Let $S$ be a closed orientable surface. Suppose $S$ has a triangulation with the property that the degree of each vertex is $5$. Show that $S$ must be a the sphere. What follows in my attempt.
Let the triangulaton, $\Delta$, of $S$ be given with the above mentioned properties. We can compute the Euler characteristic of $S$ from $\Delta$ with the equation $\chi(S) = e_0 -e_1 +e_2$, where of course, $e_i$ is the number of simplices of dimension $i$ in $\Delta$. As the degree of each vertex is $5$ we attempt to count the edges by the vertices, however for each edge there are two vertices so we arrive at the conclusion that $e_1 = \frac{5e_0}{2}$. Likewise for the faces, each vertex meets five faces, but the faces meet three vertices yielding $e_2=\frac{5e_0}{3}$. Substituting in the equation we have: $$\chi(S) = e_0-e_1+e_2=e_0 - \frac{5e_0}{2}+\frac{5e_0}{3}=\frac{e_0}{6} >0$$ Then as $S$ is closed orientable and $\chi(S) >0$ $S$ must be the sphere.
I suppose as for an answer, if incorrect I would like to see a correct proof, and if correct affirmation and any other proof one can come up with that uses different techniques.