= Solution
The <long exact sequence in homology> of the triple $A\subseteq U\subseteq X$ contains
$$
\cdots\to H_q(U,A)\to H_q(X,A)\to H_q(X,U)\to H_{q-1}(U,A)\to\cdots.
$$
Because $A\hookrightarrow U$ is a <homotopy equivalence>, $H_*(U,A)=0$. Hence the middle map is an isomorphism in every degree:
$$
H_*(X,A)\cong H_*(X,U).
$$
The <Excision theorem> says that if $Z\subseteq A\subseteq X$ and $\overline Z\subseteq\operatorname{int}A$, then inclusion induces
$$
H_*(X\setminus Z,A\setminus Z)\xrightarrow{\sim}H_*(X,A).
$$
A <good pair> $(X,A)$ has $A$ closed and a neighborhood $U$ that deformation retracts onto $A$. The <collapsing a pair theorem> states that the quotient map gives
$$
H_q(X,A)\cong\widetilde H_q(X/A).
$$
To prove it, choose such a neighborhood $U$. The first result identifies $H_*(X,A)$ with $H_*(X,U)$. Excision identifies the latter with $H_*(X/A,U/A)$, while $U/A$ is contractible because the deformation retraction can be chosen relative to $A$ using the <homotopy extension property>. The long exact sequence of the pair $(X/A,U/A)$ then identifies this relative group with $\widetilde H_*(X/A)$.
For $f:(D^n,\partial D^n)\to(D^n,\partial D^n)$, define $\deg f$ by
$$
f_*[D^n,\partial D^n]=(\deg f)[D^n,\partial D^n]
$$
in $H_n(D^n,\partial D^n;\mathbb Z)\cong\mathbb Z$. In the long exact sequence of the pair, the <connecting homomorphism>
$$
H_n(D^n,\partial D^n)\xrightarrow{\sim}H_{n-1}(\partial D^n)
$$
is an isomorphism. Naturality shows that $f_*$ and $(f|_{\partial D^n})_*$ multiply the corresponding generators by the same integer, so
$$
\deg f=\deg(f|_{\partial D^n}).
$$
Finally,
$$
(D^n\times I)/\partial(D^n\times I)\cong\Sigma(D^n/\partial D^n)\cong S^{n+1}.
$$
Under this identification, the map induced by $g(x,t)=(f(x),t)$ is the suspension of the map induced by $f$. By <degree under suspension>,
$$
\deg g=\deg f.
$$
This argument proves the needed product assertion directly from the natural suspension isomorphism in reduced homology.
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