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Stream: theory: category theory

Topic: Lawvere-Tierney topologies

Patrick Nicodemus (Jan 16 2023 at 17:04):

Nlab says of lawvere-tierney topologies:

Remark

Equivalently, the third axiom in def. can be replaced with the (internal) statement that j is order-preserving.

The equivalence amounts to the fact that, within the internal logic of topoi, one can demonstrate that every monad on the preorder of truth values is in fact strong (a special case of the fact that, for an endofunctor on some monoidal closed V, tensorial strengths are the same as V-enrichments, as described in the article on the former), and therefore automatically preserves finite meets.

This is something of an abstract nonsense argument. I cannot follow it. Is this argument correct?

Is a Lawvere-tierney topology really just the same thing as a monad on Omega?

If you assume j is monotonic then it is not hard to get $j(x \land y) \leq j(x)\land j(y)$. I do not know how to derive the other direction $j(x)\land j(y)\leq j(x\land y)$

It seems to be making use of the cartesian closed structure on $\Omega$, which suggests that I convert the problem to $j(x)\leq j(y)\Rightarrow j(x\land y)$ and use that $j$ is internally monotonic, so it would suffice to show that $j(x) \leq y \Rightarrow (x \land y)= x$? But that is a dead end.

Mike Shulman (Jan 16 2023 at 19:43):

The first observation is that $x\Rightarrow y \le j(x) \Rightarrow j(y)$. This is because $x\Rightarrow y$ is the same as the internal statement $x\le y$ (because $\Omega$ is the object of internal truth values), so this is just the fact that $j$ is internally monotonic. This is what is meant by being V-enriched. By adjunction, it tells us that $(x\Rightarrow y) \wedge j(x) \le j(y)$.

Now we use the fact at [[strong monad]] stated as the converse in the section "Left-strong monads are enriched monads": we have $x \le y \Rightarrow (x\wedge y)$, hence $x\wedge j(y) \le (y \Rightarrow (x\wedge y)) \wedge j(y) \le j(x\wedge y)$, using the observation above.

Now we just apply this twice: $j(x)\wedge j(y) \le j(j(x)\wedge y) \le j(j(x\wedge y)) = j(x\wedge y)$.

Patrick Nicodemus (Jan 16 2023 at 20:07):

Cool.

Mike Shulman (Jan 16 2023 at 20:08):

If you feel like helping future readers, you could add more detail to the nLab page!

John Baez (Jan 16 2023 at 20:20):

Yes, that's the best way to pay back the debt!

Patrick Nicodemus (Jan 16 2023 at 20:23):

Fair enough