Definition: We say that a continuos map between topological spaces $i: A \longmapsto X$ has the Homotopy Extension Property (HEP) for $Y$ if given $h: A \times I \longmapsto Y$ and $f: X \longmapsto Y$ such that $f(i(a)) = h(a,0)$ $\exists H: X \times I \longmapsto Y$ such that $\begin{cases}H(x,0)=f(x)\\H(i(a),t)= h(a,t)\end{cases}$. We called $H$ the extensio of $h$ with initial condition $f$.
Define $SX = X \times I /(X \times \left\lbrace 0 \right\rbrace \cup X \times \left\lbrace 1 \right\rbrace \cup \left\lbrace x_0 \right\rbrace \times I)$ and $\Sigma X = X \times I /\sim$ the unreduced and reduced suspension of $X$ respectively, where $\sim$ is such that $(x,0) \sim (y,0)$ and $(x,1) \sim (y,1)$.
Edit : According to the Wikipedia my professor swapped the two notions, thanks to Paul Frost for pointing that out.
I'd like to prove that the two spaces are homotopically equivalent in "reasonable spaces", such as $CW$ complex.
I think I read somewhere on MSE that a sufficient condition is to require the inclusion $* \hookrightarrow X$ to be a closed fibration. The latter should be true in $CW$ complex where we know that the inclusion of the $n-$ skeleton(even a subcomplex) is a cofibration. But how to proceed from here?
My knowledge on cofibration besides the definitions are :
Proposition $1$: Are equivalent:
- $i: A \longmapsto X$ is a cofibration.
- $i : A \longmapsto X$ has the homotopy extension property for the mapping cylinder $M_i$.
- $s:M_i \longmapsto X \times I$ has a retraction.
Proposition $2$: Let $i: A \longmapsto X$ be a closed inclusion. We have that $i$ is a cofibration $\iff X \times \left\lbrace 0 \right\rbrace \cup A \times I$ is a retract of $X \times I$.
Are those two facts enough in order to prove that $SX \simeq \Sigma X?$ Any help,hint or solution will be appreciated.