Cohomology ring of $n$-torus While developing the cup product, Hatcher gives the following example:

I understand most of it, but I am having trouble understanding what he means at the end by the first two sentences in the last paragraph (which begin "An equivalent statement is that..." and "Via the long exact sequence..."
I just don't know why that statement is equivalent or why the long exact sequence implies what he says it does.
 A: One has the following quotient map $q : (Y \times I, Y \times \partial I) \to (Y \times S^1, Y \times s_0)$ obtained from the equivalence relation $\sim$ on $Y \times I$ defined on $\partial I \times Y$ by $(y, 0) \sim (y, 1)$. One can prove that $q_*$ on cohomology is an isomorphism as follows. 
$$\require{AMScd}
\begin{CD}
H^k(Y \times I, Y \times \partial I) @<{q_*}<< H^k(Y \times S^1, s_0 \times Y)\\
@A{\cong}AA @A{\cong}AA \\
H^k(Y \times I/Y \times \partial I, pt) @<{\cong}<< H^k(Y \times S^1/s_0 \times Y, pt)
\end{CD}$$
The vertical maps are isomorphisms because $(Y \times I, Y \times \partial I)$ and $(Y \times S^1, s_0 \times Y)$ are both good pairs, as so are $(I, \partial I)$ and $(S^1, pt)$. The bottom horizontal map is obtained from the obvious homeomorphism between $Y \times I/\partial I \times Y$ and $S^1 \times Y/s_0 \times Y$, hence is also an isomorphism. The diagram commutes, hence $q_*$ is also an isomorphism. By naturality of cross product, we have the following commutative diagram,
$$\require{AMScd}
\begin{CD}
H^n(Y; R) @>{\times \alpha}>> H^{n+1}(Y \times I, Y \times \partial I; R)\\
@A\text{id}AA @A{q_*}AA \\
H^n(Y; R) @>{\times \alpha'}>> H^{n+1}(Y \times S^1, Y \times s_0 ;R);
\end{CD}$$
Where $\alpha$ is the generator of $H^1(I, \partial I; R)$ and $\alpha'$ is the generator of $H^1(S^1, s_0)$. As the top map and the two vertical maps are both isomorphisms, the bottom map is too.
For the second statement, note that we have the following split short exact sequence
$$0 \to H^{n+1}(Y \times S^1, Y \times \{s_0\}; R) \stackrel{\pi^*}{\to} H^{n+1}(Y \times S^1; R) \to H^{n+1}(Y \times \{s_0\}; R) \to 0$$
obtained from the long exact sequence for $(Y \times S^1, Y \times \{s_0\})$, via the section obtained from the induced map of the retraction $r : Y \times S^1 \to Y$ on cohomology. 
According to splitting lemma, $H^{n+1}(Y \times S^1, Y \times s_0; R) \times H^{n+1}(Y \times s_0; R) \cong H^{n+1}(Y \times S^1; R)$ where the isomorphism is given by $(\beta, \beta') \mapsto \pi^*(\beta) +r^*(\beta')$. Using the previous isomorphism, this means we have an isomorphism $$H^n(Y; R) \times H^{n+1}(Y; R) \to H^{n+1}(Y \times S^1; R)$$ given by $(\beta_1, \beta_2) \mapsto \pi^*(\alpha \times \beta_1) + r^*(\beta_2) = \alpha \times \beta_1 + 1 \times \beta_2$ by removing the $\pi^*$ because it's not just an embeddeding but an actual inclusion, and $r^*(\beta_2) = 1 \times \beta_2$ as the retraction $r$ is nothing but projection onto 1st coordinate.
