# Inclusion in cone is homotopy equivalence

Suppose $X$ is a topological space and $x \in X$. Let $CX$ be the cone of $X$, i.e. the quotient space $X \times [0,1]/{\sim}$ where $(x,1) \sim (y,1)$ for alle $x,y \in X$. I would like to show that the inclusion map $i: \{x\} \rightarrow CX$, where $x$ is mapped to the equivalence class of $(x,0)$, is a homotopy equivalence.

To show that, I have to show that there exists a continuous map $j: CX \rightarrow \{x\}$ such that $i \circ j$ is homotopic to $id_{CX}$ and $j \circ i$ is homotopic to $id_{\{x\}}$. The latter part is trivial. Now, I am stuck trying to proove the first part.

I could show that with $F: CX \times [0,1] \rightarrow CX, F([y,t],s) = [y, (1-s)t +s]$ the identity map $id_{CX}$ is homotopic to a constant map and thus the cone is contractible. Does that help in this case? Could someone give me a hint how to solve this?

Cheers,
quizzle

• I think it is correct, the homotopy you defined is a strong deformation retraction to the point $(*,1)$. The homotopy between the identity on $CX$ and $i\circ j$ can be defined to be a contatenation of the homoropy that retracts everything to $(*,1)$ and a homotopy that slides this point back to $x$. – Peter Franek Jun 23 '15 at 16:42
• But how can I define the second homotopy that slides it back to x? – Quizzle123 Jun 23 '15 at 17:20
• $\tilde{F}(s,y)=F(2s,y)$ where $F$ is "your" homotopy for $s\leq \frac{1}{2}$ and $\tilde{F}(s,y)=\gamma(s)$ (independent of $y$) where $\gamma: [\frac{1}{2},1]\to CX$ is a path starting in $(*,1)$ and ending in $x$, for $s\in [\frac{1}{2},1].$ – Peter Franek Jun 23 '15 at 17:32
• Ah, so because the cone is path-connected, the homotopy is obvious. Thank you! – Quizzle123 Jun 24 '15 at 12:06