I'm quoting from the Stacks Project:

Definition 4.3.6. A contravariant functor $C \to \text{Sets}$ is said to be representable if it is isomorphic to the functor of points $h_u$ for some object $u$ of $C$.

Let $C$ be a category and let $F: C^{op} \to \text{Sets}$ be a representable functor. Choose an object $u$ of $C$ and an isomorphism $s : h_u \to F$. The Yoneda lemma guarantees that the pair $(u,s)$ is unique up to unique isomorphism. The object $u$ is called an object representing $F$.

I am having trouble with the bolded part. I understand how to prove the Yoneda lemma as seen here in my answer.

So how do we go from $\text{Nat}(h_u, F) \simeq F(u)$ to $(u,s)$ is unique up to unique isomorphism? Please guide me as someone new to Yoneda lemma application.


We have that $s : F \simeq h_u$ is a natural isomorphism of functors. Now $u$ is uniquely determined from $h_u$ and thus $s$ clearly. If $s' : F \simeq h_v$ then $s' \circ \theta = s$ for some natural isomorphism $\theta$ I suppose. And also $u \simeq v$. I guess I need to prove that now.

Well do $\theta = s'^{-1} s$ then clearly that isomorphism is unique. We can do that since they compose: $h_u \simeq F \simeq h_v$. Now $u \simeq v$ anyone?


2 Answers 2


If we have a diagram

$$ \begin{matrix} h_u &\xrightarrow{s}& F \\ & & \ \ \ \uparrow \small s' \\ & & h_{u'} \end{matrix} $$

Then the only way to complete this diagram with a map $h_u \to h_{u'}$ to get a commutative triangle is with $(s')^{-1} s$.

The Yoneda lemma implies that there is exactly one morphism $u \to u'$ for which the induced map $h_u \to h_{u'}$ is $(s')^{-1} s$

  • $\begingroup$ Oh I see, $u \to u'$ induces the map $h_u \to h_{u'}$ . I did learn that. Thank you! $\endgroup$ May 29, 2017 at 21:34

The key is applying Yoneda lemma with $F=h_v$, which will yield ${\rm Nat}(h_u,h_v)\cong \hom(u,v)$.

Let $(v,t)$ be another such pair, i.e. $v\in Ob\,C$ and $t:h_v\to F$ natural isomorphism.

Now we have $t^{-1}\circ s:h_u\to h_v$, thus it has a unique correspondent $\gamma:u\to v$ which has an inverse, namely the correspondent of $s^{-1}\circ t:h_v\to h_u$.
(To prove they're indeed inverses, naturality conditions have to be used.)


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