Can I use Seifert-van Kampen theorem infinite times

Question

I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?

Thanks.

Answer 1

As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $\pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.

Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_{i\in I}$ is an open cover of $X$ such that $*\in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_i\cap U_j$, and $U_i\cap U_j\cap U_k$ are path-connected for all $i,j,k\in I$. Then $\pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $\pi_1(U_i,*)$ by relations which say the two maps $\pi_1(U_i\cap U_j,*)\to \pi_1(U_i,*)$ and $\pi_1(U_i\cap U_j,*)\to\pi_1(U_j,*)$ become equal.

Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.

Answer 2

What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.

In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.

Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.

Answer 3

If we have an infinite cover of a space $X$, one approach is to compute $\pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 \subseteq U_2 \subseteq \cdots \subseteq X = \bigcup_{i=1}^\infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.

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Given a space $X$ which is an increasing union of open sets $U_1 \subseteq U_2 \subseteq \ldots$, with a specified basepoint $\ast \in U_1$, we have a diagram $$\require{AMScd} \begin{CD} \pi_1(U_1,\ast) @> >> \pi_1(U_2,\ast) @>>> \ldots \\ @VVV @VVV \\ \pi_1(X, \ast) @= \pi_1(X,\ast) @= \ldots \end{CD}$$ which gives a map from the direct limit $$\Phi: \varinjlim \pi_1(U_i, \ast) \to \pi_1(X,\ast).$$

Claim: $\Phi: \varinjlim \pi_1(U_i, \ast) \to \pi_1(X,\ast)$ is an isomorphism.

Proof: To see that $\Phi$ is surjective, observe that if $\gamma: (S^1, \ast) \to (X,\ast)$ is a based loop, then $\gamma(S^1) \subseteq \cup_{i=1}^\infty U_i$ is compact, so its image is contained in a finite subcover of $\{U_i\}_{i=1}^\infty$ and thus in $U_i$ for some $i$.

In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[\gamma_1]$ and $[\gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 \times I \to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[\gamma_1] = [\gamma_2]$ in $\pi_1(U_k,\ast)$ for $k \geq j$. This shows injectivity.

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While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $\{x_1,x_2,\ldots, x_n\}$, with maps $F_n \to F_{n+1}$ given by including the generators $\{x_1,x_2,\ldots, x_n\} \subseteq \{x_1,x_2,\ldots, x_{n+1}\}$, then $$ \varinjlim F_n = F_\infty,$$ where $F_\infty$ is the free group on infinitely many generators $\{x_1, x_2,\ldots\}$.