One point union, second homotopy group is not finitely generated?

Question

Let $X$ be the one-point (wedge sum) union of the circle $S^1$ and the sphere $S^2$. What is the easiest way to see that the abelian group $\pi_2(X)$ is not finitely generated?

Answer 1

Recall that $\pi_n(X) \cong \pi_n(\widetilde{X})$ for $n > 1$, where $\widetilde{X}$ is the universal cover of $X$.

That said, universal cover of $X = S^1 \vee S^2$ is $\Bbb Z$-many $S^2$'s wedged at $\Bbb R$. From the previous remark, $\pi_2(X) \cong \pi_2(\tilde{X})$. As $\tilde{X}$ is simply connected, Hurewicz's theorem implies $\pi_2(\tilde{X}) \cong H_2(\tilde{X}) \cong \bigoplus^\infty \Bbb Z$ which is not finitely generated.

Answer 2

Instead of going via covering spaces, as is traditional, you can also use a Higher Homotopy Seifert-van Kampen theorems (HHSvKT), as in this book Nonabelian Algebraic Topology (pdf available there). The Introduction to Chapter 8 explains that if $X=S^n \vee [0,1], n>1,$ then $\pi_n(X, \{0,1\})$ is the free module on $1$ generator over the groupoid $\mathcal I= \pi_1([0,1],\{0,1\})$. When you identify $0,1$ in $X$ you get $S^n \vee S^1$ and the groupoid $\mathcal I$ becomes the group $\mathbb Z$. The HHvKT implies what you want.

I agree that the background and proof of the HHSvKT is complicated, but it does easily give so much more, such as a generalisation of the Relative Hurewicz Theorem, and nonabelian results in dimension $2$ on second relative homotopy groups. Students should also know about modules over groupoids!