Proof of Seifert-Van Kampen Theorem

[PsychonautQQ]

To help me with this question, I think you'll need to have access to the proof, it's pretty involved and technical. I'm going the proof found in John M. Lee's "Introduction to topological manifolds", but I suspect that the proof will be the same no matter where you find it.

Let $U,V$ be the open sets who's union is $X$ and $U\cap V$ is path connected.

Let $F: \pi(U\cap V)$ --> (Free product of $\pi(U)$ and $\pi(V))$

Let $P:$ (Free product of $\pi(U)$ and $\pi(V))$ --> $\pi(X)$

Let $i_{*}$ be the inclusion of $\pi(U\cap V)$ into $\pi(U)$ and $j_{*}$ be the inclusion of $\pi(U\cap V)$ into $\pi(V)$.

Let $k_{*}$ be the inclusion of $\pi(U)$ into $\pi(X)$ and $l_{*}$ be the inclusion of $\pi(V)$ into $\pi(X)$.

Let $N$ be the normal closure of the image of $F$

I don't understand the step where we must prove that $Ker(P)$ is contained $N$.

First we suppose that $P(\gamma)=1$ so that $\gamma$ is an element of $Ker(P)$
We can factor $\gamma$ as $[a_{1}*...*a_{n}]_{[X]}=1$

So that there is a path homotopy between this factorization of $\gamma$ and 1 in X.

We then use the same technique we did in proving the unique homotopy lift property, by cutting up the mapping square of the homotopy between $\gamma$ and 1 so that each square is in either U or V.

Okay, so the part that I'm confused on is the inductive step where we show that each time we move up to the next "layer" of squares in the homotopy lifting square, the two layers are equivalent mod N. I followed the proof as far as recognizing that they did a bunch of algebraic manipulation and substitution for the terms and made things work out, but it seems very much arcane.

Can somebody give me some insight as to why each of these layers on the mapping square should be equivalent mod N? Or perhaps even some insight as to why I should not be surprised that $Ker(P)$ is contained in $N$

To be honest, I am slightly confused on a step before this as well. The text reads "by taking n to be a sufficiently large power of 2, we can ensure that the endpoints of the paths $a_{i}$ in this product are of the form $i/n$, so the path obtained by restricting $H$ (our homotopy between $\gamma$ and 1) to the bottom edge of the square can be written as:

$a_{1}*a_{2}*...*a_{k}$ is path equivalent to $(a_{1,0}*...*a_{p,0)*...*(a_{r,0}*...*a_{n,0})$ (edit: don't know why LaTeX not working here)

I don't understand what's going on with "a power of 2", why is it necessary to be a power of 2? Why not just an arbitrarily large n? Furthermore, what exactly is going on here, is $(a_{1,0}*...*a_{p,0})$ in the second factorization equivalent to $a_{1}$ in the first factorization?
Thanks PF!

 

[fresh_42]

Maybe we could discuss -or do not need anymore - this proof instead
https://www3.nd.edu/~andyp/notes/SeifertVanKampen.pdf
so that we all have the same source.

 

[member 587159]

A proof is also in Hatcher's book (which is freely available on the net) but his proof has a more geometric flavour. I recently read this proof and I was glad once I finally got through it.

I doubt the OP still needs an answer to his question though.