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

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.

• 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. Commented May 24, 2023 at 23:04
• 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...? Commented May 25, 2023 at 0:01
• $X_1$ 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 $X_2$. Commented May 25, 2023 at 4:30
• @JohnPalmieri but how can we show that this is contractible? Commented May 25, 2023 at 10:16

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.
• 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. Commented May 25, 2023 at 10:15
• @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. 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?
• 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... Commented May 25, 2023 at 3:09