the most convenient case of this map is $n = 1$, as $H_1$ is just maps from a bunch of circles. you can even derive the explicit isomorphism $\pi_1^{ab} \cong H_1$ from this now.
@iwriteonbananas the general Hurewicz? yes! the general Hurewicz theorem also tells you that if $X$ is a CW-complex such that all the lower homology groups vanish, $H_n(X) \cong \pi_n(X)$. this is immensly helpful in homotopy theory.
@iwriteonbananas but it's not helpful in homology theory, see. we are using homology to compute homotopy, so it's the other way :) homology theory is still the easiest invariant you can get.
okay, it's late. i think i need to run. have fun with relative homology groups.
@iwriteonbananas Yep, I saw that I had already proved it(since it is really simple from definition) but I wasn't sure how I should write it, but I see now obviously A->B->C is good
@Semiclassical Or $a=1$ and $b=1$. If I find a closed form for this case when $t=1$ then I'm encouraged to dig up further.
I mean the particular forms also give an idea about the difficulty of the generalization. If I find terribly hard such a case I couldn't think of doing generalizations so easily.
@Semiclassical maybe the best you can get is some series representation. This discussion reminds me of a problem I found a long time ago in a book for olympiad that I never solved. Maybe it's related to some extent to it.
i think the large $t$ behavior should be a sum of decaying exponentials, and for literature reasons i think their prefactors should be modified bessel functions
and if were using somewhat different pairs of $(a,b)$, i might be able to show that explicitly
@Semiclassical I know the name of the authors, and I'm going to talk one day with some of the professors I know and try to reach them. I don't have yet the emails of the authors.
@Semiclassical I'll probably write a letter using words like "With profound humble one of the greatest admirers of yours wanna ask mercy and receive a solution to the problem XXX". :-)
@Semiclassical lol, I cannot find anything about these guys (in terms of contact emails). One of them wrote a very good textbook for the 9th grade in high school that I have here.
I read about Clifford's Theorem online, would that break this proposition if $C$ is "hyperelliptic" and $D$ is not canonical but is a hyperelliptic divisor?
oh no, maybe that's not an "if and only if" condition for equality in Clifford's
@JasperLoy By the way, this is the type of girl my bro dreams of (I told him many times he wants too much from life, really ...). Such a girl can also heal all your diseases. :D
@Chris'ssis I am feeling very depressed now, so I will talk a lot of rubbish, sorry. I just think that I only have a 1 per cent chance of getting well.
Let X be a Hausdorff space, let K be a compact subset of X, and let U1 and U2 be open subsets of X such that K ⊆ U1 ∪ U2. Then, there are compact sets K1 and K2 of X such that K1 ⊆ U1, K2 ⊆ U2, and K = K1 ∪ K2.
so the reason this works is because K is compact subset of set which we can think of it as being "finite" so that is why we can break them into "pieces"
@Karim I wouldn't call compactness as finiteness. $(0, 1)$ is finite, but not compact. On the other hand, some people would argue that it looks "infinite-like", but then I could argue about the cantor set being compact.
