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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.

jacopoburelli
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  • In the unreduced suspension you do not identify $X \times \partial I$ to a single point. – Paul Frost Jul 18 '21 at 22:18
  • @PaulFrost Is the edit correct? I thought I identify $(X \times \left\lbrace 0 \right\rbrace \cup X \times \left\lbrace 1 \right\rbrace)$, but according to the following seems a different costruction respect to my notes https://ncatlab.org/nlab/show/reduced+suspension – jacopoburelli Jul 19 '21 at 06:43
  • Also the nLlab-article uses a wrong definition f the unreduced suspension. See https://math.stackexchange.com/q/959183 and https://en.wikipedia.org/wiki/Suspension_(topology). – Paul Frost Jul 19 '21 at 09:21
  • @PaulFrost Thanks, I hope I clarified the content of the question know. – jacopoburelli Jul 19 '21 at 09:36
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    Hatcher proves (Prop 0.17) that if a pair $(Y,A)$ satisfies the homotopy extension property and if $A$ is contractible, then $Y \to Y/A$ is a homotopy equivalence. Apply when $Y$ is the unreduced suspension and $A={x_0} \times I$. – John Palmieri Jul 19 '21 at 17:45
  • @JohnPalmieri with my notation $Y$ is the reduced suspension right? – jacopoburelli Jul 19 '21 at 18:05
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    I guess? That's completely backwards from the standard definitions, but I guess. (The reduced version is a quotient of the unreduced version, with the standard definitions.) – John Palmieri Jul 19 '21 at 18:33
  • $SX$ is the reduced suspension and $\Sigma X$ the unreduced suspension. Your first sentence suggests something else. The essence is that in the unreduced suspension both $X \times 0$ and $X \times 1$ are identifed to (distinct!) points. – Paul Frost Jul 19 '21 at 22:56
  • Right, so apply my comment with $Y=\Sigma X$. – John Palmieri Jul 19 '21 at 23:32
  • @JohnPalmieri The definition of Hatcher of $HEP$ is different from mine, check the edited post. – jacopoburelli Aug 03 '21 at 13:26
  • @PaulFrost Are the two propositions enough in order to prove the $SX \sim \Sigma X$? – jacopoburelli Aug 03 '21 at 13:26
  • What is the relevance of the different versions of HEP? You should be able to verify that the pair $(\Sigma X, {x_0} \times I)$ satisfies Hatcher's version, and then use Hatcher's result + his Prop 0.17 to obtain what you want. – John Palmieri Aug 03 '21 at 15:26
  • @JohnPalmieri It is relevant to me, since I have to stick with a program of the course, your manner is not a problem, just I'd like to understand the fact with the knowledge given to me – jacopoburelli Aug 03 '21 at 15:32
  • If you want to know what your instructor intended, ask them. – John Palmieri Aug 03 '21 at 16:59
  • The two propositions are essentially enough (if you add Hatcher's Proposition 0.17). Note that your definition of HEP contains Hatcher's as special case. Hatcher considers pairs $(X,.A)$; now apply your definition to the inclusion $i : A \to X$ and replace your $f$ by $f' : X \times {0} \to Y, f'(x,0) = f(x)$. Moreover, your definition is not really more general; it is easy to see that any $i$ having the HEP in your sense is an embedding. See Hatcher Proposition 4H.1. – Paul Frost Aug 03 '21 at 17:16
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    Thus it will suffice to prove that if the inclusion of $x_0$ into $X$ has the HEP, then the inclusion of ${x_0} \times I$ into $\Sigma X$ has the HEP. – Paul Frost Aug 03 '21 at 17:16
  • @PaulFrost I see, maybe it easier the way John suggested, noticing that if $X$ is a CW complex then $\Sigma X$ is a CW complex and now we can easily apply prop $0.17$ of Hatcher – jacopoburelli Aug 04 '21 at 07:24

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