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?

  • $\begingroup$ 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 '19 at 13:44
  • $\begingroup$ Ok, but thanks anyway! $\endgroup$ – user246336 Jan 11 '19 at 14:17
  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 $\pi_1(SY,z_0) = 0$.

  • $\begingroup$ That's a very useful result, thanks! $\endgroup$ – user246336 Jan 11 '19 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.