I’ve just read the set up and the statement for van Kampen’s theorem. Here’s the version that we use in our class.
van Kampen's Theorem. Suppose we have an open cover $\left\{U_{\alpha}: \alpha \in A\right\}$ of $\left(X, x_{0}\right)$ by path-connected open sets $U_{\alpha} \ni x_{0}$ containing the basepoint. As above, let $F:= \ast_{\alpha} \pi_{1}\left(U_{\alpha}\right)$ be the free product and $\varphi: F \rightarrow \pi_{1}(X)$ be the map induced by the inclusions. Also let $K \triangleleft F$ be the normal subgroup generated by the words $i_{\alpha \beta}(\sigma) i_{\beta \alpha}(\sigma)^{-1}$ for $\sigma \in \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right) .$ Then since $K<\operatorname{ker} \varphi$
as explained above, we get a map $\bar{\varphi}: F / K \rightarrow \pi_{1}(X)$.
- If $U_{\alpha} \cap U_{\beta}$ is path-connected for all $\alpha, \beta \in A$, then $\bar{\varphi}$ (or equivalently $\varphi$ ) is surjective.
- If furthermore $U_{\alpha} \cap U_{\beta} \cap U_{\gamma}$ is path-connected for all $\alpha, \beta, \gamma \in A$, then $\bar{\varphi}$ is injective, thus an isomorphism $F / K \cong \pi_{1}(X).$
The accompanying diagram is as follows:
What is not clear to me is what happens, under $\bar{\varphi}$ (or $\varphi$), to the loops in $X$ that is not in $U_\alpha \cap U_\beta$. For example, let $X$ be covered by $U_\alpha$, $U_\beta$, and a “weird” (but still path-connected) $U_\gamma$.
The loop in red (let’s call it $\lambda$) is in $\pi_1(X)$, and it has a corresponding element $a_\gamma^j$ in $F$, where $a_\gamma$ is a generator of $\pi_1(U_\gamma)$. Under the projection $p: F \to F/K$, it is sent to the coset $a_\gamma^jK$ of $K$ in $F$. I’m inclined to think that $a_\gamma^j$ “gets its own coset” under $p$, as do other loops not in $\pi_1(U_\alpha \cap U_\beta)$ (if any), since they have nothing to do with $K$. Is there anything else we know about $a_\gamma^jK$ in this setting? What would its elements look like?
Also, the reason for this question is that when the theorem was introduced, I got the impression that the diagram would somehow give a complete picture of $\pi_1(X)$. But it seems like that is not the case, or at least some elements of $\pi_1(X)$ would not be described as precisely as others with this diagram. Would a remedy be that, instead of considering $F$ modulo just $K$, we take $F$ modulo all such normal subgroups over all pairs of covering sets, i.e. $F/\kappa$, where $\kappa$ is the normal subgroup of $F$ generated by all the $K_1,K_2,\dots,K_t$?
Best Answer
I think you are misunderstanding how van Kampen's theorm works. This is an unfortunately common problem, because I think a lot of instructors jump to overly complex examples (if they give good examples at all). Let's give a couple really really simple examples to show how van Kampen's theorem works, and hopefully that will clarify things.
Van Kampen's theorem tells us how to compute the fundamental group of $X$ given the fundamental groups of $U_\alpha$ for some open cover of $X$. For simplicity, we'll focus on the case $X$ is covered by two open sets: $U$ and $V$.
Van Kampen's theorem says something really really obvious, once you're comfy with amalgamated products:
That's it. That's the theorem. Let's see why.
Here we're covering a cylinder by $2$ cylinders. Obviously we don't need van Kampen's theorem to compute the fundamental group of this space. But that's why it's such an instructive example! We know we should get $\mathbb{Z}$ at the end. There's only one nontrivial loop.
Well, let's look at $U$. It's a cylinder, so its fundamental group is generated by a loop, say, $\gamma_U$. But $V$ is also a cylinder, and its fundamental group is generated by a loop $\gamma_V$.
But now we think about van Kampen's theorem! It says a loop in $X$ is either a loop in $U$ or a loop in $V$... So it seems like we'll learn $\pi_1(X)$ is generated by $2$ elements! What's the issue? We're double counting the loop. Inside of $X$, $\gamma_U = \gamma_V$, and so in the inclusion maps $\pi_1(U), \pi_1(V) \hookrightarrow \pi_1(X)$ we need to make sure we identify $\gamma_U$ and $\gamma_V$.
But this is exactly what amalgamation buys us! If we look at $U \cap V$, which is a cylinder too, we find a generator $\gamma_{U \cap V}$.
Now when we include $U \cap V \hookrightarrow U$, what happens to the fundamental group? Inside of $U$ (up to homotopy) $\gamma_{U \cap V}$ and $\gamma_U$ are the same loop! So the induced map on $\pi_1$ is
$$ \begin{align} \pi_1(U \cap V) &\to \pi_1(U) \\ \gamma_{U \cap V} &\mapsto \gamma_U \end{align} $$
Similarly, for $U \cap V \hookrightarrow V$, we find $\gamma_{U \cap V} \mapsto \gamma_V$. Because again, inside of $V$, they're the same loop.
So when we look at the pushout in $\mathsf{Grp}$, we see
So
$$ \begin{align} \pi(X) &= \pi_1(U) \underset{\pi_1{U \cap V}}{\ast} \pi_1(V) \\ &= \langle \gamma_U \rangle \underset{\langle \gamma_{U \cap V} \rangle}{\ast} \langle \gamma_V \rangle \\ &= \langle \gamma_U, \gamma_V \mid \iota_U(\gamma_{U \cap V}) = \iota_V(\gamma_{U \cap V}) \rangle \\ &= \langle \gamma_U, \gamma_V \mid \gamma_U = \gamma_V \rangle \\ &= \langle \gamma_U \rangle \end{align} $$
So we find $\pi_1(X) = \mathbb{Z}$, generated by either of $\gamma_U$ or $\gamma_V$ (since they're the same). But this is exactly what we would expect!
As a second example, let's think about the fundamental group of a torus. This is a famous example, and it's probably been done before on mse, but we're already here, so ¯\_(ツ)_/¯.
Now $U$ is a wedge of two circles, so its fundamental group is $F_2 = \langle a, b \rangle$. If you like, this is another application of van Kampen.
Similarly, $V$ is just a square. It's simply connected, and we know its fundamental group is $1$.
Lastly, $U \cap V$ is a loop. So its fundamental group is $\mathbb{Z}$, generated by, say, $\gamma$.
Let's look at the inclusion maps.
$U \cap V \hookrightarrow U$ induces the map $\gamma \mapsto aba^{-1}b^{-1}$, since $\gamma$ inside of $U \cap V$ is the same loop as $aba^{-1}b^{-1}$ inside of $U$.
$U \cap V \hookrightarrow V$ induces the map $\gamma \mapsto 1$, since $\gamma$ is the trivial loop inside of $V$ (indeed, every loop is trivial inside $V$).
So now, we look at our pushout in $\mathsf{Grp}$ and we get:
$$ \begin{align} \pi_1(X) &= \pi_1(U) \underset{\pi_1(U \cap V)}{\ast} \pi_1(V) \\ &= \langle a, b \rangle \underset{\langle \gamma \rangle}{\ast} 1 \\ &= \langle a, b \mid \iota_U \gamma = \iota_V \gamma \rangle \\ &= \langle a, b \mid aba^{-1}b^{-1} = 1 \rangle \\ &= \mathbb{Z}^2 \end{align} $$
Again, loops in $X$ are the same as loops in $U$ and loops in $V$, but now we "double counted" the loop $aba^{-1}b^{-1}$ and $1$. Inside of $X$ they're really the same loop, so we have to make sure we remember that.
Every van Kampen theorem problem looks more-or-less like this, and if you want to see a trickier example of this same idea, you might check out a previous answer of mine. I've also been planning to write up these examples (and a few other instructive ones) in a blog post, so I might come back and link that when I get around to it.
Edit:
@ensbana -- The idea is to subdivide. This is actually where most of the difficulty in proving van Kampen's theorem comes from. We're not saying that every loop is something in $U$ or something in $V$. We're saying every loop is some product of loops in $U$ and loops in $V$.
Since $U$ and $V$ are open and a loop is a circle is compact, we can break our loop into finitely many pieces, each of which is entirely contained in $U$ or in $V$. Then a not-too-tricky argument uses path-connectedness to argue that these pieces can be homotoped to loops so that the concatenation of these loops is homotopic to what we started with.
In particular, loops of the kind you're worried about don't really exist at small enough scales -- everything is either a loop entirely in $U$, a loop entirely in $V$, or a product of things of this form.
Of course, this same idea works when we have more sets in our open cover of $X$. Then we use $U_\alpha$ to construct an open cover of $S^1$, which has a finite subcover, and again we've broken up our loop into finitely many pieces, each of which is contained in exactly one $U_\alpha$. The same homotopy argument that I alluded to last time goes through unchanged.
I hope this helps ^_^