@TedShifrin "Some of them hated the mathematics that drove them, and some were afraid, and some worshiped the mathematics because it provided a refuge from thought and from feeling" is the full sentence, from one of the first chapters (4th or 5th I think) of The Grapes of Wrath, describing the representatives of the banks kicking the farmers out of the land
The trouble with all the "self-learning" that people on this site love to do ... is that you don't get written feedback on your writing. And without that you don't have any idea if you're actually learning.
I'll just talk to myself to clear it out: The point is the graph of groups of $A * B$ is a bipartit-ed $\Gamma(F_2)$ with vertex groups $A$ and $B$ alternately appearing. The total space, $\widetilde{K(A*B, 1)}$, is obtained from getting a $\widetilde{K(A, 1)}$ or $\widetilde{K(B, 1)}$ for each vertex and taking one-point union depending on the edge. $H \leq A * B$ acts on this graph of groups, so take the quotient graph $\Gamma(F_2)/H$, which is the underlying graph of the corresponding graph of groups whose total space is $\widetilde{K(G, 1)}/H = K(H, 1)$.
Well, they are topological spaces, so we know what bundles are, @Balarka. Or are you thinking of varying ranks along strata? This sounds more sheafical.
@TedShifrin Yeah, fiber varies stratum-to-stratum. I have the correct notion of vector bundles over stratified spaces down; over each stratum you have a vector bundle so that they have a "natural" extension to the closure of the stratum, and the bundles over the lower stratums "naturally" include in that extended bundle.
It's just a big commutative diagram and obviously the right notion because we're modelling Whitney condition (a). The union of the tangent bundles to each stratum satisfies this property
Namely, how should the fibers collapse to lower dimensional fibers as you go along an arc running into a lower dimensional stratum (ideally, try to understand how the "fiber bundle" will look over an $\overline{m}$-allowable simplex wrt the lower dimensional stratum, where $\overline{m}$ is the middle perversity)
For most examples I have, they come from proper actions of smooth Lie groups on manifolds, this is not a problem because those "collapses" are modeled on the $G$-equivariant tubular neighborhood theorem
So apparently graphons appear as "limits" of graphs as the number of vertices go to infinity, somehow. I think it's some Gromov-Hausdorff sense of convergence but I don't know.
Okay, so I'm working on showing that a group with $2p$ elements is isomorphic to $\mathbb{Z}/2p\mathbb{Z}$ if it's abelian or it's isomorphic to the dihedral group of the $p$-gon if it's non-abelian. I feel like it should be simple to show this, but I can't see the outline of how a proof would go. I feel like I should be paying attention to subgroups, though.
Also, I've been using class notes for abstract algebra. It's a pdf written out like a book with chapters and whatnot, but unfortunately, you can't buy it, topological.
It's five times as long as it should be imo. The exercises are pretty good, but there are so many of them it's hard for someone to tell how to pick and choose.
It's scary that people are taking real analysis (presumably) and can't answer the question: If you write the series $\sum a_n x^{n!}$ in the form of $\sum a_k x^k$, what is the formula for $a_k$?
Problem: Suppose $f$ and $g$ are continuous functions on $[a,b]$. Show that if $f=g$ a.e. on $[a,b]$, then, in fact, $f=g$ on $[a,b]$. Proof: Let $A := \{x \in [a,b] \mid f(x) \neq g(x)\}$. Then by hypothesis $m(A)=0$. Moreover, $A$ is open. Hence, if there were a point $x \in A$, then we could choose $r > 0$ such that $(x-r, x+r) \subseteq A$. But this means $0 \neq 2r \le m(A)$, which is a contradiction. Hence $A$ is empty.
