Hello I was studying https://www2.warwick.ac.uk/fac/sci/maths/people/staff/vincent/cohomology.pdf
On page 18 given a space $A$ and a map $f:\mathbb{S}^{n-1} \rightarrow A$ he defines the cone $X := A \cup_f D^n$
I was trying to show that $(D^n, \mathbb{S}^{n-1}) \overset{\pi i}{\hookrightarrow} (X,A)$ induces isomorphism on the homology (theorem 2.27). This problem has been taking me days and I haven't been able to do it in a way that preserves the commutativity of the diagram (all of my tries involves retractions that alter the inclusions)
$$ \require{AMScd} \begin{CD} H_n(D^n,\mathbb{S}^{n-1}) @>i_*> \cong> H_n(X,A) \\ @V \cong VV @VV \delta V \\ H_{n-1}(\mathbb{S}^{n-1}) @>> f_*> H_n(A)\\ \end{CD} $$
my idea was, as a first step, prove that
$(D^n,\mathbb{S}^{n-1}) \overset{( \pi , f)}\rightarrow ( \ \overline{D^n}, \overline{\{x \in D^n : |x| \geq 1/2 \}} \ )$
is an homotopic equivalence of pairs, where the bars mean projected inside $X$.
To do this, first prove an homotopic equivalence of pairs
$(D^n,\mathbb{S}^{n-1}) \overset{i}\rightarrow ( \ {D^n}, {\{x \in D^n : |x| \geq 1/2 \}} \ )$
and then compose with the projection $(D^n,{\{x \in D^n : |x| \leq 1/2 \}}) \overset{(\pi, f)}\rightarrow ( \ \overline{D^n}, \overline{\{x \in D^n : |x| \geq 1/2 \}} \ )$
to get the desired homotopic equivalence of pairs.
Is this route correct?
This is NOT an homotopic equivalence of pairs, rather it just induces an isomorphism in the homology.
The issue why I wasn't able to do this is that I didn't know much about homotopy of pairs. We need the following lemma:
Let $f:(X,A) \rightarrow (Y,B)$ a morphism of pairs. Then if $f:X \rightarrow Y$ and $\left.f\right|_A:A \rightarrow B$ are both homotopic equivalences, then $f:(X,A) \rightarrow (Y,B)$ induces isomorphisms in the homology. Though is not necessarily an homotopic equivalence.
Given that, name $L= \{x \in D^n : |x| \geq 1/2 \}$
First, clearly $\mathbb{S}^{n-1} \hookrightarrow L $ induces a strong deformation rectract (I wasn't able to explicitely write it out) $r:L \hookrightarrow \mathbb{S}^{n-1}$ $(\left.r\right|_{\mathbb{S}^{n-1}} = \mathbb{1}_{\mathbb{S}^{n-1}})$.
Thus $(D^n, \mathbb{S}^{n-1}) \hookrightarrow (D^n, L)$ induces isomorphism on the homology.
I think I can extend $r$ to $A \mathbin{\overset{d}\cup} L$ by setting $\left.r\right|_A= \mathbb{1}_{A}$. Then since $\left.r\right|_{\mathbb{S}^{n-1}} = \mathbb{1}_{\mathbb{S}^{n-1}}$ r induces a map $\overline{r}:\overline{L} \rightarrow \overline{{S}^{n-1}}$ (I checked that is well defined on the quotient) that is a SDR of $\overline{{S}^{n-1}} \hookrightarrow \overline{L}$.
Thus by the lemma
\begin{align*} (D^n, \mathbb{S}^{n-1}) \hookrightarrow (D^n, L)\\ (\overline{D^n}, \overline{\mathbb{S}^{n-1}}) \hookrightarrow (\overline{D^n}, \overline{L}) \end{align*}
Induce isomorphisms at the homology level.
Now consider the conmutative diagram of inclusions
$$ \require{AMScd} \begin{CD} (D^n,\mathbb{S}^{n-1}) @>e>> (D^n,L) @<c<< (\overset{\circ}{D^n}, L- \mathbb{S}^{n-1}) \\ @VgVV @VVdV @VVaV \\ (\overline{D^n}, \overline{\mathbb{S}^{n-1}}) @>f>> (\overline{D^n}, \overline{L}) @<b<< (\overline{\overset{\circ}{D^n}}, \overline{L- \mathbb{S}^{n-1}})\\ \end{CD} $$
$a$ is an homeomorphism since the characteristic map doesn't identify anything in the interor of $D^n$. $c$ and $b$ are excision maps and thus induce isomorphisms. Thus $d$ is an isomorphism. Finally $e$ also induces isomorphisms by the lema. We conclude that the inclusion $f_\circ g $ induces isomorphisms.