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!



Does $\mathbf{Cat}$ have the Cantor–Schröder–Bernstein property?
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](https://mathoverflow.net/questions/452811/does-mathbfcat-have-the-cantor-schroder-bernstein-property#comment1171201_452814).


