Suspension: if $X$ is $(n-1)$-connected CW, is $SX$ $n$-connected?

Question

If $X$ is $(n-1)$-connected CW complex, is that true that $SX$ is $n$-connected?

I'm trying to understand Freudenthal Suspension Theorem on Hatcher. We define the suspension map:

$\pi_i(X)\simeq \pi_{i+1}(C_+,X) \to \pi_{i+1}(SX,C_-X) \simeq \pi_{i+1}(SX)$

($SX$ is the suspension of $X$, $C_-X$ is the lower cone) where the middle map is given by the inclusion and the isomorphisms are given by the exact sequence of pairs. Using the long exact sequence for the pair $(C_+,X)$ we also see that it is $n$-connected, given that $X$ is $(n-1)$-connected. Using the long exact sequence for the pair $(SX,C_-X)$ we only get $\pi_n(SX,C_-X) \simeq \pi_n(SX)$. Thus if $\pi_n(SX)=0$ given $\pi_{n-1}(X)=0$ we can finally apply the Excision Theorem and conclude.

Thanks!

Answer 1

You can remark that the suspension of a $n$-sphere is a $n+1$-sphere, if $X$ is $n-1$-connected, you have a cellular approximation $X'$ of $X$ with trivial $n-1$-skeleton. The previous remark implies that the $n$-skeleton of $X'$ is trivial since you can build this suspension cell by cell, since the suspension of $X'$ is also a cellular approximation of $X$ you obtain the result.

Answer 2

Here is a different method which works when $n \ge 1$, i.e. $X$ is at least path-connected (it's obvious that $SX$ is always path-connected so that case $n=0$ is not very hard).

• By the Hurewicz theorem, if $X$ is $(n-1)$-connected, then $\tilde{H}_i(X) = 0$ for $0 \le i \le n-1$.
• It's also well-known that $\tilde{H}_i(SX) = H_{i-1}(X)$ for $i \ge 1$ (use Mayer–Vietoris for example), and $\tilde{H}_0(SX) = 0$.
• Finally, since $X$ is path-connected, by the Seifert–van Kampen theorem, $SX$ is simply connected (write $SX = C^+X \cup_X C^-X$, both cones are contractible and their intersection $X$ is path-connected so you can apply the theorem). This implies that $\pi_1(SX) = 0$.
• Applying the Hurewicz theorem again, you find that $\pi_i(SX) = H_i(SX) = 0$ for $2 \le i \le n$.