# Fundamental group of the $m$-fold suspension of a finite discrete space

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?

• 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) – Max Jan 11 at 13:44
• Ok, but thanks anyway! – user246336 Jan 11 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 $$\pi_1(SY,z_0) = 0$$.

• That's a very useful result, thanks! – user246336 Jan 11 at 14:29