How different can equivalent categories be?

Question

Some pretty big results in mathematics can be expressed as equivalence of categories (or rather, as adjunctions). The usual definition $FG\cong 1,GF\cong 1$ is easily justifiable in category theory since equality is too strict for objects of a category (functor categories in this case).

While looking at Borceux (Vol I, paragraph after prop 3.4.5), I saw for instance that equivalent categories may have non-isomorphic sets of objects (the functor $N:\mathbf 2\rightarrow \mathbf 1$). What are some explicit examples of interesting equivalent categories with non-isomorphic classes of objects?

More generally:

So exactly how different can equivalent categories be?

This MSE question seems to be very much related, both to my question and to the remark in Borceux about equivalent categories not being interchangeable "from outside". However, the level of discussion there is too high for me to really feel I understand exactly what's going on...

Intuition, instructive examples, and formal results are all welcome.

Answer 1

Equivalent categories are identical except that they might have different numbers of isomorphic "copies" of the same objects. One way of making this precise is as follows. Say a category $\mathcal{C}$ is skeletal if isomorphic objects of $\mathcal{C}$ are equal. Given any category $\mathcal{C}$, you can find an equivalent skeletal full subcategory (or "skeleton") $\mathcal{D}$ of $\mathcal{C}$ by just taking one representative of each isomorphism class of objects (the inclusion functor $\mathcal{D}\to\mathcal{C}$ is then an equivalence). Furthermore, if $\mathcal{C}$ and $\mathcal{D}$ are both skeletal, and $F:\mathcal{C}\to\mathcal{D}$ is an equivalence, then $F$ is actually an isomorphism (this is easy to see from the characterization of equivalences as fully faithful and essentially surjective; note, however, that an inverse of $F$ as an equivalence might not be an inverse of $F$ as an isomorphism). It follows that the skeleton of a category is unique up to isomorphism, and two categories are equivalent iff their skeleta are isomorphic.

Another way to make it precise is the following. Let $F:\mathcal{C}\to\mathcal{D}$ be an equivalence of categories. Then $F$ is naturally isomorphic to an isomorphism of categories $F':\mathcal{C}\to\mathcal{D}$ iff for each object $C$ in $\mathcal{C}$, the set of objects isomorphic to $C$ has the same cardinality as the set of objects isomorphic to $F(C)$. (Proof sketch: Fix representatives of each isomorphism class in $\mathcal{C}$ and $\mathcal{D}$, together with isomorphisms from each object to the representative and bijections between the corresponding classes in $\mathcal{C}$ and $\mathcal{D}$. Then modify $F$ so that it stays the same on the representatives, but on all other objects it is given by what it does on the representatives together with the data above.) So in this sense, if you know a category up to equivalence, the only thing you don't know about it is how many isomorphic copies of each object it has.

More importantly, pretty much every useful thing you can say about objects in a category is invariant under isomorphisms. So, you never care about the distinction between an object and some other object that is isomorphic to it (at least, when you have chosen a specific isomorphism between them). In the case of locally free sheaves and vector bundles, the point is not that locally free sheaves and vector bundles are literally in bijection with each other; rather, the point is that isomorphism classes of locally free sheaves are in bijection with isomorphism classes of vector bundles, in a way compatible with the maps between these objects. When you unravel what this compatibility should mean, you find that an equivalence of categories exactly gives you the compatibility you are looking for.

Answer 2

(Warning: I'm writing this from memory, and don't have the reference handy)

In Categories, Allegories, Freyd and Scedrov prove a theorem

A property of categories is preserved by equivalence if and only if it can be described on a blackboard

The key feature of a blackboard, here, is that if you draw two distinct objects, they remain distinct throughout a description; you can't insist they are equal. So we might restate the theorem as

A property of categories is preserved by equivalence if and only if it does not involve equations of objects

To prove this, Freyd and Scedrov define a formal diagrammatic language whose semantics involve the existence of functors from diagram categories. For example, limits from a category $J$ are described as follows:

• Define a category $J_1$ by adjoining to $J$ a new object $L$, arrows $\hom(L,L) = \hom(L, j) = \{ * \}$
• Define a category $J_2$ by adjoining to $J_1$ a new object $M$ and arrows $\hom(M,M) = \hom(M,j) = \{ * \}$ and $\hom(L,M) = \hom(M,L) = \varnothing$
• Define a category $J_3$ by adjoining to $J_2$ a new arrow $p : M \to L$

Then, a functor $F : J \to \mathcal{C}$ has a limit if and only if:

• There exists a a functor $F_1 : J_1 \to \mathcal{C}$ extending $F$ such that
• For every functor $F_2 : J_2 \to \mathcal{C}$ extending $F_1$
• There exists a unique functor $F_3 : J_3 \to \mathcal{C}$ extending $F_2$.

In the formal version of the claimed theorem, the relevant property is that the functors between the diagram categories must be injective on objects.

An example of a property not preserved by equivalence is as follows:

• Let $J_1$ be the category of two objects and an isomorphism between them
• Let $J_2$ be the category consisting of a single object and $\hom(*, *) = \mathbb{Z}$

And map $J_1 \to J_2$ by sending the two isomorphisms to $\pm 1$. Then, a category $\mathcal{C}$ is skeletal if and only if every functor $J_1 \to \mathcal{C}$ can be extended to a functor $J_2 \to C$.

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With the benefit of hindsight, I think the key idea underpinning this is the canonical model structure. In particular, consider the square

$$ \begin{matrix} J &\xrightarrow{f}& \mathcal{C} \\ i\downarrow\ \ & & \ \ \downarrow e \\ J' &\xrightarrow{f'} & \mathcal{C}' \end{matrix} $$

A lift, here, is a diagonal arrow $\ell : J' \to \mathcal{C}$ making the diagram commute.

Then $i$ is injective on objects if and only if, for every such diagram where $e$ is a surjective equivalence $e$, a lift exists.

Similarly, $e$ is a surjective equivalence if and only if, for every such diagram where $i$ is injective on objects, a lift exists.

The connection to the logic described above are, if $i$ is injective on objects and $e$ a surjective equivalence, then

A functor $f : J \to \mathcal{C}$ can be extended to $\ell : J' \to \mathcal{C}$ if and only if $ef : J \to \mathcal{C}'$ can be extended to $f' : J' \to \mathcal{C}'$. Furthermore, in both cases the lifts can be chosen so that they are related via $e$.

This extends to the general case, since any equivalence can be 'lifted' to a span of surjective ones; for any equivalence $e:\mathcal{C} \to \mathcal{C}'$, there is a category $\mathcal{C}''$ and surjective equivalences $e_1 : \mathcal{C}'' \to \mathcal{C}$ and $e_2 : \mathcal{C}'' \to \mathcal{C}'$ such that $e e_1 = e_2$.