Is there only one isomorphism $h_C(X) \simeq h_C(Y)$ if it exists?

Question

I'm trying to prove that if $F \simeq h_C(X)$ or "$X$ represents the functor $F$", then $X$ is unique up to unique isomorphism. I already know that if $h_C(X) \simeq F \simeq h_C(Y)$ that $s: X \simeq Y$ since Yoneda says that $h_C(X)$ is fully faithful, so reflects isomorphisms (in either direction). If $h_C(X) \simeq h_C(Y)$ is unique then I'm done as then $\psi(s) = \psi(s')$ ans so $s = s'$, where $\psi : h_C(X,Y) \to \text{Hom}_{C^{\wedge}}(h_C(X), h_C(Y))$ is the Yoneda bijection.

But how do I know that $h_C(X) \simeq h_C(Y)$ is unique?

Answer 1

There are as many isomorphisms between $h_C(X)$ and $h_C(Y)$ as there are isomorphisms between $X$ and $Y$.

If $\varphi: h_C(X)\cong F$ and $\psi : h_C(Y)\cong F$, then there is a unique isomorphism $\alpha : h_C(X) \cong h_C(Y)$ such that $\psi \circ \alpha = \varphi$ (just post-compose both sides by $\psi^{-1}$), but there are many ways of exhibiting a representation (i.e. many possible natural isomorphisms even for a given representing object $X$) so simply knowing that there is some natural isomorphism that exhibits $X$ as a representation of $F$ is not enough to uniquely pick out an isomorphism between $X$ and $Y$ where $Y$ is another object that represents $F$ via some unspecified natural isomorphism.

Consider coproducts, i.e. representations of $Z\mapsto\mathsf{Hom}(A,Z)\times\mathsf{Hom}(B,Z)$. What does a representation of this functor look like? What is the universal property of coproducts? How does that universal property relate to representability?

Answer 2

The Yoneda lemma gives you a canonical bijection $$d: \textrm{Hom}_{C}(X,Y) \rightarrow \textrm{Hom}_{\textrm{nat}}(\textrm{Hom}_C(-,X),\textrm{Hom}_C(-,Y))$$

Specifically, to each morphism $f: X \rightarrow Y$, associate the natural transformation $d(f): \textrm{Hom}_C(-,X) \rightarrow {Hom}_{C}(-,Y)$ which assigns to each object $T$ of $C$ the function

$$d(f)_T: \textrm{Hom}_C(T,X) \rightarrow \textrm{Hom}_C(T,Y)$$

$$ g \mapsto f \circ g$$

"Unique isomorphism" says in this case that if $\Phi: \textrm{Hom}_C(-,X) \rightarrow \textrm{Hom}_C(-,Y)$ is an isomorphism of functors, then there is a unique isomorphism $\phi: X \rightarrow Y$ such that $d(\phi) = \Phi$. In other words, the bijection $d$ sends nonisomorphisms to nonisomorphisms.

Answer 3

You're asking the wrong question — you're considering the wrong kind of isomorphism. Natural isomorphisms between functors are irrelevant here — the subject is about natural isomorphisms between fuctors equipped with a natural transformation from $F$

Put differently, the relevant notions of isomorphism are those of the coslice category $F / \widehat{C}$

Simplifying, you are only interested in natural isomorphisms $h_C(X) \to h_C(Y)$ that make a commutative triangle

$$ \begin{matrix} F &\to & h_C(X) \\ & \searrow & \downarrow \\ & & h_C(Y) \end{matrix} $$

and this isomorphism is unique.

Regarding the isomorphisms bewteen $X$ and $Y$, the same applies — you are only interested in those isomorphisms that make the above triangle commute. Or put differently, the notion of isomorphism in the comma category $(F, h_C)$.