$1$-skeleton of $S^2$ can't be graphs $K_5$ or bipartite graph $K_{3,3}$ 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.
 A: For $K_5$ (I leave $K_{3,3}$ out, as it's a homework): it has $10$ edges, so your polygons have together $20$ edges (every edge of $K_5$ is supposed to come from gluing of two polygons along an edge). Euler characteristic gives you that the number of polygons is $7$. If $n_1,\dots,n_7$ are the numbers of edges of the polygons, since $n_1+\dots+n_7=20$, we can't have $n_i\geq 3$ for every $i$. An $i$ where $n_i=2$ (it's not really a polygon, but whatever) would mean that two of the vertices of $K_5$ are connected by two edges - and that's not the case.
($K_{3,3}$ is very similar - but there is a little change in the reasoning)
A: 3 line proof
Suppose it is possible.$v=5,e=10$. By euler's formula (i.e $v-e+f=2$)$f=7$. Each face has $3$ (atleast) edges. For $7$ faces we need atleast $7\times3$ edges. But each edge bounds exactly $2$ faces so we need atleast '$10.5$' edges. So this means $e\geq11$. Contradiction!!!
**I'm not saying this is different from the proof above. By the way, see K3,3  for the other case **
A: The first $1$-skeleton has $5$ vertices and $10$ edges. Using the canonical $CW$ structure of $S^2$ together with the fact that the Euler characteristic is invariant under $CW$ decomposition we have, $$ 5 - 10  + F =2 $$
$$ F =7 $$ 
Each face has at least $3$ edges so we count at least $3F$ edges. Every edge is counted twice (it is on the border of two faces). Thus we have, $3F=2E$, and so $E \ge \frac32 F$. However, we see that this does not hold for the values calculated above. Thus, such a $1$-skeleton for a $CW$ structure of $S^2$ cannot exist
