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Stream: learning: questions

Topic: A sheaf isomorphic to a sheaf of continuous sections

Julius (Oct 16 2023 at 16:21):

https://tcsc.lakecomoschool.org/files/2018/06/Como2018.pdf Proposition 1.5

A sheaf is a presheaf where for a family of open sets in its underlying topological space, given they are compatible in a certain way (the sets $a_i$ they get mapped to are consistent regarding intersections of those open sets $U_i$), there will be a set $a \in F(U)$, which basically also is consistent with those restrictions, $a|_U_i = a_i$.

A presheaf is just any contravariant functor into the category of sets, but in this case, it’s thought of as from $O(X)$, the lattice of open sets of topological space X, which acts like a category.

If you have a mapping $p: Y \rightarrow X$ from a different topological space, a sheaf of sections of $p$ maps each open set $U$ to the continuous mappings from $U$ to $Y$ which act like the identity when composed with $p$. (Why are they called “sections”?)

Proposition 1.5 says that there is always some map $p: Y \rightarrow X$ such that a sheaf $F$ is isomorphic to the sheaf of sections of $p$.

They then define the stalk of a sheaf, for the proof, and this part I don’t get.

The stalk of a sheaf at a point is the “filtered colimit” of F(U).

I read that a filtered diagram is something like a functor $M: I \rightarrow C$ where for any two objects in $I$, there is always some other object they both map to; and I believe there are “equalizers” for any two parallel arrows.

https://stacks.math.columbia.edu/tag/04AX

What is the general purpose or intuition of a filtered diagram / filtered colimit? Why is it useful, what properties does it have?

How is it used in this proof?

Thank you

David Egolf (Oct 16 2023 at 17:13):

Hi Julius!
I haven't studied this stuff yet, so I don't know the answer to your question.
However, I think this might be a nice resource on some related things: https://johncarlosbaez.wordpress.com/2020/01/05/topos-theory-part-1/

(By the way, to make the LaTeX show up nicely on this zulip, you need to put two dollar signs on each side of your LaTeX expression!)

John Baez (Oct 16 2023 at 17:39):

(Why are they called “sections”?)

They're called sections because they're like sections of a bundle. If you don't know what sections of a bundle are, you can read about them here:

• Topos theory (part 2).

where I explain how any bundle gives a presheaf (which is in fact a sheaf). In the same text I also try to explain the stalk construction you're wondering about. I don't say the phrase "filtered colimit", just "colimit", but that's enough to get started on understanding what's going on here.

Luckily Kevin has a lot more good stuff to say below. All I mean by "bundle" is the map $p: Y \to X$ that he's talking about.

Kevin Arlin (Oct 16 2023 at 17:40):

For a map $p:Y\to X$, you can get an intuition for the sheaf of sections by first considered its elements over $U=X$ itself. These are the maps $s:X\to Y$ such that $p\circ s=\mathrm{id}_X.$ Such a map is called a section, or a splitting, of $p$, and in itself a split monomorphism, in any context, not just sheaf theory.

Kevin Arlin (Oct 16 2023 at 17:40):

Why "section"? Imagine that $X$ is a circle and $Y=X\times [0,1]$ is a cylinder. Then a section is essentially just a continuous map $X\to [0,1],$ and you can visualize its image as a circular "slice" out of the cylinder, hitting exactly one point at each polar coordinate. "Section" means something like "cutout", etymologically, and so I believe the terminology comes from this visual of "cutting an $X$ out of $Y$."

Kevin Arlin (Oct 16 2023 at 17:40):

A filtered diagram is one indexed by a filtered category, which is a category $J$ such that every diagram $D:K\to J,$ with $K$ finite, admits some cocone--not necessarily a colimiting cocone! Directed posets, that is, those for which every two elements admit an upper bound, are a good example. A more important example is the category whose objects are all maps $A\to G$ where $G$ is a fixed group/module/ring/whatever algebraic thing and $A$ varies over all finitely presented instances of that structure, with morphisms commutative triangles. This category is filtered, this time by using actual colimits in the category of whatevers, and $G$ is its colimit. In slogan form, every algebraic gadget is the filtered colimit of its maps from finitely presented algebraic gadgets, which allows us to understand quite a lot about infinite algebras from finitely presented ones.

Kevin Arlin (Oct 16 2023 at 17:41):

Filtered colimits have a couple of very important properties. One which is easy to state is that filtered colimits commute with finite limits: if $J$ is filtered and $K$ is finite, then for a diagram $D:J\times K\to \mathbf{Set},$ you get the same result if you take the $J$-colimit of the $K$-limits or the $K$-limit of the $J$-colimits. This is terribly false if $J$ is not filtered; consider for instance the case that $J$ and $K$ are both the two-object discrete categories.

Kevin Arlin (Oct 16 2023 at 17:41):

For the purposes of the proposition you're asking about, the key property of filtered colimits is that they have a simple explicit expression in $\mathbf{Set}.$ A general colimit of $D:J\to \mathbf{Set}$ is given by the disjoint union $\sqcup_j D(j)$ quotiented by the equivalence relation generated by the relation $x\sim y$ if $D(f)(x)\sim D(g)(y)$ for maps $f:j\to j',g:j''\to j'$ in $J$ and $x\in D(j),y\in D(j'').$ Since we have to manually make this relation transitive, it is in generally extremely difficult to tell when two elements of the colimit are equal. (It's instructive to think about a colimit over a category like $\bullet\to \bullet\leftarrow \bullet\to \cdots \bullet.$)

Kevin Arlin (Oct 16 2023 at 17:41):

For a filtered colimit, though, the original relation is already an equivalence relation! It becomes transitive because, given the situation below in $J$ with $j_1$ through $j_5$, you can always extent to the full diagram such that the routes from $j_2$ to $j_7$ are equal. This implies that $x\in D(j_1),y\in D(j_2)$ become equal in $D(j_4)$ while $y,z\in D(j_3)$ become equal in $D(j_5)$ (i.e. $x\sim z$ in the generated equivalence relation) if and only if $x,y,z$ all become equal in $D(j_7)$ (i.e. $x\sim z$ in the original not-obviously-equivalence relation.) So it's transitive!

Kevin Arlin (Oct 16 2023 at 17:41):

This means that we have an unusually good handle on filtered colimits, which has lots of consequences. For instance, filtered colimits in those algebraic categories we mentioned before are computed just as in sets, which is again terribly false for arbitrary colimits. So not only is a group the filtered colimits of finitely presented subgroups, we can actually understand that colimit!

Anyway, for your purposes, you should find that if you try to finish the proof of your proposition, it becomes critical that you actually know explicitly when two elements of the stalk are equal.