Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: deprecated: topos theory

Topic: subobjects of products

Morgan Rogers (he/him) (Jun 23 2023 at 08:15):

Since subobject lattices for objects in a topos are frames, I often think of them as behaving like open sets. This leads me to expect (by analogy with the product topology) that a subobject of a product $A \times B$ of objects $A,B$ in a topos should be a union of products of subobjects of $A$ and $B$. Since it seems like a fun problem, I thought I would post it here before thinking too hard about whether or not this is true.

Morgan Rogers (he/him) (Jun 23 2023 at 08:20):

In a topos of sheaves over a space $X$ I bet we can use the equivalence with the category of local homeomorphisms over $X$ to deduce that this must be the case (or maybe this will provide a counterexample?). But if so does the result work more generally?

David Michael Roberts (Jun 23 2023 at 08:51):

Probably in the internal logic this is a quick argument, but I didn't put any thought into it.

Reid Barton (Jun 23 2023 at 08:58):

It's not true for $G$-sets. For example, regarding $G$ as a $G$-set under left multiplication, $G \times G$ has a lot more subobjects than $G$.

Morgan Rogers (he/him) (Jun 23 2023 at 08:59):

David Michael Roberts said:

Probably in the internal logic this is a quick argument, but I didn't put any thought into it.

That perspective actually makes me lean towards it being false. If I take a signature with two sorts $A,B$ and a binary relation $R \rightarrowtail A,B$, the classifying topos for the empty theory on that signature might be a counterexample: we would need to be able to prove in the empty theory that $R$ is a union of subrelations of a particular form, and I don't think any such decomposition is provable.

Morgan Rogers (he/him) (Jun 23 2023 at 09:00):

Reid Barton said:

It's not true for $G$-sets. For example, regarding $G$ as a $G$-set under left multiplication, $G \times G$ has a lot more subobjects than $G$.

Ah great, this should have been the first place I looked.

Josselin Poiret (Jun 23 2023 at 09:03):

Reid Barton said:

It's not true for $G$-sets. For example, regarding $G$ as a $G$-set under left multiplication, $G \times G$ has a lot more subobjects than $G$.

what do you mean by "a lot more"? Since you can have arbitrary unions of products of subobjects, I don't immediately see a contradiction. I guess the diagonal subobject is the counterexample though

Morgan Rogers (he/him) (Jun 23 2023 at 09:05):

As in $G$ has only two subobjects as a $G$-set (one of which is empty); there are only $2$ subobjects expressible as a union of products of subobjects of $G$!

John Baez (Jun 23 2023 at 14:32):

While meanwhile the subobjects of $G \times G$ as a left $G$-set are in bijection with subsets of $G$, where the subset $S \subseteq G$ corresponds to

$\{(g,h): g^{-1}h \in S \} \subseteq G \times G$

Mike Shulman (Jun 23 2023 at 15:39):

However, it is true internally in any topos, by the same argument that makes it true in the category of sets: for $S\subseteq A\times B$ we have $S = \bigcup_{(x,y)\in S} \{x\} \times \{y\}$. So this is a case where the internal statement is very different from the external one.