# If $\operatorname{Hom}(X,-)$ and $\operatorname{Hom}(Y,-)$ are isomorphic, why are $X$ and $Y$ isomorphic?

Suppose you have a locally small category $\mathcal{C}$ and objects $X$ and $Y$ such that $\operatorname{Hom}(X,-)$ and $\operatorname{Hom}(Y,-)$ are isomorphic as covariant functors. How can you best see that $X$ and $Y$ are isomorphic in $\mathcal{C}$?

By Yoneda Lemma, we know $$\operatorname{Hom}(\operatorname{Hom}(X,-),\operatorname{Hom}(Y,-))\cong\operatorname{Hom}(Y,X).$$

By assumption, there exists an isomorphism of functors $\Phi\colon\operatorname{Hom}(X,-)\to\operatorname{Hom}(Y,-)$, which by Yoneda lemma corresponds to some $f=\Phi_X(\operatorname{id}_X)\in\operatorname{Hom}(Y,X)$. Again by Yoneda Lemma, we have an isomorphism $\Phi^{-1}\colon\operatorname{Hom}(Y,-)\to\operatorname{Hom}(X,-)$, which corresponds to some $g=\Phi^{-1}_Y(\operatorname{id}_Y)\in\operatorname{Hom}(X,Y)$.

My suspicion is $f\circ g=\operatorname{id}_X$ and $g\circ f=\operatorname{id}_Y$, to give the conclusion. But I'm getting completely bogged down trying to show that. Does anyone have an explicit computation or something to give the conclusion? Thanks.

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The Yoneda Lemma tells you that $C \to \widehat{C}$, $X \mapsto \hom(X,-)$ is fully faithful. Any fully faithful functor reflects isomorphism.

Proof: Let $F : C \to D$ be a fully faithful functor and let $F(X) \cong F(Y)$. Choose an isomorphism $F(X) \to F(Y)$. Choose a morphism $f : X \to Y$ inducing this isomorphism, and choose a morphism $g : Y \to X$ inducing the inverse isomorphism (fullness). Then $f \circ g$ and $g \circ f$ induce the identity on $F(Y)$ resp. $F(X)$, so that they are the identity (faithfulness). $\square$

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Thanks! ${}{}{}$ – Hailie Mathieson Mar 9 '14 at 4:39

Martin Brandenburg's answer is the best way to look at your question. But, if for some reason you do not want to refer to the notions of fullness and faithfulness explicitly, just assume that $$\eta:\operatorname{Hom}(X,-)\to\operatorname{Hom}(Y,-)$$ is a natural isomorphism and prove that $$\phi=\eta_X(1_X)$$ is an isomorphism.

This is done very easily by looking at what naturality of $\eta$ says for $\phi$: The fact that $\eta_Y$ is an isomorphism gives you an $f:X\to Y$ such that $\eta_Y(f)=1_Y$. Then the naturality square gives $\phi\circ f=1_X$.

To prove that $f\circ \phi=1_Y$, just write down what naturality says for $f$ and evaluate the naturality square at $1_X$.

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This is a repetition of the proof of the Yoneda Lemma. – Martin Brandenburg Mar 9 '14 at 11:50
Are you saying that this does not work? – ByronReust Mar 9 '14 at 12:12
I am saying that you repeat the proof of the Yoneda Lemma. – Martin Brandenburg Mar 9 '14 at 12:15
It is an answer to the OP's question. – ByronReust Mar 9 '14 at 12:28
Yes, sure. $\phantom{~~}$ – Martin Brandenburg Mar 9 '14 at 12:30

In the proof of the Yoneda Lemma we show that natural transformations $\text{Hom}(X,-)\to K(-)$ are in bijection with elements of $KX$. An element $e\in KX$ corresponds to the transformation $\eta$ which sends $g\in\text{Hom}(X,Y)$ to $(Kg)(e)\in KY$.

In particular, if $K(-)=\text{Hom}(Y,-)$, then an arrow $f\in\text{Hom}(Y,X)$ corresponds to the natural transformation $$f^*:\text{Hom}(X,-)\to\text{Hom}(Y,-)\\ g\mapsto gf$$ sending $g\in\text{Hom}(X,Z)$ to $(Kg)(f)=\text{Hom}(Y,g)(f)=gf\in\text{Hom}(Y,Z)$.
It is trivial to show that $f^*$ is injective for each object $Z$ iff $f$ is epic, and it's also not difficult to show that $f^*$ is always surjective iff $f$ is a section (i.e. it has a left inverse).

Corollary: If $\text{Hom}(X,-)\cong\text{Hom}(Y,-)$, then $X\cong Y$.
Proof: A natural isomorphism $\sigma:\text{Hom}(X,-)\cong\text{Hom}(Y,-)$ is induced by $\sigma(1_X)=:f\in\text{Hom}(Y,X)$, so $\sigma=f^*$. Since $f^*$ is an isomorphism, it is bijective, hence $f$ is an epic and also a section, which implies that $f$ is an isomorphism.

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