The question is from Hatcher's Algebraic Topology Problem 2.2.24.
Suppose we build $S^2$ from a finite collection of polygons by identifying edges in pairs. Show that in the resulting CW structure on $S^2$, the 1 skeleton cannot be either of the two graphs shown, with five and six vertices. (the graphs are $K_5$ and the bipartite graph $K_{3,3}$)
I'm not sure what Hatcher means by "identifying edges in pairs". I was assuming that each edge belongs in a unique pair of edges that are identified with each other, but $K_{3,3}$ has $9$ edges, so there would be one edge left out.
Anyways, assuming that my take on the question is correct, I don't really know how to proceed for $K_5$. I was thinking of finding a contradiction of $H_2(S^2)=H_0(S^2)=\mathbb{Z}$ and $H_1(S^2)=0$, but there's so much messy casework and I'm not confident that this will even work. The Euler Characteristic equation doesn't reduce the casework. All I get is that all the vertices are identified and we must have $\operatorname{im}\partial_2=\operatorname{ker}\partial_1= \langle e_1,\cdots, e_5 \rangle$ for the $5$ distinct edges $e_i$.
There must be a better approach.