# Representability $h_C(X) \simeq F$ of functors determines rep. $X$ up to unique isomorphism (using Yoneda Lemma).

A functor $F \in C^{\wedge}$ from $C^{op}$ to $\text{Set}$ is representable if there is an isomoprhism $\varphi \in \text{Hom}_{C^{\wedge}}(h_C(X), F)$ for some $X \in C$, we can also write that as $h_C(X) \simeq F$.

I want to prove that $X$ would be determined up to unique isomorphism.

Proof. Suppose that $\varphi: h_C(X) \simeq F$, and $\phi : h_C(Y) \simeq F$, then $\phi \varphi^{-1} : h_C(X) \simeq h_C(Y) \in \text{Hom}_{C^{\wedge}}(h_C(X), h_C(Y)) \simeq h_C(X,Y)$.

How do I complete the proof?

Thanks @Anton

That got me this far:

We have a functor from $C^{op} \to \text{Set}$ given by $h_C(X)$ which is fully faithful. By functoriality, any isomorphism in $C$ is also one under $h_C(X) \circ \text{op}$. Since $h_C$ is fully faithful, we have that an isomorphism in $h_C(X, Y)$ corresponds to one in $\text{Hom}_{C^{\wedge}}(h_C(X), h_C(Y))$ under the Yoneda bijection, $\psi$. Let the "forward" Yoneda bijection be $\varphi$. Then $h_C(X) \simeq F, \ h_C(Y) \simeq F$ gives an isomorphism $h_C(X) \simeq h_C(Y): \theta$. Then $\varphi(\theta) : X \to Y$. We need to show that $\varphi(\theta)$ is unique (the only such isomorphism). Well if $\phi$ were another isomorphism, then $\phi^{-1} \varphi(\theta)$ is an automorphism of $X$. We know that $\text{Hom}_{C^{\wedge}}(h_C(X), h_C(X)) \simeq h_C(X,X)$

But still stuck.

## 2 Answers

Note that the representing morphisms provided by the Yoneda's lemma satisfy the same relations as the natural transformations they represent, by uniqueness of the representation. This implies that the identity natural transformation is represented by the identity morphism and a composition of natural transformations corresponds to a composition of representing morphisms. Now you can turn an isomorphism between the functors into an isomorphism of representing objects.

• I don't know about uniqueness of representation. – BananaCats Category Theory App Aug 19 '17 at 3:30
• @EyesOnBud Your proof is finished, because the natural isomorphism $\theta$ admits a unique representing morphism $\phi(\theta)$ by Yoneda's lemma. You only need to prove that it is an isomorphism, which you already did: the inverse morphism represents the inverse natural transformation and the inverse laws hold by Yoneda's lemma. – Anton Fetisov Aug 19 '17 at 5:14

Since $\varphi$ is a bijection there can be only a unique $\theta$ such that $\varphi(\theta) \colon h_C(X) \to h_C(Y)$ is your natural isomorphism between the representable presheaves.

This should conclude your proof.

• I don't understand why though. Could you explicitly that? – BananaCats Category Theory App Aug 19 '17 at 22:43
• @EyesOnBud you said it yourself "Let the "forward" Yoneda bijection be $\varphi$". – Giorgio Mossa Aug 20 '17 at 11:44