How to show $q:H_2(S^1\times S^1)\longrightarrow H_2(S^1\times S^1, S^1\vee S^1)$ is an isomorphism?

Solution 1:

By your exact sequence you already have exactness of $$0 \rightarrow H_2(S^1\times S^1) \rightarrow H_2(S^1\times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1).$$ Now by your computation, the first two of these groups are isomorphic to $\mathbb Z$. Thus $q^\ast$ is multiplication by $d$ for some integer $d$, and the map $$\rightarrow H_2(S^1\times S^1, S^1 \vee S^1) \rightarrow H_1(S^1 \vee S^1)$$ has kernel $d\mathbb Z$, but since $H_1(S^1 \vee S^1)$ is free abelian, this is only possible for $d=\pm 1$ (why?) which proves that $q^\ast$ is an isomorphism (why?)

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