Does $\mathbf{Cat}$ have the Cantor–Schröder–Bernstein property?

Question

I am wondering if the category of small categories $\mathbf{Cat}$ is known to (not) have the Cantor–Schröder–Bernstein property? That is, for any two categories $\mathcal{C}$ and $\mathcal{D}$, does the statement that there exist embedding functors (monomorphisms in $\mathbf{Cat}$) $\mathcal{F}: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{G}: \mathcal{D} \rightarrow \mathcal{C}$ imply that $\mathcal{C} \cong \mathcal{D}$?

If not, what can we say about the validity of the following weaker statement?

For any two categories $\mathcal{C}$ and $\mathcal{D}$, the statement that there exist embedding functors $\mathcal{F}_1: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{F}_2: \mathcal{D} \rightarrow \mathcal{C}$ implies that there exists bimorphisms (epic embedding functors) in $\mathbf{Cat}$ $\mathcal{G}_1: \mathcal{C} \rightarrow \mathcal{D}$ and $\mathcal{G}_2: \mathcal{D} \rightarrow \mathcal{C}$.

Thank you in advance for your help!

Answer 1

One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

This example answers directly the first part of the question. But actually it answers also the second part of the question, as explained in Peter's comment.

Answer 2

This is simply a summary that includes all the details.

$\mathbf{Pos}$ usually denotes the category of partially ordered sets as objects and monotone functions as morphisms. For any partially ordered set $(A, \leq)$, we may consider its associated categorification $\mathcal{C}(A, \leq)$ (this is simply viewing $(A, \leq)$ as a category).

There is a natural functor $U: \mathbf{Pos} \rightarrow \mathbf{Cat}$ defined such that, for every partially ordered set $X$, $U(X)=\mathcal{C}(X)$ and such that, for every morphism $f$, $U(f)$ is the unique functor for which the object component of $U(f)$ equals $f$.

@Joel David Hamkins has directly answered the first part of my question by providing a simple counterexample which shows that $\mathbf{Cat}$ does not have the Cantor-Schroder-Bernstein property.

In his answer, Joel considered $(\mathbb{Q}_{\geq 0}, \leq )$ and $(\mathbb{Q}, \leq )$. Given two morphisms $f_1 : (\mathbb{Q}_{\geq 0}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ and $f_2: (\mathbb{Q}, \leq ) \rightarrow (\mathbb{Q}_{\geq 0}, \leq )$ in $\mathbf{Pos}$ (monotone maps), $U(f_1): \mathcal{C}(\mathbb{Q}_{\geq 0}, \leq ) \rightarrow \mathcal{C}(\mathbb{Q}, \leq )$ and $U(f_2): \mathcal{C}(\mathbb{Q}, \leq ) \rightarrow \mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ are easily seen to be monomorphisms in $\mathbf{Cat}$ (these functors are injective on objects since any monotone map is injective and are faithful vacuously). However, it doesn't hold that $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq ) \cong \mathcal{C}(\mathbb{Q}, \leq )$ since $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ has an initial object and $\mathcal{C}(\mathbb{Q}, \leq )$ does not.

As was mentioned by @Peter LeFanu Lumsdaine in the comments, it turns out that Joel's counterexample can be extended to completely answer my question. Namely, $U$ is a faithful functor. Thus, to show that there are no bimorphisms from $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ to $\mathcal{C}(\mathbb{Q}, \leq )$ in $\mathbf{Cat}$, we only need to show that there is no bimorphism from $(\mathbb{Q}_{\geq 0}, \leq )$ to $(\mathbb{Q}, \leq )$ in $\mathbf{Pos}$.

Peter defined, for every morphism $g : (\mathbb{Q}_{\geq 0}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ in $\mathbf{Pos}$, a morphism $\tilde{g} : (\mathbb{Q}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ defined by the rule

\begin{equation*} \forall x \in \mathbb{Q}: \tilde{g}(x)=\left\{\begin{matrix} x, & x \geq g(0)\\ x-1, & x < g(0) \end{matrix}\right. . \end{equation*}

It holds that $\tilde{g} \circ g=g$. Thus, $g$ is not an epimorphism and is thus not a bimorphism. We conclude that there exists no bimorphism in $\mathbf{Pos}$ from $(\mathbb{Q}_{\geq 0}, \leq )$ to $(\mathbb{Q}, \leq )$.