Part of proof of Yoneda's Lemma from Vakil

Question

I am trying to understand the second half of the proof of Yoneda's Lemma, which is given as a problem in Vakil's notes.

So suppose we have two objects $A$ and $A'$ in a Category $D$, and morphisms $i_C:Mor(C, A) → Mor(C,A′)$ that commute with the maps $Mor(C,A) → Mor(B,A)$, which are induced by some morphism $f:B→C$. Then I have shown that the $i_C$ (as $C$ ranges over the objects of $D$) are induced from a unique morphism $g: A → A′$ . Specifically, there is a unique morphism $g: A → A′$ such that for all $C \in D$ , $i_C$ is given by $u\mapsto g\circ u$.

Now, the next part of the problem, labelled (b), asks to show that if all the $i_C$ are bijections, then $g$ is actually an isomorphism. Now, the diagram that was useful for the first part is the following:

$$\require{AMScd} \begin{CD} \operatorname{Mor}(A,A) @>{i_A}>> \operatorname{Mor}(A,A')\\ @V{}VV @V{}VV \\ \operatorname{Mor}(C,A) @>{i_C}>> \operatorname{Mor}(C,A') \end{CD}$$

However, I am having trouble seeing how to apply this to (b). I know that $i_A(Id_A)=g\circ Id_A=g$. Further, since the sets $\operatorname{Mor}(A,A)$ and $\operatorname{Mor}(A,A')$ are in bijection with each other, $g$ maps back to $Id_A$ via postcompositon with some morphism $h:A'\rightarrow A$, giving $g$ a left inverse. Is this correct? If the right inverse for $g$ can be constructed in a similar manner, does this give the result?

Answer 1

So, by part (a) you know that there exists a unique morphism $g\colon A\to A'$, such that for all objects $C$ the map $i_C\colon \operatorname{Mor}(C,A) \to \operatorname{Mor}(C,A')$ is given by $u\mapsto g\circ u$.
The inverse maps $i_C^{-1}\colon \operatorname{Mor}(C,A') \to \operatorname{Mor}(C,A)$ again satisfy the hypotheses of (a), so there exists a unique morphism $\tilde g\colon A'\to A$ such that for all objects $C$ the map $i_C^{-1}$ is given by $v\mapsto \tilde g\circ v$.

This is all you need from (a) to conclude: Putting $C=A$ we know that the composition $$ \operatorname{Mor}(A,A) \xrightarrow{i_A} \operatorname{Mor}(A,A')\xrightarrow{i_A^{-1}} \operatorname{Mor}(A,A) $$ is the identity. Hence, $$ \operatorname{id}_A = i_A^{-1}\bigl(i_A(\operatorname{id}_A)\bigr) = i_A^{-1}(g\circ \operatorname{id}_A) = \tilde g\circ (g\circ \operatorname{id}_A) = \tilde g\circ g. $$ Similarly, putting $C=A'$, the composition $$ \operatorname{Mor}(A',A') \xrightarrow{i^{-1}_{A'}} \operatorname{Mor}(A',A) \xrightarrow{i_{A'}} \operatorname{Mor}(A',A') $$ is the identity, so that $$ \operatorname{id}_{A'} = i_{A'}\bigl(i_{A'}^{-1}(\operatorname{id}_{A'})\bigr) = i_{A'}(\tilde g\circ \operatorname{id}_{A'}) = g\circ (\tilde g\circ \operatorname{id}_{A'}) = g\circ \tilde g. $$ But $\tilde g\circ g = \operatorname{id}_A$ and $g\circ \tilde g = \operatorname{id}_{A'}$ means that $g$ is an isomorphism with inverse $\tilde g$.

Answer 2

If $i_1: h_A \to h_{A’}$ is a objectwise bijection then its an isomorphism of functors by taking the objectwise inverses, so you get $i_2 = i_1^{-1}$.

If part (a) is done then you know these correspond to $g_1, g_2$ and the question is why $g_2 = g_1^{-1}$.

Need to show $h_A(g_2g_1^{-1}) = id_{Mor(A,A)}$ implies $g_2g_1^{-1} = id_A$, or more generally that for $f\in End(A)$, then $h_A(f) = id_{End(A)}$ implies $f= id_A$. This follows from evaluating at $id_A$, ie $h_A(f)(id_A) = f\circ id_A = f$.