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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$)$.

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    $\begingroup$ 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 $B$. $\endgroup$
    – Keith Kearnes
    Commented Nov 10, 2018 at 20:11
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    $\begingroup$ 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
    Commented Nov 10, 2018 at 21:10
  • $\begingroup$ They equivalent because there is a bijective correspondence between the two? $\endgroup$
    – Brofessor
    Commented Aug 18, 2020 at 10:36

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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 $|\textrm{Hom}({\sf E}, {\sf D})| = 4$.

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  • $\begingroup$ 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
    Commented Nov 15, 2018 at 12:21
  • $\begingroup$ @BenEva: I don't know if there exist such $C$ and $D$. $\endgroup$
    – Keith Kearnes
    Commented Nov 19, 2018 at 19:30

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