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: learning: questions

Topic: locales

Vincent L (Mar 23 2021 at 20:50):

Hi, I'm watching this video :
https://www.youtube.com/watch?v=Ro8KoFFdtS4&t=1934s
In this video it is said that there is an adjunction between the category $Top$ and $Fram$, with the set of opens of a topological space $X \to \mathcal{O}(X)$ acting as the left functor. However I don't understand the right adjoint functor : it's $Hom(., [1])$ with $[1] = \{ 0, 1\}$. As far as I understand, it's an Homset of morphism in the $Fram$ category. But we want a topological space, and HomSet is "just" a set so I don't understand how this right functor fits the adjunction.

Spencer Breiner (Mar 23 2021 at 20:58):

Hi Vincent! There are two important things to remember here. First is that the direction of arrows on frames and spaces is reversed, second is that [1]={0,1} is the frame corresponding to a single point. That means that a frame morphism $p:\mathcal{O}(X) \to [1]$ is the same as a point $p\in X$, in the ordinary sense. In particular $p(U)=1$ iff $p\in U$.

Correspondingly, given any frame $F$, we can build a topological space where the points are literally maps $q:F\to[1]$. The neighborhoods of the topology come from the elements of the frame: $A\in F \leftrightarrow U_A=\{q|q(A)=1\}$.

Vincent L (Mar 23 2021 at 21:11):

thank you !

Vincent L (Mar 23 2021 at 21:16):

In topology the morphisms correspond to the membership relation ?

Vincent L (Mar 23 2021 at 21:19):

I don't understand why the frame $[1]$ is corresponding to a single point. Is it the initialobject in the category ?

Spencer Breiner (Mar 23 2021 at 21:20):

The morphisms in Top are ordinary continuous functions $f:X\to Y$.

If $1=\{*\}$ represents the point, then we can represent each point $x\in X$ as a continuous function $\tilde{x}:1 \to X$ (with $\tilde{x}(*)=x$)

Spencer Breiner (Mar 23 2021 at 21:21):

Notice that $\mathcal{O}(1) = \{\emptyset, \{*\}\}\cong[1]$.

Vincent L (Mar 23 2021 at 21:21):

ok

Vincent L (Mar 23 2021 at 21:21):

yes I see

Todd Trimble (Mar 23 2021 at 21:21):

Right: as in essentially all such dualities, the dualizing object, which here is the "topological frame" $\mathbf{2} = [1]$, is simultaneously a topological space (Sierpinski space) and a frame: in fact a frame internal to topological spaces. So when you hom a space $X$ into it, the hom-set $\hom_{Top}(X, \mathbf{2})$ acquires a frame structure whose operations are defined pointwise. When you hom a frame $A$ into it, the hom-set $\hom_{Fram}(A, \mathbf{2})$ acquires a spatial structure (as a subspace of a power of Sierpinski space; this is equivalent to what Spencer was saying).

Likewise, the classical Stone duality between Boolean algebras and compact Hausdorff totally disconnected spaces is effected by homming into a topological Boolean algebra $\{0, 1\}$ (internal to the category of compact Hausdorff spaces). I think that if you were to look inside Johnstone's Stone spaces, just about all the dualities arise according to this type of idiom.

Vincent L (Apr 04 2021 at 22:48):

I have another question coming from this video, he sais that in a poset $X$ if $f$ and $g$ are adjoint functors then $f$ can be decomposed into a morphism $P\to P_\sigma$ , an isomophism between $P_\sigma$ and $Q_\rho$ and a monomorphism $Q\to Q_\rho$, where $P_\sigma$ is the fixpoint of $\sigma = g.f$ and $\rho = f . g$. Is it a theorem ? It's not obvious for me why $P_\sigma$ and $Q_\rho$ are isomorph for instance.

Vincent L (Apr 04 2021 at 22:56):

hmm actually I think I got the isomorphism part, but not the decomposition

Fawzi Hreiki (Apr 05 2021 at 00:50):

This follows from two general facts: any adjunction between posets is idempotent, and any idempotent adjunction factors into a reflection followed by a co-reflective subcategory inclusion.

Fawzi Hreiki (Apr 05 2021 at 00:55):

It's then the case that you can restrict to the fixed points by 'going around in each direction'

Vincent L (Apr 05 2021 at 20:02):

Is there an article somewhere on idempotent adjunction ? Especially with a proof of the factorisation. Nlab mentions this but doesn't provide a proof

Mike Shulman (Apr 05 2021 at 23:25):

Maybe you can work out the proof as an exercise and add it to the nLab page. (-:

Vincent L (Apr 07 2021 at 19:27):

Actually it looks like it's the proposition 3.3 of this page https://ncatlab.org/nlab/show/Galois+connection if I understand correctly, with $I_E$ being $f$ and $V_E$ being $g$

Mike Shulman (Apr 07 2021 at 22:58):

That's the abstract proof that any adjunction between posets is idempotent. Is that what you were asking about?

Vincent L (Apr 25 2021 at 13:04):

Sorry for the delay, I finally had some time to dive in properly.

I think I was confusing several things there:

• In the presentation Joyal says that $f:C\to D$ where $C$ and $D$ are posets, and $f$ is an adjunction then it can be decomposed into 3 functors, one being an isomorphism between the sets of fixpoints of the monad/comonad, and the third one being a full and faithfull functor.
• In the ncatlab there is a more general property saying that an idempotent adjunctions can be factored into 2 adjunctions.

It looks like when restricted to poset, it's possible to transform a function f into a function "(f . g) . f(gf)* . id" , because since fg is idempotent, then f.g will map to the set of fixpoint of fg, then applying f move from the fixpoint of the monadic operator to the fixpoint of the comonadic operator, and applying g in the other direction move us back to the fixpoint of the monadic operator.

However I'm still struggling with the ncatlab proof. I proved the chain of 1 to 10 in https://ncatlab.org/nlab/show/idempotent+adjunction but I'm struggling with proving 10=> 11, or even proving all steps between 1 and 10 proves 11. I don't know what to take for E. The "intuitive" choice would be to take F1 as F or FG or FGF... (because we don't really have anything else) but I don't know how many pair of GF I need to use, and I don't know what kind of unit to take. We only have $\eta$ and $\epsilon$ so again an intuitive choice would be to use $\eta^2$ for a unit and FG as F1 and G1, but then it doesn't look like I can get an identity morphism out of $\eta^2$ and $\epsilon^2$, or a similar expression (there isn't a lot of thing that could be used to build something between Id and FGFG except the unit and counit). I don't think F1 = F and G1 = G, I mean it would mean that F2 and G2 are identity, and I don't think G is full and faithfull

Mike Shulman (Apr 25 2021 at 15:20):

Remark 1.2 on the nlab page says that E is the full image of either functor F or G.

Vincent L (Apr 25 2021 at 15:32):

ho right thank you