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.

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    $\begingroup$ 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? $\endgroup$
    – freakish
    Commented May 24, 2023 at 21:54
  • 1
    $\begingroup$ I think the standard proof would use the Seifert-Van Kampen theorem. $\endgroup$ Commented May 24, 2023 at 23:04

1 Answer 1


Its seems that you found the above "proof" in an answer to Simply connected reduced suspension on path connected X.

As you suspect, this does not work.

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.


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