Exercise 0.14

Proof. For a given triple $(v,e,f)$, we can construct the following 1-skeleton $X^1$:

Each petal is homeomorphic to $S^1$ and each segment is homeomorphic to $I=[0,1]$. This construction works since for each $n$-cell in addition to the 1 0-cell and 1 2-cell that are minimally required to construct $S^2$, each 1-cell that contributes to $e$ must be offset by either a 0-cell that contributes to $v$ or a 2-cell that contributes to $f$.

It now suffices to attach 2-cells such that the resulting space is $S^2$. We see that our 1-skeleton divides the plane up into $f$ regions, one for each interior of a petal and one encompassing the rest of the plane. Attaching a 2-cell to each petal by identifying $\partial e_i^2\simeq S_i^1$ where $S_i^1$ is one of the petals for $1\le i\le f-1$, and then attaching a 2-cell such that $\partial e_f^2\simeq X^1$, i.e., by having the boundary be identified with all of the edges in our 1-skeleton $X^1$, results in $S^2$ as desired. □