Homology group of a planar graph

Question

In my Algebraic Topology course we are studying topological and there is some point in a proof I am struggling with.

We define a graph $(G, V)$ such that $V \subset G$ is finite, $G\backslash V$ consists on finite path components $\overset{\circ}{e_1}, \ldots , \overset{\circ}{e_J}$ such that their closure $e_j$ is homeomorphic to $[0,1]$ and $e_j \backslash \overset{\circ}{e_j}$ consists on two distincts points. Now we say that the graph is planar if there exists an embedding into $\mathbb R^2$, that is if $G$ can be drawn on the plane such that edges are represented by simple continuous curves that intersect only at the vertices. Clearly each planar graph decomposes $\mathbb R^2$ into a finite number of bounded domaines and an unbounded domain. By means of stereographic projection, we cn view $G$ as a subspace of $S^2$.

Now in some proof, my teacher uses that fact that
$$H_k(S^2, G) \cong \begin{cases} \mathbb Z & \text{if } k = 0,
\\\mathbb Z^F & \text{if } k = 2, \\ 0 & \text{else,}
\end{cases}$$
for $F$ the number of faces of $G$. By considering the long exact sequence of $(S^2, G)$ we have (if we assume that $G$ is connected)
$$0 = H_2(G) \to H_2(S^2) \cong \mathbb Z \to H_2(S^2, G) \to H_1(G) = 0$$
which should give us $H_2(S^2, G) \cong \mathbb Z$. How can we get this $\mathbb Z^F$ ?

Answer 1

$G$ is planar, so we have $F-1$ bounded regions enclosed by its edges, and $1$ unbounded region. So thinking of $G$ as a complex, $H_1(G) = \mathbb{Z}^{F-1}$. Thus your exact sequence is: $$ \underbrace{H_2(G)}_{\cong \, 0} \to \underbrace{H_2(S^2)}_{\cong\, \mathbb{Z}} \to H_2(S^2, G) \to \underbrace{H_1(G)}_{\cong\, \mathbb{Z}^{F-1}} \to \underbrace{H_1(S^2)}_{\cong \, 0}, $$ so $H_2(S^2, G) \cong \mathbb{Z}^F$, as the map into $H_2(S^2, G)$ is injective hence has image $\mathbb{Z}$.

Edit: It's probably easiest to fall back on an example. Suppose your graph is defined via $V = \{(0,0), (1,0), (0,1)\}$, and let your edges be the line segments connecting each pair of points in $V$, yielding a triangle. Clearly this is planar, and has two faces: inside the triangle and out (i.e. $F = 2$). Sure we can compute $H_1(G) = \mathbb{Z}$ by showing that its homeomorphic to a circle, but we can also just view this as (the geometic realization of) the simplicial complex: $$\big\{\{v_1\}, \{v_2\}, \{v_3\}, \{v_1,v_2\}, \{v_2, v_3\}, \{v_1, v_3\} \big\}, $$ and compute its simplicial homology directly.

Sense Check: If you know about the idea of a good pair you can show $H_2(S^2,G) \cong H_2(S^2/G)$. What does this quotient look like? If we think of our triangle on the surface of $S^2$ like a drawstring to be cinched shut, when we do so our sphere gets pinched into two spheres, jointed at a point (i.e. a wedge sum of spheres). Thus $H_2(S^2/G) \cong H_2(S^2) \oplus H_2(S^2) \cong \mathbb{Z}^2 = \mathbb{Z}^F$ as claimed.