I'm using the statement from Hatcher. I really don't understand the statement of the theorem, let alone the proof, and I especially don't understand what the normal subgroup $N$ generated by $i_{\alpha\beta}(w)i_{\beta\alpha}(w)^{-1}$ refers to.

First, I know that the two functions are inclusion from $A_{\alpha}\cap A_{\beta}$ to $A_\alpha$ and $A_\beta$ respectively.

Then, what I get from Van Kampen's theorem is that if you take any loop $w$ in the fundamental group of this intersection, and you include that loop into $\pi_1(A_\beta,x_0)$, reverse that loop, then concatenate it with the same loop $w$ included into $\pi_1(A_\alpha,x_0)$, you get a loop $f$ homotopic to the trivial loop at the basepoint $x_0$. But I don't get where this loop lies. Is this loop really in the full fundamental group of the entire space $X$? Or does one part of the loop really lie in $A_\alpha$ and the other lies in $A_\beta$?

To help illustrate my grave confusion, I'll attempt to compute the fundamental group of the ${\bf RP^2}$, the example I think I'm closest to understanding.

First, I will view $X={\bf RP^2}$ as the disk with the antipodal points on the boundary identified, so something like this picture. rp2

So I'm going to break up $X$ into two pieces $U$ and $V$. The open set $U$ will be $X-{x}$ for some point other than $x_0$, the basepoint. Then $V$ will be the open disk around $x$. Then the intesection $U\cap V$ is the open punctured disk around $x$.

First, the easy part is that $\pi_1(V)=0$ since the open disk is simply-connected. Then, we just need $\pi_1(U)$ and the normal subgroup $N$ as in the theorem.

For $\pi_1(U)$, the answer is that it is isomorphic to ${\bf Z}$. I can't come up with a good reason, though it intuitively makes sense. I read that it is because $U$ is a deformation retract of the $S^1$. I can picture pushing $U$ to a circle around $x$ so I buy this enough for now.

Now for the fun part. We need to consider any loop $w$ in the intersection of $U$ and $V$. For illustrative purposes, let's pick $w$ to be a loop which wraps around the hole in $U\cap V$ one time.

I take this loop and include it into $V$ first and then reverse its direction. By doing so, this adds back the point $x$ to the space in which the loop lies. Thus, since all loops in $V$ are homotopic to the trivial loop, this loop by inclusion to $V$ is trivial, so we ignore it.

Then I take the loop and include it into $U$. I cannot homotope this to the constant loop because the hole at $x$ still remains. So this loop maps to a nontrivial loop. But I can homotope this loop to the edge of X, although I'll have to move the basepoint around to the boundary also. Then, this loop has the form $a^2 = 1$ since it follows the edges. Then $N$ is the normal subgroup generated by $a^2$, which I've read is written $\langle a^2 \rangle$.

Then, by Van Kampen's theorem, $\pi_1({\bf RP^2}) = {\bf Z}/\langle a^2 \rangle$ which is isomorphic to ${\bf Z}/{\bf 2Z}$.

Can someone correct the errors I've made in my solution, and clear up the confusion I have with Van Kampen's theorem in general?


Tyler, Nice to see an undergrad from my grad alma mater. You did just fine. A few picky corrections: The $S^1$ is a deformation retract of the punctured disk, not vice versa (think about radial projection from the puncture). I would probably choose the basepoint to be in both open sets, but this isn't critical. Last, when you include $w$ into $U$ you get $a^2\in\pi_1(U)$. Thus, $\pi_1(X) \cong \langle a\rangle / \langle a^2\rangle \cong\mathbb Z/2\mathbb Z$.

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    $\begingroup$ Ah, I see. It's incorrect to say that by inclusion I get $a^2 = 1$ in $\pi_1(U)$; that equality is only true in $\pi_1({\bf RP^2})$ because in that case the loop $a^2$ can be homotoped to the trivial loop. Thanks for your response. $\endgroup$ – Tyler Brabham May 9 '13 at 22:10

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