How to prove $\Sigma X \cong S^{1}\wedge X$

Question

$\Sigma X$ is the reduced suspension of X, where X is a space based. if anyone has a hint or suggestion of how I can show this result, I would greatly appreciate it

Answer 1

Hint: consider the map $f:S^1 \times X \to \Sigma X$ given by $f([t],x) = [(t,x)]$, where $S^1 = I / \partial I$, $\Sigma X = (I \times X)/((\partial I \times X) \cup (I \times \{x_0\}))$, and $[\cdot]$ denotes equivalence classes. This induces a map $S^1 \wedge X \to \Sigma X$, and this map has an inverse induced by $g:I \times X \to S^1 \wedge X$, $g(t,x) = [([t],x)]$. I'll leave it to you to check that a) $f$ is well-defined and b) $f,g$ do descend to their respective quotients.