The Yoneda Lemma and coends

Question

I am trying to understand the proof of Proposition 2.2 of nlab's page on the co-Yoneda lemma. I don't understand the last part of the argument, and I was hoping somebody could help. In particular, we have a presheaf $\mathcal{C}^{\text{op}} \overset{F}{\rightarrow}\text{Set},$ and we have an isomorphism $$\text{Nat}\left(\int^cF(c)\times\mathcal{C}(\_,c),G\right)\cong\text{Nat}(F,G)$$ natural in $G.$ Now the site says we can "use Yoneda" to get $$\int^cF(c)\times\mathcal{C}(\_,c) \cong F,$$ but I don't understand how. It would be really great if somebody could explain this last step as explicitly as possible.

Answer 1

When category theorists say "by Yoneda" or "by the Yoneda lemma", we often actually mean "by an argument in the same fashion as the usual proof of the Yoneda lemma". Specially, this refers to chasing identity morphisms through natural bijections.

For instance, suppose we have bijections $$\mathcal{A} (X, A) \cong \mathcal{A} (Y, A)$$ that are natural in $A$. Then we may set $A = X$ and ask which morphism $g : Y \to X$ corresponds to $\textrm{id} : X \to X$. Or we may set $A = Y$ and ask which morphism $f : X \to Y$ corresponds to $\textrm{id} : Y \to Y$. Or we could look at the naturality squares: $$\require{AMScd} \begin{CD} \mathcal{A} (X, X) @>{\cong}>> \mathcal{A} (Y, X) \\ @V{f_*}VV @VV{f_*}V \\ \mathcal{A} (X, Y) @>>{\cong}> \mathcal{A} (Y, Y) \end{CD}$$ $$\require{AMScd} \begin{CD} \mathcal{A} (X, Y) @>{\cong}>> \mathcal{A} (Y, Y) \\ @V{g_*}VV @VV{g_*}V \\ \mathcal{A} (X, X) @>>{\cong}> \mathcal{A} (Y, X) \end{CD}$$ Chasing $\textrm{id}_X$ around the first diagram tells us that $f \circ g$ is the morphism $Y \to Y$ that corresponds to $f : X \to Y$, i.e. $\textrm{id}_Y$. Chasing $\textrm{id}_Y$ around the second diagram tells us that $g \circ f$ is the morphism $X \to X$ that corresponds to $g : Y \to X$, i.e. $\textrm{id}_X$. So $f : X \to Y$ and $g : Y \to X$ are an inverse pair of isomorphisms, and in particular $X \cong Y$.

Now set $\mathcal{A} = [\mathcal{C}^\textrm{op}, \textbf{Set}]$, $X = \int^c F (c) \times \mathcal{C} (-, c)$, $Y = F$, and $A = G$ to get the result you want.

Answer 2

The Yoneda lemma tells us that if we have two representable functors, say $\text{Hom}(c,-)$ and $\text{Hom}(d,-)$, then natural transformations between them are in one-to-one correspondence to morphisms $d\rightarrow c$.

Now natural transformations are of course again just morphisms in the category of (small) categories, and hence this natural transformation you are given must correspond to a morphisms (here natural transformation)

$$F\rightarrow \int^c F(c)\times \mathcal{C}(-,c)$$

Then since your original transformation is an isomorphism, we also find the reverse morphism

$$\int^c F(c) \times \mathcal{C}(-,c) \rightarrow F, $$

and these must be inverse to each other by construction.