Why is the (reduced) suspension of a genus $g$ surface homotopy equivalent to $(\vee_{2g}S^2)\vee S^3$? Why is the (reduced) suspension of a genus $g$ surface $X_g$ homotopy equivalent to $(\bigvee_{2g} S^2)\vee S^3$?
An attempt:
We can construct $X_g$ by attaching an $D^2$ to $\bigvee_{2g} S^1$ along the `obvious' map given by the relation in the fundamental group.  So we get a cofibration $S^1\rightarrow \bigvee_{2g} S^1 \rightarrow X_g$.  In particular we have a cofibration sequence,
$S^1\rightarrow \bigvee_{2g} S^1 \rightarrow X_g \rightarrow S^2 \rightarrow \bigvee_{2g} S^2 \rightarrow \Sigma X_g \rightarrow S^3\rightarrow \bigvee_{2g} S^3 \rightarrow \Sigma^2 X_g \rightarrow \cdots$
I'm not sure how to proceed from here.
 A: $\bigvee_{2g}S^2\to \Sigma X_g\to S^3$  is a cofibration sequence.
Moreover, $H_2(\Sigma X_g) = H_1(X_g) = \mathbb Z^{2g}$ and $H_3(\Sigma X_g)= H_2( X_g) = \mathbb Z$,  and the higher $H_n$'s are zero
$\Sigma X_g$ is simply connected, and so is the space you want to compare it to, so it suffices to find a map which induces an isomorphism in homology. 
It's clear that the map $\bigvee_{2g}S^2\to \Sigma X_g$ already induces an isomorphism on $H_2$, so it suffices to find a map $S^3\to \Sigma X_g$ which induces an isomorphism on $H_3$. This map will not come from a map $S^2\to X_g$ by suspension, because $X_g$ has no higher $\pi_i$'s (if I remember correctly, orientable surfaces are always $K(\pi,1)$'s)
So for this, we may want to use the better version of Hurewicz, which not only says that $\pi_2\to H_2$ is an isomorphism (because $\Sigma X_g$ is simply-connected), but also that $\pi_3\to H_3$ is an epimorphism (see here, at "absolute version")
This in particular means that there is $S^3\to \Sigma X_g$ such that $H_3(S^3)\to H_3(\Sigma X_g)$ sends $1$ to $1$, in particular it's an isomorphism on $H_3$, and we get what we wanted
