsto Y^{\mathbf{y}{V}}$ is the identity functor, and other things change too.

Jens Hemelaer (Oct 14 2021 at 19:00):

Yes I agree. I still think it is a bit surprising intuitively that the product of a point and an edge is two points...

Jacques Carette (Oct 14 2021 at 21:36):

This whole discussion is a splendid illustration of why I wanted explicit examples!

Jens Hemelaer (Oct 15 2021 at 09:24):

Jacques Carette said:

This whole discussion is a splendid illustration of why I wanted explicit examples!

Yes exactly... Here is the easiest example that I could come up with. Let $\mathcal{E}$ be the category that has as objects the triples $(f,A,B)$ where $A$ and $B$ are sets and $f : A \to B$ is a function. We look at the object $j : \varnothing \to \{\ast\}$ in $\mathcal{E}$. For this object, you can compute that the functor $X \mapsto X^j$ is given as follows: if $f : A \to B$ is an object of $\mathcal{E}$, then $f^j$ is the identity map $B \to B$. This exponential functor has a right adjoint (the amazing right adjoint), which sends an object $g : C \to D$ in $\mathcal{E}$ to the identity function $C \to C$.

Some theoretical background (using the terminology and results from the nlab page here): $\mathcal{E}$ is the topos of presheaves on the category $\mathcal{C}$ with two objects and precisely one non-identity morphism (from one object to the other). Because $\mathcal{E}=\mathbf{PSh}(\mathcal{C})$ is a sheaf topos, the exponential $(-)^j$ has a right adjoint if and only if it preserves colimits, so if and only if $j$ is internally tiny. Further, because $\mathcal{C}$ is a category admitting finite products, the internally tiny and externally tiny objects in $\mathcal{E}$ agree (Proposition 2.4). Now from Example 2.2 it follows that $j$ is tiny, because it is a representable presheaf. Conversely, you can use Theorem 2.14 here to show that $j$ and the terminal object are the only tiny objects in $\mathcal{E}$.

John Baez (Oct 15 2021 at 12:20):

Very nice explanation! Presheaves on this particular category $\mathcal{C}$ are like graphs, only simpler. I sometimes think of them as graphs where every edge has a source but no target.

John Baez (Oct 15 2021 at 12:22):

It's cool that $\mathcal{C}$ has finite products. One object is terminal, and the product of the other object with itself is itself.

John Baez (Oct 15 2021 at 12:24):

One way to think about this is that $\mathcal{C}$ is a poset, and a poset has finite products if every pair of objects has a greatest lower bound.

John Baez (Oct 15 2021 at 12:29):

Viewed as a poset $\mathcal{C}$ is an ordinal, the ordinal called "2", and any ordinal has finite products.

Steve Awodey (Oct 17 2021 at 04:05):

Here's a very simple example, the "cartesian cubical sets" that @Reid Barton already mentioned. Let $B$ be the category of finite , strictly bipointed sets and $C = B^{op}$ the cartesian cube category - which is the free finite product category on an interval $1 \rightrightarrows I$. Write the objects in the form $[n]$, so $I = [1]$ and $[n] \times [m] = [n+m]$. Presheaves on $C$ are called (Cartesian) cubical sets, written $cSet$. There is a functor $S : C \to C$ taking $[n]$ to $[n+1]$ , which induces $S^* : cSet \to cSet$ by restriction, and therefore has both left $S_!$ and right $S_*$ adjoints. One can easily calculate that

$S!X = X\times I$

so its adjoint must be

$S^*X = X^I$.

The further right adjoint $S_*$ then also exists and is therefore an "amazing right adjoint" $X^I \dashv S_*X$. We write it as $X_I$ and call it the "$I^{th}$ root". In particular, $I$ itself is therefore "tiny".

Steve Awodey (Oct 17 2021 at 04:31):

this is essentially the same as Reid's example.

Jacques Carette (Oct 18 2021 at 13:17):

Can one exhibit / compute that further right adjoint? What does that functor actually do?

Steve Awodey (Oct 18 2021 at 14:49):

in cubical sets:
$X^I (n) = [ I^n , X_I ] = [ (I^n)^I , X ] = [ (I^I)^n, X ] = [(I + 1)^n, X]$
$= [ \Sigma_{k=0, ..., n}C(n,k) I^{k} , X ]$
$= \Pi_{k=0, ..., n}[ I^{k} , X ]^{C(n,k)}$
$= \Pi_{k=0, ..., n}X(k)^{C(n,k)}$,
where $C(n,k)$ is the binomial coefficient.

Jacques Carette (Oct 18 2021 at 15:17):

I think I can reconstruct all the justifications, except for $\left[I^k,X\right]$ to $X(n)$. [I'm probably missing something simple, yet I am missing it!]

Jacques Carette (Oct 18 2021 at 15:18):

It would make more sense if it were $X(k)$, but even then, I'm not sure.

Jens Hemelaer (Oct 18 2021 at 20:06):

Indeed, it should be $[I^k, X] = X(k)$. Then $[I^k, X] = X(k)$ is an application of Yoneda lemma to the category of presheaves
$cSet \simeq \mathbf{PSh}(C).$
We have $I^k = \mathbf{y}([k])$ where $\mathbf{y}$ is the Yoneda embedding.

Jacques Carette (Oct 18 2021 at 22:03):

Aha, Yoneda, of course! Now, the above computation exhibits $X^I$ quite explicitly, and $X_I$ shows up in there as well -- but I was kind of hoping for $X_I$ to be the item being isolated, not $X^I$. Or maybe it's just a typo and the very first thing should be $X_I(n)$ ?

Steve Awodey (Oct 18 2021 at 23:39):

Jacques Carette said:

It would make more sense if it were $X(k)$, but even then, I'm not sure.

thanks - that was indeed a typo!