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Recall that the suspension of a topological space $X$ is the space $SX$ resulting by identifying $X\times\{0\}$ and $X\times\{1\}$ to single points of the "cylinder" $X\times[0,1]$. Now let $X_m$ be a discrete space consisting of $m$ points, and denote by $S^nX$ the $n$-fold suspension, i.e., $S^nX=SS^{n-1}X$.

What is the fundamental group of $S^nX_m$ for $n,m\geq1$?

Here is my attempt. The case $m=2$ is easy since $SX_2$ is a circle, so the fundamental group is $\mathbb Z$, and $S^2X_2$ is the sphere, whose fundamental group is trivial. Further, the suspension of the $n$-sphere is the $(n+1)$-sphere so $\pi_1(S^nX_2)$ is trivial for $n\geq2$.

Now consider $m>2$. Then $SX_m$ is homotopy equivalent to an "$(m-1)$-fold eight"

$$\underbrace{\bigcirc\hspace{-.1cm}\!\bigcirc\hspace{-.1cm}\!\bigcirc\cdots\bigcirc}_{\text{$m-1$ circles}}$$

so the fundamental group is $F_{m-1}$, the free group on $m-1$ generators. If I am not mistaken two homotopy equivalent spaces have homotopy equivalent suspensions, so $S^2X_m$ is homotopy equivalent to the wedge product of $m-1$ spheres, and consequently $\pi_1(S^2X_m)$ is trivial. I am having trouble, however, to visualise the space $S^nX_m$ for $n>2$. How can $\pi_1(S^nX_m)$ be computed?

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  • $\begingroup$ I had to delete my answer, because it's actually reduced (or pointed) suspension that preserves wedges, and though for spheres reduced and unreduced suspension agree, I don't see an easy argument as to why it should be the same for wedges of spheres (though it seems to be the case in low dimension so it's probably true; but that's really not a good argument) $\endgroup$ – Max Jan 11 at 13:44
  • $\begingroup$ Ok, but thanks anyway! $\endgroup$ – iqcd Jan 11 at 14:17
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  1. $m = 1$: All suspensions of $X_1$ are contractible.

  2. $m > 1$:

2.1. $n = 1$: As you stated correctly, $\pi_1(SX_m) = F_{m-1}$.

2.2. $n > 1$: Let us prove the following theorem.

If $Y$ is path connected, then $SY$ is simply connected.

This applies to $Y = S^{n-1}X_m$.

Let $p : Y \times I \to SY$ denote the quotient map. Since $Y$ is path connected, so are $Y \times I$ and $SY$. As a basepoint for $SY$ choose $z_0 = p(y_0,1/2)$, where $y_0 \in Y$ is an arbitrary point. The sets $U = p(Y \times (0,1])$ and $V = p(Y \times [0,1))$ are open in $SY$, and we have $z_0 \in U \cap V = p(Y \times (0,1)$ which is pathwise connected. Since $U$ and $V$ are contractible, we get $\pi_1(U,z_0) = \pi_1(V,z_0) = 0$. Now apply the Seifert-van Kampen-theorem to see that $\pi_1(SY,z_0) = 0$.

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  • $\begingroup$ That's a very useful result, thanks! $\endgroup$ – iqcd Jan 11 at 14:29

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