1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ The number of isomorphisms between two isomorphic objects is the number of automorphisms of either (exercise). Fortunately this is not what "unique isomorphism" means. $\endgroup$ – Qiaochu Yuan Aug 29 '17 at 0:41
  • $\begingroup$ @QiaochuYuan your reputation precedes you :) $\endgroup$ – BananaCats Category Theory App Aug 29 '17 at 3:46
1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $(F, h_C)$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.