# Hatcher 0.14

Hatcher, Algebraic Topology, Chapter 0

14. Given positive integers $v$, $e$, and $f$ satisfying $v-e+f=2$, construct a cell structure on $S^2$ having $v$ 0-cells, $e$ 1-cells, and $f$ 2-cells.

Proof. To begin, consider the triple $(v,e,f)$ of integers satisfying $v-e+f=2$ and note1 that the following system of equations holds:

$\begin{array}{rcl}v&=&1+(v-1)+0 \\ e&=&0+(1-v)+(1-f) \\ f&=&1+0+(f-1)\end{array}$.

In particular, introducing the parameters $m=v-1$ and $n=f-1$, it follows that the triple $(v,e,f)$ can be written as $\{1,m,n\}$-linear combinations of the triples $(1,0,1),(1,-1,0),(0,-1,1)$ as follows:

$(v,e,f) = (1,0,1) + m(1,-1,0) + n(0,-1,1).\,\,\,\,\,\,\,\,\,\,(1)$

From the beginning, the problem could have been reduced to the related problem of describing in detail how to modify the base CW-structure of $S^2$ to accommodate for the addition of 0-, 1-, or 2-cells; given this new parameterization, the problem will be solved by showing how to accommodate unit increases of $m$ and $n$ independently given a base CW-structure for $S^2$ and by reverting to induction for the general case.

Consider two cases.

Case 1: Increasing $m=v-1$ by 1 while leaving $n$ fixed.
Notice that in equation $(1)$ above, $m$ is being applied to the triple $(1,-1,0)$, whereby the increase $m\mapsto m+1$ equates to adding a vertex and an edge. For this construction, consider the sphere $S^2$ with the cell structure corresponding to one 0-cell $e^0_1=x_0$ and one 2-cell $e^2_1$, whereby $S^2$ is formed via $S^2=e^0_1\sqcup_\alpha D^2$ and the attachment map $\varphi_\alpha:\partial D^2\to x_0$. The goal is to “start the CW construction over” now with two 0-cells, one 1-cell, and the above 2-cell and to show that $S^2$ can have this CW-structure.

Let the added components be denoted as $e^0_2=x_1$ and $e^1_1$ for the 0- and 1-cells, respectively, and without loss of generality, suppose that $e^1_1$ is attached to $e^0_2$ in such a way that $e^0_1\in e^1_1$ as shown in Figure 1 below. This describes completely the 1-skeleton $X^1$ of the CW-structure being constructed.

Figure 1

Next, attach the 2-cell $e^2_2$ in the obvious way, namely by identifying the boundary $\partial D^2$ of a disc $D^2$ with the entirety of $X^1=e^1_1\cup e^0_1\cup e^0_2\cong S^1$. More precisely, if $X$ denotes the structure at hand, $X=X^1\sqcup_\alpha D^2$ where $\varphi_\alpha:\partial D^2\to X^1$. The result, then, is that $X$ is “almost a $D^2$ in a precise way” as shown in Figure 2 below.

Figure 2

Obviously, then, $X$ can be deformed continuously into a hemisphere by reaching under the space $X$ (think about $X$ as a 2-dimensional space in $\mathbb{R}^3$ so that “under” makes sense) and pushing up until $e^2_2$ is rounded. Of course, the map $f:\mathbb{C}\to\mathbb{C}$ for which $z\mapsto z^2$ is a continuous mapping of a hemisphere into a sphere (presuming our hemispheres/spheres are unit hemispheres/spheres, which can be assumed), whereby it follows that the space $X$ can be continuously deformed into $S^2$. This means $S^2$ can be given such a cell structure.

Case 2: Increasing $n=f-1$ by 1 while leaving $m$ fixed.
As in Case 1, notice that equation $(1)$ above indicates the effect of increasing $n$. In particular, $n$ is applied to the triple $(0,1,-1)$, and so the map $n\mapsto n+1$ is equivalent to adding an edge and a face (i.e., a 1-cell and a 2-cell). For this construction, the CW-model of $S^2$ that’s most convenient is the one having two cells in each of dimensions 0, 1, and 2. Letting $e^0_i$, $e^1_i$, and $e^2_i$, $i=1,2$, denote these cells, the CW-structure of $S^2$ is shown in Figure 3 below. Note that the cells $e^2_i$, $i=1,2$, can be thought of as hemispheres being glued to the equator circle formed by $X^1=e^0_i\cup e^1_i$, $i=1,2$.

Figure 3

As per the above remark, increasing $n$ means adding a 1-cell $e^1_3$ and a 2-cell $e^2_3$ without adding any 0-cells. In particular, the most logical way to do so would be to add $e^1_3$ as a loop from one of the vertices $e^0_i$, $i=1,2$, and to glue the 2-cell $e^2_3$ to this 1-cell by identifying the boundary $\partial D^2$ of a disc $D^2$ to the (homeomorphic to) $S^1$ that is $e^1_3$. A picture of the components of this construction is shown in Figure 4 below.

Figure 4

Using the above identification of the 2-cells $e^2_i$, $i=1,2,3$, with hemispheres, it’s easy to see that the space $X$ constructed in this way is homotopic to the space consisting of the wedge sum $S^2\vee H$ where $H$ is a hemisphere of $S^2$. In particular, then, applying the map $z\mapsto z^2$ takes the space $X = S^2\vee H$ to the space $X'=S^2\vee S^2$, and because the wedge of two 2-spheres is again a 2-sphere, it follows that this space $X$ can describe a viable CW-structure on $S^2$.

To conclude, here’s how to form the general case: Say we’re given $v,e,f$ and told that $v-e+f=2$. Some sub-collection $v',e',f'$ (not necessarily proper) must generate a copy of $S^2$ because the collection $v,e,f$ must satisfy $v\geq 1$, $f\geq 1$. Using the sphere generated by the cell-complex with $v',e',f'$ 0- 1-, and 2-cells (respectively), consider the values of the integers $m$ and $n$ to accommodate the differences $v-v',e-e',f-f'$: This will yield values $m\geq 0,n\geq 0$. Following the procedure in cases 1 and 2 above, “count up” to $m$ and $n$ by considering $m'=0=n'$ using the pseudo-algorithm:

Do $m'\mapsto m'+1$, $n'$ fixed, until $m'=m$. Do $n'\mapsto n'+1$ until $n'=n$.

The result will be a cell structure on $S^2$ satisfying the desired parametrization.    $\square$

1. The motivation for this technique was borrowed from the very terse solution given here.

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## 2 thoughts on “Hatcher 0.14”

1. Your equation for e (line 5) is incorrect; I think the original source also made this mistake. The equation should read (v,e,f) = (1,0,1) + m(1,1,0) + n(0,1,1), no minus signs.

• Thanks for letting me know! I’ll look into this as soon as I’m back in Algebraic Topology mode!