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?

  • $\begingroup$ 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
  • 4
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

You proof is correct and I do not believe that there are many alternatives.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.