If a topological space $X$ is not path-connected, then the suspension is not simply connected.

Question

A while ago, on a homework sheet we had a problem about the suspension $\Sigma X$ of a topological space $X$, where $\Sigma X = (X \times I)/\sim$ with $$ (x,s) \sim (y,t) \mspace{24mu} \colon\iff \mspace{24mu} s=t \;\land \; (x=y\;\lor\;s\in\{0,1\}). $$ Next to proving $\Sigma X$ is path-connected, which I had no problem with, we were asked to show that if $X$ isn't path-connected, then $\Sigma X$ isn't simply connected.

I have read online that this can be quickly done using Van Kampen's theorem, but at this time we just started preparing for this theorem, so we were asked for a solution without it.

Now for my idea: Take $x,y \in X$ such that there is no path from $x$ to $y$. Consider the paths $\alpha$ and $\beta$ in $\Sigma X$, where $\alpha$ linearly (in the second component) connects $[(x,\frac{1}{2})]$ with $[(x,1)] = [(y,1)]$, and then goes to $[(y,\frac{1}{2})]$ in the same way, and where $\beta$ does the same but instead goes through $[(x,0)]$. Written out that is $$ \alpha(t) \; \colon= \begin{cases} [(x, \frac{1}{2}+t)], & \text{for } t\in [0,\frac{1}{2}]\\ [(y, \frac{1}{2}-t)], & \text{for } t\in (\frac{1}{2},1]\\ \end{cases} \mspace{48mu} \beta(t) \; \colon= \begin{cases} [(x, \frac{1}{2}-t)], & \text{for } t\in [0,\frac{1}{2}]\\ [(y, \frac{1}{2}+t)], & \text{for } t\in (\frac{1}{2},1]\\ \end{cases}$$ In a simply connected space, these would be homotopic. If such homotopy exists, my idea is to construct a path $\gamma$ in $\Sigma X$ from it, that avoids the two outer points $[(x,0)], [(x,1)]$. A such path should exist, because of the continuous homotopy. $\gamma$ could then be "projected" to $(X\times \{\frac{1}{2}\})/\sim \;\;\cong X$, which yields a path from $x$ to $y$ in $X$. This is a contradiction, so $\Sigma X$ cannot be simply connected.

My problem is the actual construction of this path $\gamma$. Nothing I tried really worked, so any help with that would be appreciated. I also wondered if there is a more elegant way of doing this, by using fundamental groups and algebraic properties, although I think this wouldn't work without Van Kampen's theorem.

Answer 1

van Kampen's theorem allows to prove that the suspension of a path-connected space is simply connected, but I do not see how it could be used to prove the converse.

Here is at least a partial result:

If $X$ is not connected, then $\Sigma X$ is not simply connected.

If $X$ is not connected, then there exists a continuous surjection $f : X \to D = \{0,1\}$. Here $D$ has the discrete topology. Choose any function $i : D \to X$ such that $f \circ i = id$. Since $D$ is discrete , $i$ is continuous.

The maps $\Sigma i : \Sigma D \to \Sigma X$ and $\Sigma f : \Sigma X \to \Sigma D$ have the property $\Sigma f \circ \Sigma i = \Sigma id = id$, thus $(\Sigma i)_* : \pi_1(\Sigma D) \to \pi_1(\Sigma X)$ is injective. Since $\Sigma D \approx S^1$, we conclude that $\pi_1(\Sigma X) \ne 0$.

Answer 2

It is correct, but I can only give a proof based on homology theory.

1. For the reduced homology groups we have $\tilde H_n(X) \approx \tilde H_{n+1}(\Sigma X)$. See Reduced homology of the suspension. Since $\tilde H_m(Y) = H_m(Y)$ for $m > 0$, we see that $\tilde H_0(X) \approx H_1(\Sigma X)$.

2. If $Y$ is path-connected, then the Hurewicz homomorphism $h : \pi_1(Y,y_0) \to H_1(Y)$ is surjective. See for example here.

3. If $X$ is not path-connected, the reduced homology group $\tilde H_0(X)$ is non-zero. In fact, the unreduced homology group $H_0(X)$ is a free abelian group with one generator for each path component of $X$. Since $H_0(X) \approx \tilde H_0(X) \oplus \mathbb Z$, the claim follows.

From 1. - 3. we conclude that $\pi_1(\Sigma X) \ne 0$.

I believe there will also be an elementary proof, but it could be very technical.