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$\newcommand{\PShv}{\text{PShv}}$ $\newcommand{\Fun}{\text{Fun}}$ $\newcommand{\C}{\mathcal{C}}$ $\newcommand{\Hom}{\text{Hom}}$ $\newcommand{\ra}{\rightarrow}$ $\newcommand{\op}{\text{op}}$ $\newcommand{\Set}{\mathsf{Set}}$ Let $\C$ be a small category and let $\PShv(\C) := \Fun(\C^\op, \Set)$ be the category of presheaves. I would like to show that for two presheaves $F, G$ we have the natural isomorphism $$ \Hom_{\PShv(\C)}(F, G) \cong \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G)) $$ where $(* \Rightarrow F)^\op$ is the (opposite) comma category, $P : (* \Rightarrow F)^\op \ra \PShv(\C)$ is given by $(x, y) \mapsto Y(x)$ ($Y: \C \ra \PShv(\C)$ is the Yoneda embedding $ x \mapsto \Hom_\C(-, x)$) and $$ \Delta : \PShv(\C) \ra \Fun((* \Rightarrow F)^\op , \PShv(\C))$$ is the diagonal/constant functor (used in the adjoint definition/formulation of (co)limits).

The proof I am reading starts by taking a natural transformation $\eta : F \Rightarrow G \in \Hom_{\PShv(\C)}(F, G)$, and then defines the correspoding natural transformation $\zeta \in \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G))$ by $$ \zeta_{(x, y)}(g) = G(g) (\eta_x(y)) \qquad (*)$$ for any object $(x, y) $ of $(* \Rightarrow F)^\op$ and any $g \in \Hom_\C(x', x)$. So here $\zeta : P \Rightarrow \Delta(G)$ and its component at $(x, y)$ is a map with domain and codomain $$ \zeta_{(x, y)} : P(x, y) = Y(x) \ra \Delta(G)(x, y). $$ My first question is: how can $\zeta_{(x, y)}$ possibly take a map $g \in \Hom_\C(x', x)$ as its argument if the domain of $\zeta_{(x, y)}$ is $P(x,y)=Y(x)=\Hom_\C(-, x) \in \PShv(\C)$?

The proof then considers the RHS of the isomorphism that is to be proved, namely it states that any $\zeta \in \Hom_{\Fun((* \Rightarrow F)^\op , \PShv(\C))}(P, \Delta(G))$ must be of the form $(*)$ for a unique $\eta \in \Hom_{\PShv(\C)}(F, G)$ by definition of naturality, since $\zeta$ is itself a natural transformation. I'm trying to show this explicitly but I'm not quite sure how to.

I've started by writing out $\zeta$ as a natural transformation; we have $\zeta : P \Rightarrow \Delta(G)$ natural so for $(x_1, y_1), (x_2, y_2) \in (* \Rightarrow F)^\op$ and a morphism $f: (x_1, y_1) \ra (x_2, y_2)$ we have the naturality square (in equations): $$ \Delta(G)(f) \circ \zeta_{(x_1, y_1)} = \zeta_{(x_2, y_2)} \circ P(f) $$ I'd be grateful if anybody could explain how to recover something of the form $(*)$ from what we have above. Thank you.

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  • $\begingroup$ You probably know this, but that comma category is typically referred to as the category of elements of $F$. $\endgroup$ – Derek Elkins May 13 at 22:50
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For your first question, it's a matter of sloppiness : what should be written is $(\zeta_{(x,y)})_{x'}(g)$, they just removed the $x'$ for ease of notation. I won't do this in my answer though.

For you to find $\eta$ you need to unravel the Yoneda lemma (it's no wonder the result you're trying to prove is also sometimes called the Yoneda lemma). I will change your notations slightly, and write $(x,f)$ for a generic element of $\int_CF= (*\implies F)^{op}$, that is $x\in C, f\in F(x)$, so that I can use $y$ for other generic elements of $C$ (and $g$ for a generic element of $F(y)$)

Suppose you have a natural transformation $\zeta : P\to \Delta (G)$. So in particular for $(x,f)\in\int_CF$ you have a map $\zeta_{(x,f)} : Y(x)\to G$.

Now recall what the Yoneda lemma tells us precisely about the shape of an arrow $\theta : Y(x)\to G$. It tells us that $\theta_z(f) = G(f^{op})(\theta_x(id_x))$ : this is again by the "stupid" square :

$\require{AMScd} \begin{CD} \hom(x,x) @>{\theta_x}>> G(x)\\ @V{\hom(f,x)}VV @VV{G(f^{op})}V\\ \hom(z,x) @>>{\theta_z}> G(z) \end{CD}$

But then, with $g:x'\to x$, $(\zeta_{(x,f)})_{x'}(g)= G(g^{op})((\zeta_{(x,f)})_x(id_x))$

We have our "$\zeta_{(x,f)}(g)$" (removing the $x'$-index, as you did) and our $G(g)$ (I denoted it $G(g^{op})$ for clarity but of course it's the same).

Let's look at what $(\zeta_{(x,f)})_x(id_x)$ does. It is quite clearly an element of $G(x)$, and if you move $f\in F(x)$, it moves with it. Sounds to me like this is a nice way to define $\eta$ : define $\eta_x : F(x)\to G(x)$ by $\eta_x(f) := (\zeta_{(x,f)})_x(id_x)$.

By what came just before, $(\zeta_{(x,f)})_{x'}(g)= G(g^{op})(\eta_x(f))$, which is what we wanted. We now only have to show that $\eta$ is natural.

So suppose you have a map $\alpha : x\to y$, and you want to look at the square (I'm removing the $^{op}$'s, hopefully it's still clear)

$\require{AMScd} \begin{CD} F(y) @>{\eta_y}>> G(y)\\ @V{F(\alpha)}VV @VV{G(\alpha)}V\\ F(x) @>>{\eta_x}> G(x) \end{CD}$

Start with $g\in F(y)$, and look at $f:= F(\alpha)(g) \in F(x)$. Then $\alpha : (x,f)\to (y,g)$ is a morphism in $\int_CF$, by definition, therefore we have a commutative diagram

$\require{AMScd} \begin{CD} Y(x) @>{\zeta_{(x,f)}}>> G\\ @V{P(\alpha)}VV @VV{id}V\\ Y(y) @>>{\zeta_{(y,g)}}> G \end{CD}$ (I don't know how to do triangles)

But then, if you evaluate this in $x$ and then $id_x$ you get $(\zeta_{(y,g)})_x(\alpha)=(\zeta_{(x,f)})_x(id_x) = \eta_x(f)$.

And now recall again the explicit version of the Yoneda lemma, which tells us precisely that $(\zeta_{(y,g)})_x(\alpha) = G(\alpha)((\zeta_{(y,g)})_y(id_y)) = G(\alpha)(\eta_y(g))$ so that in the end, $\eta_x(f) = G(\alpha)(\eta_y(g))$. But now remember the definition of $f$ : $\eta_x\circ F(\alpha)(g) = G(\alpha)\circ \eta_y(g)$.

This holds for any $g$, so the diagram commutes, so $\eta$ is natural; and we have found our $\eta$.

It remains to show that this $\eta$ is unique; but this is easy, as $\eta_x(f) = (\zeta_{(x,f)})_x(id_x)$ by the formula applied for $x'=x, g=id_x$ (I'm just realizing now that of course evaluating in $g=id$ would have given us $\eta$ without having to look for it as I did above - but at least it's confirming what I did, and it shows that in these examples there aren't 100 ways to do it)

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  • $\begingroup$ Thank you very much for your detailed answer, this is very helpful! One quick question: when you say we have $\theta_z(f) = G(f^{op})(\theta_x(id_x))$ by the square (Yoneda's Lemma) why exactly is this? So Yoneda says $$ \hom_{\text{PShv}(\mathcal{C})}(Y(x), G) \cong G(x)$$ which gives us the square you drew and I suppose I see why $\theta_z(f) = G (\cdots)$ because of Yoneda, but why do we have $G(f^{op})(\theta_x(\text{id}_x))$ specifically? Edit: Ah, I think it's because of the explicit map of the Yoneda embedding, namely $\theta \mapsto \theta_x(\text{id}_x)$. $\endgroup$ – mathphys May 14 at 13:56
  • $\begingroup$ Recall the proof of the Yoneda lemma : the square commutes by naturality of $\theta$, and when you evaluate it on $id_x \in \hom (x,x)$ : going down then right gives $\theta_z(\hom(f,x)(id_x))$ which is, by definition $\theta_z(f)$ (recall what $\hom(-,x)$ is ). Going right then down gives $G(f^{op})(\theta_x(id_x))$, which is the desired equality $\endgroup$ – Max May 14 at 13:58
  • $\begingroup$ Yes you beat me to it (I remembered after I asked haha). Thanks again for your help! $\endgroup$ – mathphys May 14 at 14:01
  • $\begingroup$ Yes, and I think one moral to this story (one thing to remember) is that you need to remember which isomorphisms you use to identify two things. I insisted on that in an answer to a question "why can you identify things that are isomorphic ?" and the point is that you can if you know the isomorphism because then you have a recipe to translate every statement about one to a statement about the other : just add $f$'s and $f^{-1}$'s to make it well-typed. The Yoneda lemma is no exception to this $\endgroup$ – Max May 14 at 16:59

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