Hatcher Theorem 2.13 - is the subspace $X$ of its cone $CX$ a deformation retract of some neighborhood in $CX$?

Question

Hatcher's Theorem 2.13 says

If $X$ is a space and $A$ is a nonempty closed subspace that is a deformation retract of some neighborhood in $X$, then there is an exact sequence $$\cdots \to \widetilde{H}_n(A) \xrightarrow{i_*} \widetilde{H}_n(X) \xrightarrow{j_*} \widetilde{H}_n(X/A) \xrightarrow{\partial} \widetilde{H}_{n-1}(A) \to \cdots \to \widetilde{H}_0(X/A) \to 0$$where $i$ is the inclusion $A\hookrightarrow X$ and $j$ is the quotient map $X\to X/A$.

I am currently working on a problem where i would like to apply this exact sequence to the pair $(CX,X)$. I know that $CX$ is contractible, but i couldn't figure out whether this implies that $X$ is a deformation retract of some neighborhood in $CX$.

Can someone help me on this?

Answer 1

The cone $CX$ is $X \times [0,1]$ modulo identifying the entire end $X \times \{1\}$ to a single point. Therefore a neighborhood of $X$ in $CX$ is $X \times [0,1/2)$, which deformation retracts onto $X$ via $(x,t,u) \mapsto (x,(1-u)t)$ for $x \in X$, $0 \leq t < 1/2$, and $0 \leq u \leq 1$.

Answer 2

Consider the cone $CX$. It is got by taking $X\times[0,1]$ and identifying $X\times\{1\}$ to a point $P$. If we remove $P$ from $CX$ we get the open subset $X\times[0,1)$ which deformation retracts to $X\times\{0\}$, the standard homoeomorph of $X$ embedded in $CX$.

Answer 3

$CX$ is the quotient of $[0,1] \times X$ by identifying $\{0\} \times X$ to a point. Let's use $q : [0,1] \times X \to CX$ as the quotient map.

Using the embedding of $X$ in $[0,1] \times X$ is $\{1\} \times X$, the standard embedding of $X$ in $CX$ is $q(\{1\} \times X)$.

Note that $(0,1] \times X \subset CX$ is a saturated open subset of $\{1\} \times X$, and $q((0,1] \times X)$ is therefore an open subset of $q(\{1\} \times X)$.

Since $(0,1] \times X$ clearly deformation retracts to $\{1\} \times X$, it follows that $q((0,1] \times X)$ deformation retracts to $q(\{1\} \times X)$.