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I'm asking myself: if $C$ and $D$ are two equivalent categories, each one is embedded in the other, that is to say: there exists a fully-faithful functor from $C$ to $D$ and reciprocally. Indeed, an equivalence of categories (and its inverse) is in particular fully faithful.

Is the converse true, that is: if two categories can be embedded into the other reciprocally, are they necessarily equivalent?

Loulou
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    At least in the special case of discrete categories, this is equivalent to the Cantor-Schröder-Bernstein theorem, which requires classical logic. – Naïm Camille Favier Sep 30 '24 at 09:25
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    Relevant, if slightly different: https://mathoverflow.net/questions/452811/does-mathbfcat-have-the-cantor-schr%C3%B6der-bernstein-property – Naïm Camille Favier Sep 30 '24 at 09:27
  • Thank you, I will check it out. – Loulou Sep 30 '24 at 10:34
  • The counterexample provided on that question still works for this question. The posetal categories arising from the partial orders (say) $\mathbb{R}+$ and $\mathbb{R}+^*$ embed into each other, but they are not isomorphic. Since they are skeletal, this implies they are not equivalent. – Jean Abou Samra Sep 30 '24 at 11:23
  • @JeanAbouSamra This is an answer to the question, not just a comment. Comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore, please post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. – Martin Brandenburg Sep 30 '24 at 15:07
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    @martinbrandenburg Will do. – Jean Abou Samra Sep 30 '24 at 15:10
  • Sorry I already posted one now because I wanted to add something about finite categories. – Martin Brandenburg Sep 30 '24 at 16:05

1 Answers1

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The answer is no, already for linear orders, considered as thin categories. Consider the real intervals $(0,1)$ and $[0,1]$, for example. There are injective monotonic maps* $(0,1) \hookrightarrow [0,1]$ and $[0,1] \xrightarrow{\sim} [\frac{1}{4},\frac{3}{4}] \hookrightarrow (0,1)$, but no isomorphism of linear orders. See also A "Cantor-Schroder-Bernstein" theorem for partially-ordered-sets.

But it is worthwile mentioning that, even though it is not true in general, it is true for finite categories.

Let $\mathcal{C}$, $\mathcal{D}$ be finite categories with fully faithful functors $F : \mathcal{C} \to \mathcal{D}$ and $G : \mathcal{D} \to \mathcal{C}$. Let $\mathcal{C}/{\cong}$ and $\mathcal{D}/{\cong}$ denote the sets of isomorphism classes in these categories. Then $F$ and $G$ induce injective maps between these. Therfore, $\mathcal{C}/{\cong}$ and $\mathcal{D}/{\cong}$ are finite sets with the same number elements. But then the injective map $F : \mathcal{C}/{\cong} \to \mathcal{D}/{\cong}$ must be also surjective, which precisely means that $F$ is essentially surjective.

*Notice that a fully faithful functor between partially ordered sets, considered as categories, is not the same as an injective monotonic map. Instead, it is a monotonic map (preserves the order) that also reflects the order. It is then automatically injective, but in general, an injective monotonic map does not need to reflect the order (consider the bijection $\{a,b\} \to \{c,d\}$ with no relation between $a$ and $b$, but $ c \leq d$). But for linear orders, the condition is equivalent to being an injective monotonic map.

Martin Brandenburg
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  • Monotonic and order reflecting maps between posets are also called order embeddings. – Jochen Oct 01 '24 at 16:15
  • I'm just not convinced by your statement about finite categories. How do you deduce that $\mathcal{C}$ and $\mathcal{D}$ are equivalent from the fact that the two quotients are isomorphic? – Loulou Mar 15 '25 at 15:33