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 2


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.

  • $\begingroup$ I don't know about uniqueness of representation. $\endgroup$ Aug 19, 2017 at 3:30
  • 2
    $\begingroup$ @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. $\endgroup$ Aug 19, 2017 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.

  • $\begingroup$ I don't understand why though. Could you explicitly that? $\endgroup$ Aug 19, 2017 at 22:43
  • $\begingroup$ @EyesOnBud you said it yourself "Let the "forward" Yoneda bijection be $\varphi$". $\endgroup$ Aug 20, 2017 at 11:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .