Second homotopy of $S^1\vee S^2 \vee T^2$

Question

How can I prove that the second homotopy group of $S^1\vee S^2 \vee T^2$ is infinitely generated?

I know that the second homotopu groups of $S^2$ and $T^2$ are finitely generated, so is kind of contra-intuitive for me.

Answer 1

if $p:E\to B$ is a covering, then $\pi_2(p):\pi_2(E)\to\pi_2(B)$ is an isomorphism.

consider the following covering with base $S^1\vee S^2\vee T^2$: let the total space be a line $\mathbb R$ with attached copies of $S^2$ and $T^2$ in every integer point; the map $\mathbb R\to S^1, x\mapsto e^{2\pi i x}$ extends on the whole space in the obvious way.

finally, this total space is homotopy equivalent to bouquet of infinite copies of $S^2$ and $T^2$.