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This is from pg93, ex 4.1.27.

Let $\mathcal{A}$ be a locally small category, and let $A,A' \in \mathcal{A}$ with $\operatorname{Hom}(-,A) \cong \operatorname{Hom}(-,A')$. Prove directly that $A \cong A'$.

My thoughts:

Let $\eta$ denote the natural isomoprhism between the two functors. We have $$H(A,A) \xrightarrow{ \eta_A} H(A,A')$$ So $\eta_A(id_A): A \rightarrow A'$ is a candidate. Now I want to construct inverse. $$ H(A',A') \xrightarrow{\eta_{A'}} H(A',A)$$ Then the map $\eta^{-1}_{A'} (id_{A'}):A' \rightarrow A$. So I'd like to show that the composition is identity. By the naturality condition of $\eta$, from $H_A(A) \rightarrow H_{A'}(A')$ (and similarly on other direction) $$ \eta_A(id_A) \circ \eta^{-1}_{A'}(id_{A'}) = \eta_{A'}(\eta_{A'}^{-1}(id_{A'}))= id_{A'} $$ We deduce these two maps are inverses. Hence $A' \cong A$.

Is this correct? Or is there a neat way to see this?

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  • $\begingroup$ I'm not sure but this seems to be incorrect. Let $\mathcal{A}$ be a category with only two objects $X,Y$ and two morphisms: identities. Then surely $Hom(-,X)$ is naturally isomorphic to $Hom(-,Y)$ but these are not isomorphic objects. Or am I missing something? $\endgroup$ – freakish Jun 28 '18 at 13:21
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    $\begingroup$ @freakish $Hom(-,X)$ and $Hom(-,.Y)$ are not naturally isomorphic: $Hom(X,X)$ contains the identity, $Hom(X,Y) $ is empty. $\endgroup$ – Paul Frost Jun 28 '18 at 13:43
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    $\begingroup$ You have a typo : the second arrow in your quotation block should be $$ H(A',A') \overset {\eta_{A'}^{-1}} \to H(A',A) $$ Also using just $H$ to denote hom-sets are quite unusual : either use $\operatorname{Hom}(X,Y)$ or $\mathcal A(X,Y)$. $\endgroup$ – Pece Jun 28 '18 at 15:48
  • $\begingroup$ Related : math.stackexchange.com/questions/704891/… $\endgroup$ – Arnaud D. May 28 '19 at 15:47
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You proof is correct and I do not believe that there are many alternatives.

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  • $\begingroup$ Does one also need to verify that the second composition equals $id_A$, or does it follow automatically from something? $\endgroup$ – user634426 Jul 13 '19 at 20:16
  • $\begingroup$ You must verify it, but it is the "same" proof. $\endgroup$ – Paul Frost Jul 13 '19 at 22:27
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According to the Yoneda lemma an arrow $f:A\to A'$ exists such that for every object $B$ the map $\eta_B:\mathsf{Hom}(B,A)\to\mathsf{Hom}(B,A')$ is prescribed by $h\mapsto f\circ h$.

Likewise an arrow $g:A\to A'$ exists such that for every object $B$ the map $\eta_B^{-1}:\mathsf{Hom}(B,A')\to\mathsf{Hom}(B,A')$ is prescribed by $h\mapsto g\circ h$.

Then $f\circ g=\eta_A\circ\eta_A^{-1}(1_A)=1_A$ and $g\circ f=\eta_{A'}^{-1}\circ\eta_{A'}(1_{A'})=1_{A'}$ proving that $A$ and $A'$ are isomorphic.

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