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

Proof: Take a look at this suspension drawing, and use the Seifert-van Kampen theorem. Particularly, look at Wikipedia's computation of $\pi_1(S^2)$.

Can someone please explain this in more detail? We have proven $\pi_1(S^2) = 0$, but I don't see what open covers are chosen to apply van Kampen.

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    $\begingroup$ Use the Seifert-van Kampen theorem on $X\times [0,1)$ and $X\times (0,1]$. Then it follows that the generators of the fundamental groups are trivial. In particular, a loop in $X\times (0,1]$ can always be deformed to the end point $X\times \{1\}$ which has now been collapsed to a point. $\endgroup$
    – Shrugs
    Commented May 24, 2023 at 23:04
  • $\begingroup$ Sorry, I still don't get it @AHappyMathematician . We have $X_1=X\times [0,1), X_2=X\times (0,1], X_0=X\times (0,1)$. Our version of van Kampen simply states that $\pi_1(X)$ is a pushout of $\pi_1(X_0), \pi_1(X_1), \pi_1(X_2)$. But how can I calculate these fundamental groups...? $\endgroup$ Commented May 25, 2023 at 0:01
  • $\begingroup$ $X_1$ is the quotient of $X \times [0,1)$ where you collapse the "$0$" end to a point: it's the cone on $X$. Same for $X_2$. $\endgroup$ Commented May 25, 2023 at 4:30
  • $\begingroup$ @JohnPalmieri but how can we show that this is contractible? $\endgroup$ Commented May 25, 2023 at 10:16

2 Answers 2


The key observation is that the images of $X \times [0, 1)$ and $X \times (0, 1]$ in the suspension are contractible, eg for $[0,1)$: $$([x,t], s) \mapsto [x, st]$$ provides a contraction. In particular they are simply connected. Further, the image of $X \times (0, 1)$ is path-connected. So writing $\Sigma X$ as the union of the images of $X \times [0, 1)$ and $X \times (0, 1]$, and applying Seifert-van Kampen, you get that $\pi_1(\Sigma X)$ is the free product of two copies of the trivial group, amalgamated along something ($\pi_1(X)$, but that is irrelevant). It should be easy to see that this group is trivial.

  • $\begingroup$ Thank you. I see all the other things you say, but not why the images of $X\times (0,1]$ and $X\times [0,1)$ are contractible. I tried to proove that they are homeomorphic to some $\mathbb R^n$ but I cannot find a homeomorphism. I think my problem is that I just don't see why changing $[0,1]$ to $(0,1]$, so removing a point, makes this big difference. I also tried to adjust the stereographic projection for this, but it didn't work. $\endgroup$ Commented May 25, 2023 at 10:15
  • $\begingroup$ @mathematics-and-caffeine added an explicit contraction. The issue is the same as $S^2$ vs $S^2 - \{\text{north pole}\}$. Every point other than the poles is on a unique longitude, so can be contracted along that line to the south pole, but this map cannot be continuous extended to the north pole. $\endgroup$
    – ronno
    Commented May 25, 2023 at 10:37

Note that $S^2$ is the suspension of the path-connected space $S^1$. Try to study your proof that $S^2$ is simply connected, which presumably used the Seifert-van Kampen theorem! How could you generalize that proof to apply to the suspension of a different path-connected space?

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    $\begingroup$ I vaguely recall there's a proof that $S^2$ is simply connected by first showing that any curve is homotopic to a non space-filling one... $\endgroup$
    – Zhen Lin
    Commented May 25, 2023 at 3:09

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