Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 114 / 2 / Solution

The long exact sequence in homology of the triple $A\subseteq U\subseteq X$ containsBecause $A\hookrightarrow U$ is a homotopy equivalence, $H_{\ast }(U,A)=0$. Hence the middle map is an isomorphism in every degree:

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesA 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 givesTo prove it, choose such a neighborhood $U$. The first result identifies $H_{\ast }(X,A)$ with $H_{\ast }(X,U)$. Excision identifies the latter with $H_{\ast }(X/A,U/A)$, 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 ${\tilde{H}}_{\ast }(X/A)$.

For $f:(D^n,\partial D^n)\to (D^n,\partial D^n)$, define $\mathrm{deg}f$ byin $H_n(D^n,\partial D^n;\mathbb{Z})\cong \mathbb{Z}$. In the long exact sequence of the pair, the connecting homomorphismis an isomorphism. Naturality shows that $f_{\ast }$ and $(f{|}_{\partial D^n}{)}_{\ast }$ multiply the corresponding generators by the same integer, so

Finally,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,This argument proves the needed product assertion directly from the natural suspension isomorphism in reduced homology.