Simpler proof of van Kampen's theorem? I've been trying to understand the proof of van Kampen's theorem in Hatcher's Algebraic Topology, and I'm a bit confused why it's so long and complicated.
Intuitively, the theorem seems obvious to me. Given a path $p$ in $A \cup B$, we can split it up into paths $p_1p_2...p_n$ that alternate between $A$ and $B$. So we have $\pi_1(A, x) * \pi(B, x)$, except that certain paths from $A$ and $B$ are equivalent (the ones in $A \cap B$), so we need to quotient by $\pi_1(A \cap B, x)$.
I'm confused about what the proof in Hatcher's book is doing... Is it just a more detailed version of that idea? Or is there something I'm missing?
Thank you for your help.
 A: Well...the first thing you're missing is that your $n$ might not be finite unless you have some nice assumptions. For instance, if $A$ is the closed upper half-plane, and $B$ is the closed lower half-plane, then the graph of
$$
f(x) = \begin{cases}
x^2 \sin \frac{1}{x} & x \ne 0\\
0 & x = 0
\end{cases}
$$
has infinitely many arcs in $A$ and in $B$. "But," I hear you cry, "we have that $A$ and $B$ are open!" Sure you do...and is that enough to prove finiteness of $n$? Well...yes, kind of, but there's work to do...and there's a half-page of your life gone.
"But after that," you say, "The only subtle thing is getting rid of the equivalent things ... that "amalgamation" stuff, right?" Well... yes and no. Showing that curves that are equivalent modulo the relevant stuff in the amalgamated product are in fact homotopic is subtle. But you also have to prove that those are the ONLY ones that are, and that's subtle too. And there's another page or two.
You might want to remember that once you have the van Kampen theorem, you can prove the Jordan Curve Theorem, for which a correct (and understood-to-be-correct) proof took a very long time. So...it's not likely to be an easy proof.
A: If your space is nice enough to have a universal cover, there is a very elegant proof using covering spaces due to Grothendieck.
