# Hatcher: S^2 having a CW complex of 5/6 vertices

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 $\operatorname{im}\partial_2=\operatorname{ker}\partial_1=<e_1,\cdots, e_5>$ for the $5$ distinct edges $e_i$.

There must be a better approach.

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It sounds like if you resulted in a $K_{3,3}$ 1-skeleton then you'd have started with 18 edges? –  Aaron Mazel-Gee Apr 10 '11 at 18:31
$K_5$ and $K_{3,3}$ are the obstructions to planarity (the one skeleton of $S^2$ must be a planar graph) –  yoyo Apr 10 '11 at 19:24
using euler characteristic, you have to have 5 faces for $K_{3,3}$ and 7 faces for $K_5$. $K_{3,3}$ is 3 regular and $K_5$ is 4 regular, so you might be able to count faces that way –  yoyo Apr 10 '11 at 19:43

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)

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Oh okay. Got it. I feel a little stupid after realizing this is a tremendously easy question. I was trying to picture things intuitively, which made it hard to read the question correctly. Thanks! –  user9402 Apr 10 '11 at 20:12
This answer is false. –  user88576 Feb 6 '14 at 21:53
@amoreacceptablename Why? –  user8268 Feb 7 '14 at 16:24

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 *

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