3
$\begingroup$

I've been trying to prove the following claim, but am now unsure about its truth. Is it true, and if so, where can I find a proof?

Claim: For any categories C, D, E such that C and D are equivalent,

(i) The set $Hom($C, E$)$ of functors from C to E is in bijective correspondence with the set $Hom($D, E$)$ of functors from D to E, i.e. $Hom($C, E$)$ $\simeq$ $Hom($D, E$)$.

(ii) The set $Hom($E, C$)$ of functors from E to C is in bijective correspondence with the set $Hom($E, D$)$ of functors from E to D, i.e. $Hom($E, C$)$ $\simeq$ $Hom($E, D$)$.

$\endgroup$
3
  • 10
    $\begingroup$ These claims are not true. You can refute them using 1- or 2-object categories having the property that for any objects $A$ and $B$ there is a unique morphism from $A$ to $B$. $\endgroup$ Nov 10, 2018 at 20:11
  • 1
    $\begingroup$ What is true is that the categories $\mathrm{Hom}(C,E)$ and $\mathrm{Hom}(D,E)$ (and vice versa) are equivalent. Are there size issues here? $\endgroup$
    – MTyson
    Nov 10, 2018 at 21:10
  • $\begingroup$ They equivalent because there is a bijective correspondence between the two? $\endgroup$
    – Brofessor
    Aug 18, 2020 at 10:36

1 Answer 1

5
$\begingroup$

I will turn my comment into an answer.

If $X$ is any nonempty set, then let ${\mathcal C}_X$ be the category which has $X$ as its class of objects and, for each pair $(x_1,x_2)\in X^2$, has exactly one morphism $\varphi_{x_1,x_2}:x_1\to x_2$. Every morphism in ${\mathcal C}_X$ is necessarily an isomorphism.

Any function $f: X\to Y$ is the object part of a unique functor $F: {\mathcal C}_X\to {\mathcal C}_Y$, and any two such functors are naturally isomorphic. In particular, ${\mathcal C}_X$ and ${\mathcal C}_Y$ are categorically equivalent for any nonempty $X$ and $Y$.

Hence for $X = \{0\}$, $Y = \{0,1\}$, and ${\sf C} = {\mathcal C}_X$, ${\sf D} = {\sf E} = {\mathcal C}_Y$ we have
(i) $|\textrm{Hom}({\sf C}, {\sf E})| = 2$ and $|\textrm{Hom}({\sf D}, {\sf E})| = 4$, while
(ii) $|\textrm{Hom}({\sf E}, {\sf C})| = 1$ and $|\textrm{Hom}({\sf E}, {\sf D})| = 4$.

$\endgroup$
2
  • $\begingroup$ Fantastic, thanks, this is very helpful. As a small follow up, what are some interesting examples of equivalent but non-isomorphic categories $C$ and $D$ which do satisfy both (i) and (ii) for any $E$? $\endgroup$
    – King Kong
    Nov 15, 2018 at 12:21
  • $\begingroup$ @BenEva: I don't know if there exist such $C$ and $D$. $\endgroup$ Nov 19, 2018 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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