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I have never been truly comfortable with Yoneda's Lemma but have noted the following consequence which seems too good to be true:

$Hom(A,-)\cong Hom(B,-)\Rightarrow A\cong B.$

Is not saying all we need is a nontrivial homomorphism $f:A\to B$ such that the above holds and then f becomes an isomorphism?

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  • $\begingroup$ I am not sure I understand what you mean, but note that, a priori, there is no morphism $f : A \to B$ inducing the isomorphism $\hom(A,-) \cong \hom(B,-)$ (but the Yoneda lemma tells you that in fact there is). It is also true that if $f : A \to B$ induces an isomorphism $\hom(A,-) \cong \hom(B,-)$, then $f$ is an isomorphism -- again, this is given by the Yoneda lemma. If you want to see applications of the Yoneda lemma, simply search for the word "Yoneda" on this website. $\endgroup$ – Najib Idrissi Sep 18 '15 at 9:08
  • $\begingroup$ I didnt quite mean f induces $Hom(A,-)\cong Hom(B,-)$. Rather, if we forget about f and say we only have $Hom(A,-)\cong Hom(B,-)$ for some A, B, then it seems too good to be true to say $A\cong B$ without knowing at least knowing something about how A and B interact. This is why I was asking if we need at least a homomorphism $f:A\to B$ so that Yoneda does the rest and tells us we have an isomorphism. $\endgroup$ – Sam Williams Sep 18 '15 at 9:54
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    $\begingroup$ No, you don't need a morphism, that's the whole point. It's constructed in the proof. Look for example at $\operatorname{id}_B \in \hom(B,B)$, then it gets sent to some $f : A \to B$ under the isomorphism $\hom(A,-) \cong \hom(B,-)$: that's your morphism. $\endgroup$ – Najib Idrissi Sep 18 '15 at 11:18
  • $\begingroup$ related: math.stackexchange.com/questions/977267/… $\endgroup$ – Jakob Werner Sep 18 '15 at 11:53

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