# Suspension is simply connected if $X$ is path connected

The suspension of a path connected topological space is simply connected.

I have questions about the following proof:

Note that $$\Sigma X = X \times I / ( X \times \partial I \cup \{x_0\} \times I )$$, where $$x_0$$ is assumed to be the base point of $$X$$ and $$\partial I = \{0,1\}$$.

Taking any loop $$f: [0,1] \to \Sigma X$$, one can define the following homotopy: $$H(x, t) = [(f(x), 1-t)]$$ where for any $$\alpha \in X \times I$$, $$[\alpha]$$ denotes its equivalence class in $$\Sigma X$$.

Use the universal property of quotient spaces to show that this map is continuous, and check that it is also a homotopy. This will show that every loop is homotopic to the constant loop, and thus, $$\pi_1(X) = 0$$. As for connectedness of $$\Sigma X$$, just note that $$X$$ is connected, so $$X \times I$$ is connected, and quotients of connected spaces are connected, therefore $$\Sigma X$$ is connected.

Why is this $$H$$ a homotopy between a loop $$f$$ and the constant loop? We have $$H_0(x)=[(f(x),1)]$$ which is always the same point for all $$x$$ by the definition of the suspension. So it is constant. Same for $$H_1(x)$$. However, I don't understand how this helps us. For showing simple connectedness, we must show $$\pi_1(\Sigma X,z)=1 \, \forall z\in \Sigma X$$. So take an arbitrary $$z=(x,i) \in \Sigma X$$. Then we need that a loop $$f$$ with basepoint $$z$$ is homotop to $$c_z$$ where $$c_z$$ is the constant loop on $$z$$. But $$H_0(x)$$ is not constant on $$z$$, it is constant on one of the endings of the suspension. Also, for having a homotopy between $$f$$ and $$c_z$$, we would need $$H_1(x)=f(x)$$, which also isn't the case. Both $$H_0$$ and $$H_1$$ are constant loops on the endings of the suspension.

• Honestly I don't understand the definition of $H$ at all. $f(x)$ is already an element of $\Sigma X$. Then what is $[(f(x),1-t)]$ supposed to mean? Where did you get that proof from? Commented May 24, 2023 at 21:54
• I think the standard proof would use the Seifert-Van Kampen theorem. Commented May 24, 2023 at 23:04

Another problem is that you consider the reduced suspension of a pointed sapce $$(X,x_0)$$. For the unreduced suspension there is a fairly simple proof based on the Seifert - van Kampen theorem. See my answer to Fundamental group of the $m$-fold suspension of a finite discrete space. Unfortunately this proof does not work for the reduced suspension.
If $$(X,x_0)$$ is well-pointed (which means that the inclusion $$\{ x_0\} \to X$$ is a cofibration), then it is well-known that the quotient map from the unreduced to the reduced suspension is a homotopy equivalence. Thus for a well-pointed path connected $$(X,x_0)$$ the reduced suspension is simply connected.
I do not know whether this is true also for non-well-pointed $$(X,x_0)$$. The other answer to Simply connected reduced suspension on path connected X invokes the Freudenthal suspesnsion theorem, but this only applies to pointed CW-complexes.