Here is a part of my notes in the algebraic topology class:
Let $A$ be a space, $k\geqslant 2$. $f:S^{k-1}\rightarrow A$ be a map. Then consider the adjunction space $$ A\cup_f e_k=A\cup_f CS^{k-1}$$ It is easy to check $$\iota: (CS^{k-1},S^{k-1})\rightarrow (A\cup_f e^k,A)$$ induces an excision isomorphism $$\iota_*: H_q(CS^{k-1},S^{k-1})\overset{\cong}{\rightarrow} H_q(A\cup_f e^k,A)$$
I am confused with it. I have no idea how to use the Excision theorem (maybe plus an argument of relative deformation retract), because in the excision theorem one need an open set $U\subseteq A$ with $\bar{U}\subseteq \mathrm{Int}(A)$, but how to construct such $U$?
Any hints or solutions are welcomed. Thanks!
Suppose the cone point is given by $S^{k-1}\times 0$ and $(x,1)\sim f(x)$ in $Cf = A\cup_f CS^{k-1}$. Let $U=A\cup S^{k-1}\times (\frac 12,1]$ and $j\colon CS^{k-1}\to CS^{k-1}, j(x,t) = (x,\frac12 t)$. Then the composite $$H_q(CS^{k-1},S^{k-1})\xrightarrow{j_\ast} H_q(CS^{k-1},S^{k-1}) \xrightarrow{i_\ast} H_q(Cf,A)\to H_q\left(Cf,A\cup S^{k-1}\times\left[\frac 12,1\right]\right)$$ is an isomorphism by excision (excise $U$ from the pair $\left(Cf,A\cup S^{k-1}\times\left[\frac 12,1\right]\right)$). Here the right hand arrow is given by the inclusion and is a homotopy equivalence and $j$ is homotopic to the identity. It follows that $i_\ast$ is an isomorphism as well.
You can also use the following result: If $(X,A)$ is a good pair or the inclusion $A\subset X$ is a cofibration (and for any good pair the inclusion is a cofibration), then the quotient map $X\to X/A$ induces an isomorphism $H_q(X,A)\cong H_q(X/A,\ast)$. For good pairs this is proven in Hatcher 2.22 and for cofibrations in tom Dieck 10.4.5. It is easily seen that $(CX,X)$ and $(Cf,Y)$ are always good pairs, where $Cf$ is the mapping cone of $f\colon X\to Y$. In the case at hand if $f\colon S^{k-1}\to A$ is a map we use that $f$ induces a homeomorphism $CS^{k-1}/S^{k-1}\cong Cf/A$, hence also an isomorphism $H_q(CS^{k-1}/S^{k-1},\ast)\cong H_q(Cf/A,\ast)$.