Homotopy between $X\setminus\{x\}$ and $A$

Question

Let $X$ be a space obtained from $A$ by attaching a n-cell $e$. If $x\in e$ how to prove that the inclusion map $i_{A\to X\setminus\{x\}}: A\to X\setminus\{x\}$ is a homotopy equivalence?

Answer 1

First note that it does not matter if we remove the point $x$ before or after the gluing because the restriction of the quotient map $q$ to the complement of $x$ is again a quotient map.

Then note that a pushout square (A,X, and B are arbitrary spaces)
$\ $ gives rise to a pushout square$\ $

because the quotient map $q:X\sqcup B\to X\cup_f B$ induces a quotient map $q\times 1:X\times I\sqcup B\times I\to(X\cup_f B)\times I$.
This means that a pair of homotopies $F_t:X→Y$, $G_t:B→Y$, such that $F_ti=G_tf$ for all $t\in I$ induces a homotopy $H_t:X∪_fB→Y$

Now, if $A\subseteq D\subseteq X$ and $D$ is a deformation retract of $X$, then $D\cup_f B$ is a deformation retract of $X\cup_f B$. For if $R_t:X→X$ is the deforming homotopy, then the homotopies $\bar fR_t:X→X∪_fB$ and $\bar i:B\hookrightarrow X∪_fB$ induce a homotopy $H_t:X∪_fB→X∪_fB$ between the identity and the retraction onto $D∪_fB$

In your case, set $B=A,\ X=\bar e-\{x\}=D^n-\{x\},\ A=S^{n-1},$ and $f$ the gluing map. Then the deformation retraction of $D^n-\{x\}$ onto $S^{n-1}$ induces by the above result a deformation retraction onto $S^{n-1}∪_fA=A$.

Answer 2

Hint: Let $B^n$ be the open $n$-ball in $\mathbb R^n$. Can you show that $\overline{B^n}\setminus\{0\}$ and $\partial B^n = S^{n-1}$ are homotopy equivalent? Does it matter which $x\in B^n$ we choose? Can you extend this homotopy equivalence of an $n$-cell with a point missing and its boundary to the homotopy equivalence you are looking for?