# Using Yoneda lemma, the pair $(U, s)$ representing presheaf $F$ are unique up to unique isomorphism.

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.

## Attempt

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?

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 $(s')^{-1} s$

• Oh I see, $u \to u'$ induces the map $h_u \to h_{u'}$ . I did learn that. Thank you! 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.)