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Of course, the answer to this question depends on what we mean by suspension. If we work with based spaces and take the reduced suspension, the answer seems to be NO:

Take $X = \mathbb N$ (a discrete countably infinite space) with basepoint $0 \in \mathbb N$; and $Y = \{1/n \ | n \in \mathbb N\} \cup \{0\}$ with basepoint $0$. Then the obvious bijection $X \rightarrow Y$ is a weak equivalence, but ($S$ is the reduced suspension) $S X$ is a wedge of circles, whereas $S Y$ are the Hawaiian earrings.

So my question is the following:

Is the statement true if we take the unreduced suspension instead?

Remark: If the spaces are connected, the statement seems to be true, no matter which suspension we take: This follows from the fact that the suspension of the given map is a homology equivalence between simply connected spaces, hence the Hurewicz theorem applies to conclude that its cofiber is contractible.

Improve this question
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    $\begingroup$ See Qiaochu Yuan's comment here: mathoverflow.net/questions/148963/… . Tyler Lawson says in an answer to that question that "[t]he unreduced suspension is more homotopically well-behaved, and in particular preserves weak equivalence because it only collapses along cofibrations". Not sure if this is what you wanted. $\endgroup$
    – user62675
    May 13, 2015 at 1:35
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    $\begingroup$ You can view the unreduced suspension $\Sigma X$ as the homotopy pushout of $\ast \leftarrow X \rightarrow \ast$ - you can prove this and the statement you are interested in by the Theorem on p.80 in May's concise course on topology. $\endgroup$
    – Lennart Meier
    May 13, 2015 at 2:12

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