To answer my own question (in the setting of CW-complexes and thus also for smooth manifolds):
According to Ganea Link to Ganea‘s paper the homotopy fiber of $X\vee Y\to X\times Y$ is homotopy-equivalent to $\Omega X*\Omega Y$ if $X,Y$ are CW-complexes.
If $X,Y$ are aspherical, then their loop...
I have two spheres, take a disk on each of those spheres, puncture those disks and then glue the two spheres together along these punctured disks. In formal terms and the previous notation, I'm identifying $\varphi(x)$ with $\varphi^{\prime}(x)$ instead of $\varphi^{\prime}((1-||x||)x)$
Ah ok. Well if you're just trying to show that there's a well-defined map whose input is two manifolds-with-embedded-discs and spits out a manifold, I'll leave you to it
what the construction accomplishes is getting an orientation on the connected sums that is compatible with the orientations on each of the summands and that's what will be violated by not doing the gluing the proper way
Notice that when you draw the tube, the ends have opposite orientations as the boundary of the tube. Now are they both the orientation you get as the boundary of what's left in the sphere when you remove the disk?
The r means I precomposed with a reflection before including the disc.
Cong means "are diffeomorphic". When the disc doesn't agree with the orientation on your manifold, as observed, there's no natural orientation on the sum.
@MikeMiller Here's the definition of connected sum I have: Let $M,M^{\prime}$ be two manifolds of dimension $n$ and take two charts $\varphi\colon B^n\rightarrow M$, $\varphi^{\prime}\colon B^n\rightarrow M^{\prime}$. We define the connected sum $M\# M^{\prime}$ as the identification space obtained by gluing together $M\setminus\varphi(0)$ and $M^{\prime}\setminus\varphi^{\prime}(0)$ via identifying $\varphi(x)$ with $\varphi^{\prime}((1-||x||)x)$. Ted told me that the reason for this $(1-||x||)$ factor (which turns the ball inside out, essentially) is to ensure that the connected sum has a…
If I glue in a backwards disk instead of a forwards disk, the resulting space fails to be orientable, so how can it be homeomorphic to a sphere? Ugh. I remember wrestling with all this stuff in grad school.
Which cohomology? I only know de Rham cohomology, but I do know the wedge product gives a ring structure on the graded cohomology algebra in that case.
The crucial thing is that if you blow up a point in $\Bbb CP^2$, you end up with an exceptional divisor ($\Bbb CP^1$) with self-intersection $-1$. So you have to compare $\Bbb CP^2\#\Bbb CP^2$ and $\Bbb CP^2\#\overline{\Bbb CP^2}$.
Every manifold $M$ and $N$ in what follows is oriented, and $M \# N$ means connected sum in the way with the factor where you're careful for orientation's sake.
For $M$ a connected, oriented, manifold of dimension $n$, $H^*(M;R)$ is [notation I'm making up on the spot] a connected oriented graded ring of dimension $n$: it's a graded ring with fixed isomorphisms $1: H^0(M;R) \to R$ and $[M]: H^n(M;R) \to R$.
Given connected oriented graded rings of dimension $n$, say $A$ and $B$, you can define their connected sum as follows: $(A \# B) = (A \oplus B)/\left(1_A = 1_B, [A] = [B]\right).$
You identify the chosen generators in bottom and top degree.
Then one has $H^*(M;R) \# H^*(N;R) \cong H^*(M \# N; R)$.
I'm writing this like an asshole, but I'm hoping it's clear enough.
I should take $A\oplus B$ to just mean taking direct sum in each degree for the graded structure, right? And then we essentially just reduce bottom and top degree to one dimension again.
Compute this for $M = \Bbb{CP}^2, N = \Bbb{CP}^2$, and then compute this for $M = \Bbb{CP}^2, N = \overline{\Bbb{CP}^2}$. Prove that the resulting rings are non-isomorphic
@Fargle I'm not sure I understand. You're saying if we change the interval of $x$ from $\left[ 0,1\right]$ to $\left[ \frac{1}{4}, \frac{3}{4} \right]$, for example, the function should change?
That is, $H^*(\Bbb{CP}^2 \# \Bbb{CP}^2; R) \not\cong H^*(\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}; R)$. That's definitely true for $R$ a ring containing $\Bbb Z$, though I think true as soon as $2 \neq 0$ in $R$.
The point is to look at the quadratic form $q(x,y) = [x \smile y]$ for $x,y \in H^2$.
The point is that the orientation on each factor determines what the intersection product does (since you need the orientation to determine signs), so if you change the orientation on exactly one side (by not using that $1-|x|$ thing) you change the intersection product for exactly one side of the manifold.
But the intersection form is an invariant, so you get qualitatively different results.
This annoying and pathological function, therefore, is not integrable on $[0,1]$.
To circle back, that's what it looks like when you conceivably have another choice. If you conceivably have another distinct choice of area, then you didn't really have any notion of area to begin with.
(It's proven that there is only one for all $n\ge1$, that much I know.)
Yeah, that makes sense.
So then tracing back to the original point, the area was necessarily $\frac {b^3}{3}$ before being proven because $n$ being included would not make sense.
That introduction is a teeeeeeeeeeeeny bit handwavy, at any rate. There is an important subtlety: the fact that the rectangles are of equal width is not a requirement. Apostol is ignoring a lot of possible upper/lower sums by not considering every possible crazy way he could have cut the interval up.
But it will turn out to be the case that he's justified in ignoring a lot of the chaff.
I mean in the guy's defense, it's the very introductory part, I just happen to be nitpicky because I wanna absorb as much as I can in the most intuitive way possible.
Understandable. I had a similar problem with Rudin, a ways back. One of the early proofs in that involves a choice that seems completely out of left field.
Here, at least, you can look at the formulae Apostol gives for $s_n$ and $S_n$ long enough and say, "Okay, I get why he guesses that the area is $\frac{b^3}{3}$."
Maybe the issue isn't as big as I make it appear to be, but a core aspect of my frustrations (or more precisely, what I struggle with the most), is understanding the intuition for why the mathematician arrived to where he was, and how I could potentially do the same.
As fascinating as some proofs and derivations are, if they start pulling crazy concepts out of thin air, it loses me.
And yeah, you're right @Fargle, thanks for the help.
If you're doing it by contradiction, then you have to work harder than this. If $A>b^3/3$, you have to show that you can find an $n$ so that $A>S_n$ as well. Similarly if you have $<$.
The right way to do this, as I explained twice, is this: You observe that $b^3/3$ satisfies the inequality. Because $S_n-s_n$, which you can calculate, goes to $0$ as $n$ gets bigger, there cannot be two different numbers that satisfy the inequality. (Think about this pictorially. If you had $A$ and $A'$ both satisfying it, then $|A-A'|$ would have to be $\le S_n-s_n$ for all $n$. Draw a number line to see this.)
To answer my own question (in the setting of CW-complexes and thus also for smooth manifolds):
According to Ganea Link to Ganea‘s paper the homotopy fiber of $X\vee Y\to X\times Y$ is homotopy-equivalent to $\Omega X*\Omega Y$ if $X,Y$ are CW-complexes.
If $X,Y$ are aspherical, then their loop...
