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**aliceinwonderland** Consider this long exact sequence,

$\displaystyle \cdots \rightarrow H_2(S^1 \vee S^1) \rightarrow H_2(S^1 \times S^1) \rightarrow$$\displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1) \rightarrow H_1(S^1 \times S^1) \cdots$.

By the Hurewicz theorem, the first homology group of a torus $\displaystyle H_1(S^1 \times S^1)$ is isomorphic to the fundamental group of a torus $\displaystyle \pi_1(S^1 \times S^1) = \mathbb{Z} \times \mathbb{Z}$. By applying the Mayer-Vietoris sequence, we have $\displaystyle H_2(S^1 \vee S^1) = 0$ and $\displaystyle H_1(S^1 \vee S^1) = \mathbb{Z} \times \mathbb{Z}$.

Now we see that $\displaystyle H_2(S^1 \times S^1) \rightarrow H_2(S^1 \times S^1, S^1 \vee S^1) $ is an isomorphism in the above long exact sequence. Since $\displaystyle H_2(S^1 \times S^1)$ is isomorphic to $\displaystyle H_2(S^1 \times S^1, S^1 \vee S^1) \cong H_2(S^2) = \mathbb{Z} $, the quotient map $\displaystyle S^1\times S^1 \to S^2$ collapsing $\displaystyle S^1 \vee S^1$ to a point is not null homotopic.

Let $\displaystyle f:S^2 \to S^1\vee S^1$. The universal cover for $\displaystyle S^1\vee S^1$ is a $\displaystyle \mathbb{Re} \times \mathbb{Re}$ such that $\displaystyle g:\mathbb{Re} \times \mathbb{Re} \rightarrow S^1\vee S^1$ is a covering map. By using the lifting criterion (Hatcher p61), we have a lifting map $\displaystyle h:S^2 \rightarrow \mathbb{Re} \times \mathbb{Re}$ such that $\displaystyle f=gh$. Since $\displaystyle \mathbb{Re} \times \mathbb{Re}$ is contractible, h is nulhomotopic. It follows that $\displaystyle f:S^2 \to S^1\vee S^1$ is nullhomotopic.