Hatcher 0.25: Wedge sum of suspensions of components of $X$ is suspension of $X$

Question

I can't figure out how to do problem 0.25 from Hatcher's Algebraic Topology. It says

If $X$ is a CW complex with components $X_{\alpha}$ , show that the suspension $SX$ is homotopy equivalent to $Y \bigvee_{\alpha} SX_{\alpha}$ for some graph $Y$.

I tried to use induction and reduced it to the case of $X$ with $2$ connected components (and I suspect $Y$ should be a $S^1$ in that case), but couldn't really get anywhere from there.

Answer 1

Here $\Sigma$ is the reduced suspension, which is homotopically equivalent to the suspension $S.$ $Y$ turns out to be the green circle, as you suspected. The case of more $X_i$ is not different. (You can't use induction for $\{ \alpha\}$ is not given to be finite.)