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?

Question

I want to solve exercise 2.2.12 in Hatcher's algebraic topology book.

Exercise: Show that the quotient map $S^1\times S^1\longrightarrow S^2$ collapsing $S^1\vee S^1$ to a point is not nullhomotopic showing it induces an isomorphism in $H_2$.

Partial Solution: Since $(S^1\times S^1, S^1\vee S^1)$ is a good pair we have $$H_2(S^1\times S^1, S^1\vee S^1)\simeq H_2\left(S^1\times S^1/S^1\vee S^1\right)=H_2(S^2).$$ Then we have $q_*:H_2(S^1\times S^1)\longrightarrow (S^1\times S^1, S^1\vee S^1)$. Consider the long exact sequence of the pair $(S^1\times S^1, S^1\vee S^1)$:$$0\longrightarrow H_2(S^1\times S^1)\longrightarrow H_2(S^1\times S^1, S^1\vee S^1)\longrightarrow H_1(S^1\vee S^1)\longrightarrow H_1(S^1\times S^1).$$ I saw somewhere $H_1(S^1\vee S^1)\longrightarrow H_1(S^1\times S^1)$ is an isomorphism, but I can't see why. Hence $q_*$ is an isomorphism so that it can be nullhomotopic for if that was the case $q_*=0$.

Doubt: How to explain why $H_1(S^1\vee S^1)\longrightarrow H_1(S^1\times S^1)$ is an isomorphism?

Answer 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?)