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So we have a category $C$ and objects $X$ and $Y$. Given that Mor$_C(-,X)$ and Mor$_C(-,Y)$ are isomorphic functors, then prove that $X\cong Y$ (as objects in $C$).

So by the condition I know that there exists a natural transformation $\eta$ such that for each object $A\in C$ we obtain the map $\eta_A$ such that $\eta_A:$Mor$_C(A,X)\rightarrow$ Mor$_C(A,Y)$. Further, we have that each $\eta_A$ is an isomorphism. My doubts comes on what $\eta$ does to Mor$_C(A,X)$, does it take an arrow from $A$ to $X$ and it makes it now point to $Y$? My intuition tells me that if we let $A=X$, then we have two arrows (call them $f$ and $g$) in Mor$_C(X,X)$ such that $fg=1_X$, and somehow I feel that $\eta_X$ should take $f$ and $g$ to arrows that do the same thing.

Could someone please tell me whether my intuition is correct and perhaps show me how to do this with rigor?

Thanks.

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See also the proof of the Yoneda lemma which generalizes the idea of this proof. –  Adeel Dec 9 '12 at 19:36
    
@Adeel, is it possible to prove this theorem as a consequence of Yoneda? –  user51427 Dec 9 '12 at 19:43
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@sunflower Yes: the Yoneda lemma implies that the functor $A \mapsto \mathcal{C}(-, A)$ is a fully faithful embedding of $\mathcal{C}$ into $[\mathcal{C}^\textrm{op}, \textbf{Set}]$, and any fully faithful functor reflects isomorphisms. –  Zhen Lin Dec 9 '12 at 19:51
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up vote 4 down vote accepted

$\newcommand{\mc}{\mathcal}$Yes, suppose we have the natural isomorphism $$\eta : \mc C(-,A) \to \mc C(-,B)$$ then we have, for arbitrary $f : B \to A$ the naturality condition: $\mc C(f,B) \circ \eta_A = \eta_B \circ \mc C(f,A)$ which, applying both sides to $1_A$ and by definition of the Hom-functor, gives $f \circ \eta_A(1_A) = \eta_B(f)$. Setting $f = \eta^{-1}_B(1_B)$ we find that $\eta_A(1_A) \circ \eta^{-1}_B(1_B) = 1_B$ so (after a similar argument to show that it is also the left inverse) $\eta_A(1_A) : A \to B$ is an isomorphism.


Recall that $\mc C(-,C) : \mc C^\text{op} \to \text{Set}$ is a contravariant functor. When $f : X \to Y$ is a morphism in $\mc C$ we have $\mc C(f,C) : \mc C(Y,C) \to C(X,C)$ defined, for $g \in C(Y,C)$ by $\mc C(f,C)(g) = g \circ f$ because $X \xrightarrow{f} Y \xrightarrow{g} C$.

Let us see the full calculation with this $$ \begin{eqnarray}{} \mc C(f,B) \circ \eta_A &=& \eta_B \circ \mc C(f,A) \\ \mc C(f,B)(\eta_A(1_A)) &=& \eta_B(\mc C(f,A)(1_A)) \\ \mc \eta_A(1_A) \circ f &=& \eta_B(1_A \circ f) \\ \mc \eta_A(1_A) \circ f &=& \eta_B(f) \\ \end{eqnarray}$$

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The key observation is that $\eta_A : \mathcal C(A,A) \to \mathcal C(A,B)$ gives you can easy way to get a function $A \to B$ and we just need to show it's an isomorphism - so naturality gives us that. –  user51427 Dec 9 '12 at 19:35
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so on the equation that you have Mor$_C(f,B)\circ \eta_A(1_A)=\eta_B\circ $Mor$_C(f,A)(1_A)$, you use contravariance to obtain on the right hand side $\eta_B\circ (1_A\circ f)=\eta_B(f)$, but on the left hand side if you have Mor$_C(f,B)\circ \eta_A(1_A)$ shouldnt you have $\eta_A(1_A)\circ f$? –  Daniel Montealegre Dec 9 '12 at 22:40
    
@DanielMontealegre, thanks for your comment, I had the order backwards by accident. I fixed it now and added full details of the calculations. –  user51427 Dec 9 '12 at 22:56
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