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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeSep 10th 2012

Guys (<– not meaning to exclude females!), I’m really sorry, but I’m having trouble following what’s being said at algebraic lattice: “The category of topological spaces is the 2-image of the projection from the comma category $(Set / Forgetful)$ to $Set$, where $Forgetful : AlgLat \rightarrow Set$ is the obvious forgetful functor.” Probably there’s a bubble in my brain or something, but I can’t make much sense of this.

Let me see: an object of $Set/Forgetful$ consists of a set $S$, and algebraic lattice $L$, and a function $S \to U(L)$ where $U$ is shorthand for $Forgetful$. The projection to $Set$ takes such a triple $(S, L, U(L)$ to the set $S$. The 2-image is the category whose objects are triples $(S, L, S \to U(L))$, and where a morphism $(S, L, S \to U(L))$ to $(S', L', S' \to U(L')$ is a function $f: S \to S'$ such that there exists $g: L \to L'$ making the evident square commutative (i.e., so that $(f, g)$ is a morphism of $Set/Forgetful$).

So first, how do I get a topological space out of $(S, L, S \to U(L))$? I guess $S$ is meant to be the underlying set, but then what?

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeSep 11th 2012

bump. I’m probably going to remove that sentence if no one can explain it to me.

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeSep 12th 2012

I don’t see it. It was written by Sridhar Ramesh, FWIW.

• CommentRowNumber4.
• CommentAuthorJon Beardsley
• CommentTimeSep 12th 2012

Is this some kind of weird Stone duality thing? Don’t suppose the algebraic situation implies distributivity or anything…? At least, I can’t see it.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeSep 12th 2012

I’m not sure what Sridhar had in mind, but I went ahead and edited at algebraic lattice so that the material being referred to makes more sense to me.

• CommentRowNumber6.
• CommentAuthorSridharRamesh
• CommentTimeSep 13th 2012
• (edited Sep 13th 2012)

I wish I’d been more clear even for my future self… Here’s my attempt to reconstruct my thoughts, though everyone else’s confusion has me worried I must have screwed something up:

I must have meant for the topology on $S$ to be such that a subset of $S$ is open just in case it is the preimage of an open subset of $L$ (with respect to the Scott topology). In other words, the topology is chosen to make the map from $S$ to $L$ a topological pre-embedding.

Since the opens are classified by $(\Omega, \Omega, 1_{\Omega})$, we automatically have that any $f : S \to S'$ which is a morphism in our 2-image category is continuous with respect to the above topology.

So we have a faithful functor from our 2-image category to $Top$ (faithful because we took the 2-image).

This functor is essentially surjective (that is, every topological space admits a pre-embedding into an algebraic lattice): given any topological space $T$, we can form the triple $(T, P{|\mathcal{O}(T)|}, x \mapsto \{U \in \mathcal{O}(T): x \in U\})$, as Todd now indicated in the article, whose image under the above faithful functor will be $T$ again. [This is because the Scott open subsets of $P{|\mathcal{O}(T)|}$ are the unions of upwards closures of finite sets of opens; thus, their preimages will be the unions of finite intersections of opens of $T$, which is to say, the opens of $T$… (hopefully, I’ve gotten that right; we could also use an algebraic lattice of filters, rather than arbitrary subsets of $\mathcal{O}(T)$)]

Finally, this functor is full, using the fact that algebraic lattices are continuous lattices, and continuous lattices (with the Scott topology) are injective objects with respect to topological pre-embeddings. [Hopefully, I’ve gotten THAT right…]

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeSep 13th 2012

Sridhar, I’m not finding anything wrong with what you wrote in #6. (I can’t tell whether I should be embarrassed by not thinking of this myself, but anyway it was simply a case of my not being clever enough on this occasion.) I’ll go through it one more time, and if all still seems well, I’ll reinstate what you wrote. Sorry (and thanks)!!

• CommentRowNumber8.
• CommentAuthorSridharRamesh
• CommentTimeSep 13th 2012

No worries. I should have spelt it out originally, rather than leave it to others (including my future self) to decipher my argument.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeSep 15th 2012

I have now reinstated Sridhar’s remarks from the earlier revision of algebraic lattice (more precisely, I rewrote them in more expanded form).

• CommentRowNumber10.
• CommentAuthorSridharRamesh
• CommentTimeSep 15th 2012

Thanks!

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeOct 1st 2012

I added to algebraic lattice the observation that in a locally finitely presentable category, we get algebraic lattices by considering subobject lattices, quotient lattices (taking equivalence classes of epis), and congruence lattices. I plan to add a little more to that article later.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 15th 2015

Added to algebraic lattice some material on congruence lattices that I’ve only just heard.