In the Yoneda Lemma, how is there an isomorphism $\text{Nat}[\text{Hom}(x,-), F(-)] \cong Fx$ when only $\text{Hom}(x,x)$ gets mapped to $Fx$?

Question

I've been trying to understand the Yoneda Lemma. I have done all of the diagram chasing and understand that the naturality is dependent on one element $id_x$ and all arrows all naturally extended from this point. But what I don't get is how there can be a bijection from $\text{Nat}[\text{Hom}(x,-), F(-)]$ to $Fx$. What does make sense at this point for me is that there can be a bijection between $\text{Hom}(x,x)$ and $Fx$. But when you look at $\text{Hom}(x,y)$, this would get mapped to $Fy$ and there would be no corresponding point in $\text{Hom}(x,x)$ so I cannot see how it can have a corresponding point in $Fx$...

The fact that the elements of $Fx$ is in one to one correspondence with all natural tranformations between the two functors is very amazing yet still mysterious to me. What am I missing here?

Answer 1

The first thing you want to see is that any $a\in F(X)$ determines a unique natural transformation. Define $\phi:\mathrm{Hom}(X,-)\to F$ to be the natural transformation whose component at $Y$, $\phi_Y$, is the function $f\in\mathrm{Hom}(X,Y)\mapsto F(f)(a)\in F(Y)$. To see that this gives a natural transformation, take any $g:Y\to Z$ and see that $$\phi_Z\circ \mathrm{Hom}(X,g)(f)=\phi_Z(g\circ f)=F(g\circ f)(a)=F(g)(F(f)(a))=F(g)\circ\phi_Y(f).$$ In particular, notice our definition of $\phi$ implies that $\phi_X(id_X)=a$.

Now let $\psi$ be any other natural transformation $\mathrm{Hom}(X,-)\to F$ with $\psi_X(id_X)=a$. What does $\psi_Y(f)$ do to a morphism $f:X\to Y$? Because the action of a Hom functor on morphisms is defined by composition, $f=\mathrm{Hom}(X,f)(id_X)$; and by naturality, $\psi_Y\circ\mathrm{Hom}(X,f)=F(f)\circ\psi_X$; so $$\psi_Y(f)=F(f)\circ\psi_X(id_X)=F(f)(a).$$ But this means that each $\psi_Y=\phi_Y$ for any $Y$, or that $\phi$ and $\psi$ are the same natural transformation. So a natural transformation $\mathrm{Hom}(X,-)\to F$ is uniquely determined by the image of $id_X$.

So for any $a\in F(X)$, let's denote by $\theta:F(X)\to[\mathrm{Hom}(X,-),F]$ the map taking $a$ to the natural transformation $\mathrm{Hom}(X,-)\to F$ that takes $id_X$ to $a$. We have just proved that the function $a\mapsto \theta(a)$ is injective; but obviously, every natural transformation sends $id_X$ to some member of $F(X)$, so it's also surjective.

Answer 2

It seems you are confused by all the widgets at play in the statement of Yoneda's lemma. I will not prove anything, but I will state clearly what we are after.

First there is this functor $$y : \mathbf C \to [\mathbf C,\mathsf{Set}]^{\rm op}$$ that maps $x \in \mathbf C$ to the covariant presheaf $\hom_{\mathbf C}(x,-)$. Next if you take some $F$ in $[\mathbf C,\mathsf{Set}]$ there is this other functor $$\hom_{[\mathbf C,\mathbf{Set}]}({-},F) : [\mathbf C,\mathbf{Set}]^{\rm op} \to \mathsf{Set}$$ that maps any other covariant presheaf $G$ to the set of natural transformations between $G$ and $F$. We can now compose those two functors to get a functor $$ \mathbf C \xrightarrow y [\mathbf C,\mathbf{Set}]^{\rm op} \xrightarrow{\hom_{[\mathbf C,\mathbf{Set}]}({-},F)} \mathsf{Set} $$ Let us call $\underline F$ this new functor. This is a functor $\mathbf C \to \mathsf{Set}$ so it makes sense to ask wether it is isomorphic to another functor of the same type, for example $F$. And the Yoneda lemma says presicely that:

There is an isomorphisms $\underline F \simeq F$ in $[\mathbf C ,\mathsf{Set}]$.

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Now the proof of Yoneda's lemma is quite enlightening so let me try to give you the means to understand it. The proof starts by defining a morphism $\underline F \to F$ and then proves it is an isomorphism. So how do we define such a morphism $\alpha : \underline F \to F$: it lives in $[\mathbf C,\mathsf{Set}]$ so we need to define it on every component $x\in\mathbf C$ and check that is is indeed natural. So we need a map $\alpha_x : \underline F(x) \to F(x)$ for each $x\in\mathbf C$. Start by detailing what is $\underline F(x)$: by definition it is $\hom_{[\mathbf C,\mathsf {Set}]}(y(x),F)$. So we need to define $\alpha_x$ that takes as input morphisms $y(x) \to F$ in $[\mathbf C,\mathsf{Set}]$ and renders as output an element of $F(x)$. Now detail a little what is such an input: a morphism $f:y(x) \to F$ in $[\mathbf C,\mathsf{Set}]$ the data for each $z\in\mathbf C$ of a map $f_z : y(x)(z) \to F(z)$ such that they verify the naturality condition, or by rewriting the domain of $f_z$ it is the data of maps $f_z : \hom_{\mathbf C}(x,z) \to F(z)$ for each $z$. Among all those $f_z$ that form the data of $f$, there is in particular $f_x:\hom_{\mathbf C}(x,x) \to F(x)$ that will become handy. Ok so now I am ready to propose some $\alpha$ to you: I propose the morphism $\alpha$ which is defined on every component $x\in\mathbf C$ by $$ \begin{aligned}\alpha_x :\hskip 50pt \underline F(x) \hskip 10pt&\to \hskip 10pt F(x) \\ (f:y(x)\to F(x))&\mapsto f_x({\rm id}_x)\end{aligned}$$ I leave to you to prove it is natural. So we have defined a map $\alpha : \underline F \to F$. Now we have to prove it is a isomorphism in $[\mathbf C,\mathsf{Set}]$. If you think strongly about what it means for a minute ("there is a morphism $\beta$ in the other direction such that $\alpha\circ \beta$..."), you should understand that we need to prove that each of the $\alpha_x$ are bijective. I leave that to you again !