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Stream: learning: questions

Topic: Isomorphic categories of presheaves

Evan Patterson (Jun 01 2020 at 18:28):

If two small categories $\mathcal{C}$ and $\mathcal{D}$ are isomorphic, then their categories of presheaves $[\mathcal{C}^{\mathrm{op}}, \mathbf{Set}]$ and $[\mathcal{D}^{\mathrm{op}}, \mathbf{Set}]$ are also isomorphic. Is the converse true?

(I know that it is "better" to think about equivalence of categories, but I like isomorphism too.)

Fabrizio Genovese (Jun 01 2020 at 18:31):

Doesn't this follow from Yoneda Embedding?

Evan Patterson (Jun 01 2020 at 18:34):

I thought maybe it should, but I don't quite see how to make it work, since an isomorphism of functor categories doesn't have to preserve the embedded copies of the original categories.

John Baez (Jun 01 2020 at 18:41):

Working with isomorphism of presheaf categories is a very delicate technical business.

You can certainly find examples of non-equivalent categories that have equivalent presheaf categories, and this phenomenon is important.

Evan Patterson (Jun 01 2020 at 18:45):

Thanks. In fact, it was trying to understand Morita equivalence of categories (still a work in progress :) that go me thinking about this question.

Jens Hemelaer (Jun 01 2020 at 22:18):

I think a good example to look at would be the category of reflexive graphs (directed graphs such that each vertex has an identity arrow): https://ncatlab.org/nlab/show/reflexive+graph.

You can view this as a presheaf topos in two different ways.

As presheaves on the category $\mathcal{C}$ with two objects $V$ and $E$, and (besides the identity arrows) one arrow $i : E \to V$, two arrows $s,t : V \to E$ and two arrows $E \to E$ (the idempotents $si$ and $ti$); with $is = 1$ and $it = 1$.

And also as presheaves on the monoid $\mathcal{D}$ with three elements $1$, $s$, $t$, such that $sx=s$ and $tx=t$ for all $x$ in the monoid.

In this case the equivalence between the two categories of presheaves is very clean and non-tricky. So I think for any reasonable definition of isomorphism of categories, the two categories would be isomorphic here.

John Baez (Jun 01 2020 at 22:33):

That sounds like a good strategy!

Evan Patterson (Jun 01 2020 at 22:42):

Thanks, I haven't seen this interesting second presentation of reflexive graphs. Let me think about this.

Jens Hemelaer (Jun 02 2020 at 07:35):

Jens Hemelaer said:

In this case the equivalence between the two categories of presheaves is very clean and non-tricky. So I think for any reasonable definition of isomorphism of categories, the two categories would be isomorphic here.

On second thought, the categories could be non-isomorphic. Starting with a presheaf on $\mathcal{C}$, you can get a presheaf on $\mathcal{D}$ by forgetting the vertices and only keeping the edges. But then there is no way to go back. You know that the vertices are bijective with the edges $e$ satisfying $si(e)=ti(e)=e$, so you can reconstruct the set of vertices up to natural isomorphism. But that's it. You don't remember the actual set.

This makes me believe your original claim a bit more, that you can reconstruct a category from the isomorphism class of its category of presheaves. But there are a lot of obstacles: for some people presheaf categories are only defined up to equivalence of categories, and the people who define presheaf categories up to isomorphism might disagree on which isomorphism class to pick exactly. Similarly, there are probably people who say that isomorphism doesn't make sense for large categories, or other people who disagree with each other on what "isomorphism" means exactly in this case.

Paolo Capriotti (Jun 02 2020 at 08:23):

Evan Patterson said:

If two small categories $\mathcal{C}$ and $\mathcal{D}$ are isomorphic, then their categories of presheaves $[\mathcal{C}^{\mathrm{op}}, \mathbf{Set}]$ and $[\mathcal{D}^{\mathrm{op}}, \mathbf{Set}]$ are also isomorphic. Is the converse true?

(I know that it is "better" to think about equivalence of categories, but I like isomorphism too.)

If we take $\mathbf{Set}$ to be the category of sets in a Grothendieck universe, and assume choice, I think the answer is no. Consider the terminal category and the walking isomorphism. They are equivalent, but not isomorphic. Their categories of presheaves are respectively isomorphic to $\mathbf{Set}$ and the category $\mathcal{C}$ of pairs of sets with a bijection. Let $\kappa$ be the cardinality of the universe.

For every cardinal $\alpha < \kappa$, let $\mathbf{Set}_\alpha$ be the objects of $\mathbf{Set}$ of cardinality $\alpha$, and define $\mathcal{C}_\alpha$ similarly. If $\alpha \neq 0$, both $\mathbf{Set}_\alpha$ and $\mathcal{C}_\alpha$ have cardinality $\kappa$, since $\kappa$ is inaccessible. If $\alpha = 0$, then both sets have cardinality 1. In both cases, they are isomorphic as sets. Fix a bijection them, sending every set $X$ into a bijection $X_1 \cong X_2$. Now choose, for every $X \in \mathbf{Set}_\alpha$, a bijection $\psi_X$ between $X$ and $X_1$.

The union of the various $\phi_\alpha$ defines a bijection on objects $\mathbf{Set} \to \mathcal{C}$. For any function $f: X \to Y$, let $f_1: X_1 \to Y_1$ be defined as $\psi_Y \circ f \circ \psi_X$, and let $f_2: X_2 \to Y_2$ be the unique function that makes the square commute. This gives a fully faithful action on object, and completes the definition of the isomorphism between $\mathbf{Set}$ and $\mathcal{C}$.

You can probably play a similar game with only Morita equivalent base categories, like in the reflexive graph example mentioned before, but I haven't checked.

However, if $\mathbf{Set}$ is taken to be the whole large category of sets, I'm not sure what the answer is.

John Baez (Jun 02 2020 at 16:25):

Great!

Evan Patterson (Jun 02 2020 at 17:56):

Thanks for this answer! From a pragmatic standpoint, my takeaway is that I should be wary of arbitrary isomorphisms between categories of presheaves because they can be too, well, arbitrary :)

John Baez (Jun 02 2020 at 18:24):

I think isomorphism of categories is a notion best avoided whenever possible. To me the more more robust, less "tricky" fact is that nonequivalent categories can have equivalent presheaf categories. This is where the "Karoubi envelope" or "Cauchy completion" of a category comes into play. I'm not finding it stated on the nLab, but I believe the idea is that:

1) every category has a Cauchy completion, whose category of presheaves is equivalent to the category of presheaves on the original category

2) if two Cauchy complete categories have equivalent categories of presheaves, they are equivalent.