Isomorphism $\varphi_\star^e :H_n(\mathbb{D}_{\lambda}^n,\mathbb{S}_{\lambda}^{n-1})\longrightarrow H_n(X,X \setminus e)$ Let's suppose that $X$ is obtained from $A$ attaching $n$-cells. We already have an isomorphism $\Phi : \oplus_\lambda H_n(\mathbb{D}_{\lambda}^n,\mathbb{S}_{\lambda}^{n-1})\longrightarrow H_n(X,A)$. The problem is how to describe the inverse of $\Phi$.
Let $e = e_\lambda^n$ be an $n-$cell and consider the homomorphism induced by the inclusion $p_\star^e : H_n(X,A) \longrightarrow H_n(X,X\setminus e)$.
I'd like to state that there's an isomorphism $$\varphi_\star^e :H_n(\mathbb{D}_{\lambda}^n,\mathbb{S}_{\lambda}^{n-1})\longrightarrow H_n(X,X \setminus e)$$ so that the composition $(\varphi_\star^e )^{-1} \circ p_\star^e$ is the inverse of the $\lambda$ component in $\Phi$.
Can this isomorphism $\varphi_\star^e$ be deduced by excision theorems ? If not, what's the way to approch it ? Any reference, help or solution would be appreciated.
 A: We need some ingredients:

*

*If $(Y,B)$ is cofibred pair (which means that $B \hookrightarrow Y$ is a cofibration)
and $B$ is contractible, then the quotient map $Y \to Y/B$ is a homotopy equivalence.
See the accepted answer to Are the fundamental groups of $X$ and $X/A$ isomorphic when $A$ is contractible? for a proof.


*If $(Y,B)$ is cofibred pair, then the quotient map $p : (Y,B) \to (Y/B,*)$ induces isomorphisms $p_* : H_n(Y,B)  \to H_n(Y/B,*)$.
To see this observe that $(Y \cup CB, CB)$ is a cofibred pair. This follows from the fact that pushouts of cofibrations are again cofibrations. See A cofibration induces a cofibration and How to prove that if $(X,A)$ has the homotopy extension property, then so does $(X\cup CA,CA)$?
Now proceed as in A question on relative homology and quotients.


*If $X$ is obtained by attaching an $n$-cell to $A$, then $(X,A)$ is cofibred pair. This is true because  $A \hookrightarrow X$ is the pushout of $S^{n-1} \hookrightarrow D^n$ which is a cofibration.
Hence the quotient maps $p : (D^n,S^{n-1}) \to (S^n,*)$ and $q : (X, X \setminus e) \to (X/ (X \setminus e),*) \approx (S^n,*)$ induce isomorphisms in homology. Now look at
$\require{AMScd}$
\begin{CD}
H_n(D^n,S^{n-1}) @>{\varphi^e_*}>> H_n(X, X \setminus e) \\
@V{\approx}VV @V{\approx}VV \\
H_n(S^n,*) @>{\approx}>> H_n(S^n,*) \end{CD}
This explains why $\varphi^e_*$ is an isomorphism and what its inverse looks like. Note the inverse is not induced by a map $(X, X \setminus e) \to (D^n,S^{n-1})$.
