# Is the suspension of a weak equivalence again a weak equivalence?

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.

• 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. – user62675 May 13 '15 at 1:35
• 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. – Lennart Meier May 13 '15 at 2:12