This is question 16 of section 1.2 in Hatcher's Algebraic Topology. I have to show that the fundamental group of the space $X$ is free on an infinite number of generators. picture from the book

So here is my approach. Using van-Kampen, I can see that $\pi_1(X)=\pi_1(U)\star\pi_1(V)$. Also $U$ and $V$ are homeomorphic. So I've to calculate $\pi_1(U)$. I have that the fundamental group of $U_g$ (torus of genus $g$ with two points removed) is free on 3g $2g+1$ generators.

Now in some sense, $U=\lim_{g\to\infty}{U_g}$ and so $\pi_1(U)$ is free on infinite generators. But how do I justify this? Is it even true?

My second approach is, using van-Kampen, $\pi_1(U)=\pi_1(U_1)\star\pi_1(U)$. This is an infinite recursion. But again I cannot proceed further.

Any help is appreciated. Thanks!


You are incorrect in your calculation of $\pi_1(U_g)$; it's free on $2g+1$ generators. What you need now is that the map $U_g \to U_{g+1}$ induces the map $F_{2g+1} \to F_{2g+3}$ sending the $i$th generator of the first to the $i$th generator of the second. Now we have...

If a space $X$ is the union of a directed set of subspaces $X_\alpha$ ordered by inclusion with the property that each compact set in $X$ is contained in some $X_\alpha$, and each $X_\alpha$ contains a given point $x_0$, then the natural map $\varinjlim \pi_1(X_\alpha,x_0) \to \pi_1(X,x_0)$ is an isomorphism.

Surjectivity follows from the compactness hypothesis; the image of a representing map $f: S^1 \to X$ of an element of $\pi_1(X,x_0)$ lies in some $X_\alpha$, so is in the image of some $f' \in \varinjlim\pi_1(X_\alpha, x_0)$. Suppose $f \in \varinjlim \pi_1(X_\alpha, x_0)$ maps to zero; then we have a null-homotopy $f_t: S^1 \times I \to X$. Since the image of this is compact, this defines a null-homotopy $f_t: S^1 \times I \to X_\alpha$ for some $\alpha$, so $f$ is zero in $\varinjlim \pi_1(X_\alpha, x_0)$ as well.

(This fact, and its proof, is an adaptation of Proposition 3.33 in Hatcher's algebraic topology book.)

Taking the limit of $\pi_1(U_g,x_0)$ gives us that $\pi_1(U) \cong F_\infty$, the free group on countably many generators. Similarly for $\pi_1(V,x_0)$.

Now $U \cap V \cong S^1$, and the inclusion $S^1 \to U, S^1 \to V$ induces the map $F_1 \to F_\infty$ sending the generator of the first to a chosen generator of the second. Thus by the van Kampen theorem, $$\pi_1(U \cup V, x_0) \cong \langle u_1, v_1, u_2, v_2, \dots \, \mid \, u_1 = v_1\rangle,$$ giving $\pi_1(X,x_0) = \pi_1(U \cup V, x_0) \cong F_\infty$ as desired.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.