Calculate $\pi_2(S^2 \vee S^1)$

Question

I am trying to calculate $\pi_2(S^2 \vee S^1)$ and having trouble fitting the pieces together.

I know that the universal cover of $S^2 \vee S^1$ is just $\mathbb{R}$ with spheres attached at integral points.

• Attempt 1: Let $\tilde{X}$ be the universal cover as above. We have that $\pi_2(S^2 \vee S^1) = \pi_2(\tilde{X})$. Now, since $\pi_1(\tilde{X}) = 0$, we have that $\tilde{X}$ is 1-connected, and thus by Hurewicz Theorem, $\pi_2(\tilde{X}) = H_2(\tilde{X})$.

• Attempt 2: Consider a map $f: S^2 \rightarrow S^2 \vee S^1$. This map lifts to a map $\tilde{f}:S^2 \rightarrow \tilde{X}$.

• Attempt 3: $H_2(S^2 \vee S^1) = H_2(S^2 / S^0) = H_2(S^2, S^0)$ gives a long exact sequence: $$ \dots \rightarrow H_2(S^0)\rightarrow H_2(S^2)\rightarrow H_2(S^2,S^0)\rightarrow H_1(S^0)\rightarrow \dots$$ $$= \dots \rightarrow 0\rightarrow \mathbb{Z} \rightarrow H_2(S^2,S^0)\rightarrow 0\rightarrow \dots$$ So, $H_2(S^2 \vee S^1) = H_2(S^2, S^0) \simeq \mathbb{Z} $, but I am not sure how/if this helps, since $\pi_1(S^2 \vee S^1) \neq 0$.

Any advice would be appreciated.

Answer 1

Your first attempt works and is probably the easiest way to go. The $\tilde{X}$ you found is homotopy equivalent to an infinite bouquet of 2-spheres: $\tilde{X} \simeq \bigvee_{n \in \mathbb{Z}} S^2$. There are various ways of computing $H_2(\tilde{X})$:

• $\tilde{X}$ is the filtered colimit $$* \subset S^2 \subset \bigvee_{n=-1}^1 S^2 \subset \bigvee_{n-2}^2 S^2 \subset \dots \subset \operatorname{colim}_{k \ge 1} \bigvee_{n=-k}^k S^2 = \tilde{X}$$ and since homology preserves filtered colimits, $$H_2(\tilde{X}) = \operatorname{colim}_{k \ge 1} \mathbb{Z}^{2k+1} = \bigoplus_{n \in \mathbb{Z}} \mathbb{Z} = \mathbb{Z}^{(\infty)}$$ is a direct sum of an infinite number of copies of $\mathbb{Z}$.
• Using cellular homology: the cellular complex of $\tilde{X}$ has $\mathbb{Z}$ in degree zero, $\mathbb{Z}^{(\infty)}$ in degree 2, and $0$ elsewhere. There cannot be any nontrivial differential for degree reasons, and so $H_2(\tilde{X}) = \mathbb{Z}^{(\infty)}$.

As for the intuition, recall that $\pi_1(X)$ always acts on $\pi_n(X)$ for $n \ge 1$ by "prepending" a loop or something similar. The inclusion $S^2 \subset S^2 \vee S^1$ gives an element $\alpha \in \pi_2(X)$, which generates a cyclic subgroup.

But for the loop $\gamma$ that goes around the $S^1$ factor once, $\gamma \cdot \alpha \in \pi_2(X)$ is another element that is not homotopic to any other element of the cyclic subgroup generated by $\alpha$. This $\gamma \cdot \alpha$ generates a cyclic subgroup too. And it continues: for every $n \in \mathbb{Z}$, $\gamma^n \cdot \alpha$ is an element of $\pi_2(X)$, and they all generate disjoint cyclic subgroups in $\pi_2(X)$. And now the argument above shows that these are all the elements of $\pi_2(X)$.