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As pointed out by David White in when mapping cone is contractible there exists an acyclic CW-complex $X$ which is not contractible but whose suspension is contractible. Namely, let $a$ and $b$ be the two loops in $X=S^1\vee S^1$ and glue in two $2$-cells along the words $a^5b^{−3}$ and $b^3(ab)^{−2}$.

My question is: Is this possible if the suspension, $\Sigma X$, of $X$ is homotopy equivalent to the unit interval $I=[0,1]$ rel ends?

More precisely: if $f:\Sigma X \to I$ is a homotopy equivalence with inverse $g$ and homotopies $h_1:fg\simeq id_I$ and $h_2:gf\simeq id_{\Sigma X}$ such that $g(i) = [X\times i]$ for $i=0,1$, $h_1$ is a homotopy rel {$0,1$} and $h_2$ is a homotopy rel {$[X\times 0]$, $[X\times 1]$} then does $X$ have to be contractible?

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up vote 8 down vote accepted

$\newcommand{\set}[1]{\lbrace #1 \rbrace}$I will assume that the notation $\Sigma X$ in the question denotes the unreduced suspension of the space $X$.

Quick answer: The notion of homotopy equivalence $\Sigma X\to I$ rel ends described in the question is actually equivalent to the contractibility of $\Sigma X$, since the inclusions of the "ends" $\set{0,1}$ into $\Sigma X$ and $I$ are both cofibrations. Thus the cited example of David White is also a counter-example for this question.

Slow answer: There is the canonical map $i:\set{0,1}\to \Sigma X$ given by $i(t)=[x,t]$ for any $t\in\set{0,1}$ and $x\in X$. Moreover, we have the inclusion $j:\set{0,1}\hookrightarrow I$. The question describes the notion of a homotopy equivalence under $\set{0,1}$ between $I$ and $\Sigma X$ (relative to the preceding maps $i$ and $j$), sometimes also called a cofibre homotopy equivalence. It is asked if the existence of such a cofibre homotopy equivalence implies that $X$ is contractible.

The answer is no, as follows from:

  1. There exists a space $X$ which is not contractible, yet $\Sigma X$ is contractible. Any non-contractible, acyclic CW-complex can be used here, such as the example given by David White in the link provided in the question.

  2. If $\Sigma X$ is contractible, then there is a cofibre homotopy equivalence $\Sigma X \to I$ of spaces under $\set{0,1}$.

To prove statement (2), it suffices to observe that the maps $i:\set{0,1}\to \Sigma X$ and $j:\set{0,1}\to I$ are both cofibrations: for example, the NDR condition is very simple to check in these cases (see, for example, section 6.4 of Peter May's book "A concise course in algebraic topology" for a description of NDR pairs). Then it is a well-known fact that any homotopy equivalence $f:\Sigma X\to I$ such that $f\circ i = j$ is actually a cofibre homotopy equivalence (see section 6.5 of the aforementioned book "A concise course in algebraic topology"). In conclusion, if $\Sigma X$ is contractible, then the projection map $f:\Sigma X \to I$ — defined by $f([x,t])=t$ for $x\in X$, $t\in I$ — is a homotopy equivalence and, by the above result, is therefore a cofibre homotopy equivalence of spaces under $\set{0,1}$.

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Thanks Ricardo! Yes I did mean unreduced suspension with $\Sigma X$. – Spiros Adams-Florou May 16 '13 at 12:49
@Spiros: You're welcome. – Ricardo Andrade May 16 '13 at 15:48

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