morita equivalence for categories

Question

Suppose we have two categories such that presheaves on them are equivalent. What can be said in this situation?

Answer 1

That their Cauchy completions are equivalent.

Answer 2

Finn's answer, as expanded by Tom, is great. However, since the question is so broad ("what can be said?") let me point out that there is a different thing that can also be said. If $C$ and $D$ have equivalent categories of presheaves, then $C$ and $D$ are equivalent objects in the bicategory of profunctors. In other words, there are profunctors $C \to D$ and $D\to C$ whose composites in either direction are naturally isomorphic to identities. This is also a necessary and sufficient condition.

Both Finn's and my answer generalize to enriched categories and enriched presheaves; the only difference is that the notion of "Cauchy completion" is different. In general, it means completion under all absolute colimits.

Answer 3

Let $k$ be a field. If $A$ and $B$ are (small) $k$-linear categories with $K$-linearly equivalent categories of $k$-linear functors to $k$-vector spaces, then $A$ can be obtained from $B$ by a sequence of "contractions and expansions» from $B$. See Theorem 4.7 in this paper by Claude Cibils and Andrea Solotar.

Answer 4

In short: it happens exactly when there is a bigger category and disjoint full embeddings of the two categories $\mathcal A,\mathcal B$ into it, such that every arrow is a composition of arrows of the form $a\to b$ or $b\to a$.

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It's easy to see that the Eilenberg-Watts theorem holds for monoid actions, moreover, it also holds for categories, presheaves and profunctors in place of rings, modules and bimodules:

Theorem.
Let $\mathcal A$ and $\mathcal B$ be categories, and $F:\mathcal{Set}^{\mathcal A}\to\mathcal{Set}^{\mathcal B}$ has a right adjoint, then there is a profunctor $U:\mathcal A\not\to\mathcal B$, such that $F\simeq (H\mapsto H\otimes_{\mathcal A}U)$.

In other words, in the realm of modules (even over categories), the only adjunctions are the tensor-hom adjunctions.
[As a note, $U$ can be obtained from $F$ by setting $U(a,b)=F(\hom_{\mathcal A}(a,-))\,(b)\ $.]

If $F$ is an equivalence of categories, then its inverse $G$ is both its left and right adjoint, so by the above theorem, $F\simeq(-\otimes U)$ and $G\simeq(-\otimes V)$ for some profunctors $U:\mathcal A\not\to\mathcal B$ and $V:\mathcal B\not\to\mathcal A$.
And, by $F$ and $G$ being inverses to each other, we obtain $U\otimes V\simeq \hom_{\mathcal A}$ and $V\otimes U\simeq\hom_{\mathcal B}$, as the hom functors are the identites for profunctor composition $\otimes$, meaning that $U$ and $V$ form an equivalence in the bicategory of categories and profunctors.
As for such, there exist an adjoint equivalence with arrows $U$ and $V$ in $\mathcal{Prof}$, that is, profunctor isomorphisms $\eta:\hom_{\mathcal A}\to U\otimes V$ and $\varepsilon:V\otimes U\to\hom_{\mathcal B}$ satisfying the zigzag identities.
Writing up these identities for the inverse of $\eta$, we receive a structure called Morita context: $$ 1_U\otimes\varepsilon=\eta^{-1}\otimes 1_U \quad \text{ and }\quad\varepsilon\otimes 1_V=1_V\otimes\eta^{-1}\quad\ (*)\,,$$ which can be very neatly interpreted as follows:

First, the elements of the sets $U(a,b)$ for a profunctor $U:\mathcal A^{op}\times\mathcal B\to\mathcal{Set}$ can be regarded as heteromorphisms $a\to b$ which we can add to the disjoint union $\mathcal A+\mathcal B$, and their compositions is described by the action of $U$ on the morphisms.
Thus, we obtain a bigger category $\mathcal U$ containing both $\mathcal A$ and $\mathcal B$ and no arrows of the form $b\to a$.
Composition ($\otimes$) of profunctors can be viewed as having formal compositions $\langle u,v\rangle$ of these heteromorphisms $u\in U(a,b),\ v\in V(b,a')$, quotienting out by $\langle u\beta,\,v\rangle\sim\langle u,\,\beta v\rangle$ for arrows $\beta\in\mathcal B$.

Thus, the above maps $\varepsilon:U\otimes V\to\hom_{\mathcal A}$ and $\eta^{-1}:V\otimes U\to\hom_{\mathcal B}$ extend the composition operations to $\mathcal U\cup\mathcal V$, assigning certain arrow $a\to a'\,\in\mathcal A$ to every $\langle u,v\rangle\in(U\otimes V)(a,a')$ and similarly for compositions $b\to b'$ in $\mathcal B$, and the above two equations $(*)$ state exactly the altogether associativity of these operations.
In other words, we obtain a bigger category, which is called a bridge $\mathcal A\rightleftharpoons\mathcal B$, that is a category containing $\mathcal A$ and $\mathcal B$ as disjoint full subcategories and no more objects (thus, it is a symmetric version of a profunctor, in a sense).

One direction of my initial statement then follows, since the composition maps $\varepsilon$ and $\eta^{-1}$ are in particular surjective.
For the other direction, their surjectivity in the presence of equations $(*)$ implies invertibility.

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If $\mathcal M:\mathcal A\rightleftharpoons\mathcal B$ is a bridge with surjective zig-zag compositions, then in particular, for every object $a\in Ob\mathcal A$, its identity $1_a$ also factors as $uv$ for some $u:a\to b,\ v:b\to a,\ b\in Ob\mathcal B$, but then $vu:b\to b$ is an idempotent,
so if we take the idempotent completion $\mathcal M^{id}$ of $\mathcal M$, that will be a bridge $\mathcal A^{id}\rightleftharpoons\mathcal B^{id}$, in which every object of $\mathcal A$, and thus also each of its idempotents, is isomorphic to some object of $\mathcal B^{id}$ and vice-versa.
The existence of such a bridge $\mathcal C\rightleftharpoons\mathcal D$ is actually equivalent to the equivalence of $\mathcal C$ and $\mathcal D$ (at least in the presence of choice): fixing isomorphism for each $a$ yields an equivalence functor $\mathcal C\to\mathcal D$, and the other direction can be proved by a similar argument as above.

So we obtain that $\mathcal A^{id}\simeq\mathcal B^{id}$.

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For more on the subject, refer to this paper.