The suspension $\Sigma X$ of a topological space $X$ is defined as the quotient space $$ \Sigma X=\dfrac {X\times \left[0,1\right]}{\sim}$$
where $(x,t)\sim (y,s)$ if and only if $s=t=0$, or $s=t=1$, or $(x,t)=(y,s)$.
Is $\Sigma X$ connected or simply connected for every space $X$? Prove it or construct a counterexample.
Intuitively, suspension would pinch the up and down sides of $X \times [0,1]$ into two points, and $[0,1]$ is connected. So I think it'd be possible that it'll be connected and simply connected. But I don't know how to prove it in mathematical language.
Best Answer
For any topological space $X$, the suspension $\Sigma X$ is path-connected and hence connected. To see this, suppose $[(x_1, a_1)], [(x_2, a_2)] \in \Sigma X$ where the square brackets denote the equivalence class of a pair from $X \times [0, 1]$ under the equivalence relation $\sim$.
The path $p_1 : [0, 1] \to \Sigma X$, $p_1(t) = [(x_1, (1-t)a_1 + t)]$ joins $p_1(0) = [(x_1, a_1)]$ and $p_1(1) = [(x_1, 1)]$.
The path $p_2 : [0, 1] \to \Sigma X$, $p_2(t) = [(x_2, 1 - t + ta_2)]$ joins $p_2(0) = [(x_2, 1)]$ and $p_2(1) = [(x_2, a_2)]$.
As $p_1(1) = [(x_1, 1)] = [(x_2, 1)] = p_2(0)$, the concatenation of the paths $p_1$ and $p_2$ gives a path $p := p_1\ast p_2$ joining $[(x_1, a_1)]$ and $[(x_2, a_2)]$.
On the other hand, $\Sigma X$ need not be simply connected. For example, if $X = S^0 = \{-1, 1\}$ (equipped with the discrete topology), then $\Sigma S^0 = S^1$ which is not simply connected. In general, $\Sigma S^n = S^{n+1}$. However, if $X$ is path-connected, then $\Sigma X$ is simply connected; this follows from the Seifert-van Kampen Theorem.