Show the quotient map $S^1 \times S^1 \rightarrow S^2$ collapsing the subspace $S^1 \vee S^1$ to a point is not nullhomotopic Hatchers problem from section 2.2 #12 b says to show that the quotient map $S^1 \times S^1 \rightarrow S^2$ collapsing the subspace $S^1 \vee S^1$ to a point is not nullhomotopic by showing that it induces an isomorphism on $H_2$.

I'm confused by this question. What is the subspace $S^1 \vee S^1$ he is referring to ?

 A: With $S^1$ I  mean the one dimension sphere and with $H_1$, $H_2$, ... I mean the  homology groups.
Obviously the torus $S^1\times S^1$ is a $CW$-complex with one $0$-cell, two $1$-cells and one $2$-cell. Moreover the wedge sum $S^1\vee S^1$ is a subcomplex of $S^1\times S^1$: indeed $S^1\vee S^1$ is isomorphic to the $CW$-complex that we find if we don’t paste the $2$-cell to $S^1\times S^1$.
Conclude $(S^1\times S^1,S^1\vee S^1)$ is a good pair and then we can find the long exact sequence (see the Theorem 2.13  Hatcher Algebraic Topology https://pi.math.cornell.edu/~hatcher/AT/AT.pdf)
$\dots\to H_2(S^1\vee S^1)\to H_2(S^1\times S^1)\xrightarrow{\rho} H_2(S^1\times S^1/S^1\vee S^1) \xrightarrow{\delta} H_1(S^1\vee S^1)\xrightarrow{i} H_1(S^1\times S^1)\to H_1(S^1\times S^1/S^1\vee S^1)\to\dots$,
where $ \rho:H_2(S^1\times S^1)\to H_2(S^1\times S^1/S^1\vee S^1)$ is the arrow that we want to show it is an isomorphism. But $S^1\times S^1/S^1\vee S^1$ is iso to $S^2$ and the groups, which appear in the sequence, are well known.
The sequence becomes
$\dots \to 0\to \mathbb{Z}\xrightarrow{\rho} \mathbb{Z}\xrightarrow{\delta}\mathbb{Z}\times\mathbb{Z}\xrightarrow{i}\mathbb{Z}\times\mathbb{Z}\to 0\to\dots $.
Then $\rho$ is injective and the arrow $i:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$ is surjective.
$\underline{Claim}$: $ker(i)=0$ ($i$ is injective$\iff$ $i$ is isomorphism, since $i$ is surjective too).
Suppose the $\underline{Claim}$ holds, then $0 = ker(i) = Im(\delta)$ which implies $\rho$ is surjective. So, $\rho$ is isomorphism (since it is injective too).
$\textit{Proof}~(\underline{Claim})$: $\mathbb{Z}\times\mathbb{Z}$ is a free $\mathbb{Z}$-module and $i$ is a $\mathbb{Z}$-homorphism. We can see $\mathbb{Z}\times\mathbb{Z}$ like an $\mathbb{Z}[X]$-module via $i$, that is for each $\eta(X)$ belong to $\mathbb{Z}[X]$ and for each $a$ belong to $\mathbb{Z}\times\mathbb{Z}$ define $\eta(X)*a=\eta(i(a))$ (the polynomial evaluation is defined componetwise). Let $I=(X)$ the obviously ideal in $\mathbb{Z}[X]$, since $i$ surjective $I(\mathbb{Z}\times\mathbb{Z})=\mathbb{Z}\times\mathbb{Z}$. Then by the Nakayama’s lemma (https://stacks.math.columbia.edu/tag/07RC) there exists an element $\gamma(X)\in I$ such that for each $a\in$ $\mathbb{Z}\times\mathbb{Z}$ $\gamma(X)*a=a$.
Let $k\in ker(i)$, then $k=\gamma(X)*k=\gamma(i(k))=\gamma(0)=0~\implies~ m=0\implies ker(i)=0 \implies i$ injective.
