Here’s a simple example of a planar bipartite graph:
a
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/ \
b--c--d
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e
It has three vertices, $a,c$, and $e$, in one part and two, $b$ and $d$, in the other. It has three faces: one is the triangular region bounded by the cycle $abcd$, another is the triangular region bounded by the cycle $bcde$, and the third is the external face bounded by the cycle $abed$. Each of these faces has $4$ edges and $4$ vertices.
To see why $$2|E|=\sum_{i\ge 4}iF_i\;,$$ note that each edge of a planar graph is adjacent to two faces, one on each side of it. Suppose that the graph has faces $f_1,\ldots,f_n$, and face $f_k$ has $e_k$ edges for $k=1,\ldots,n$. Then $e_1+\ldots+e_n$, the sum of the numbers of edges of the faces, counts each edge twice, once for each of the two faces adjacent to it. Thus, $e_1+\ldots+e_n=2|E|$. But $e_1+\ldots+d_n$ is the same thing as $\sum_{i\ge 4}iF_i$. $F_i$ is the number of faces with $i$ edges, so $iF_i$ is the total number of edges for all of the faces with $i$ edges, and when we sum over $i$, we get the total number of edges of all of the faces of the graph. The degrees of the individual vertices don’t enter into it.
In the example above, $F_4=3$, and $F_i=0$ for $i\ne 4$, so $\sum_{i\ge 4}iF_i=4\cdot3=12=2\cdot6$, which is as it should be, since there are $6$ edges.
The see why the inequality
$$4\sum_{i\ge 4}F_i\le\sum_{i\ge 4}iF_i$$
holds, notice that if $i\ge 4$, then $4F_i\le iF_i$. Since we’re summing only over values of $i$ greater than or equal to $4$, we therefore have
$$\sum_{i\ge 4}4F_i\le\sum_{i\ge 4}iF_i\;:$$
each term on the left is less than or equal to the corresponding term on the right, so the sum on the left must be less than or equal to the sum on the right. And of course
$$4\sum_{i\ge 4}F_i=\sum_{i\ge 4}4F_i\;.$$
Finally, $\sum_{i\ge 4}F_i$ is simply $|F|$, the total number of faces, since every face has at least $4$ edges: we’re just adding up the number with $4$ edges, the number with $5$ edges, and so on. Thus, $4\sum_{i\ge 4}F_i=4|F|$. Combining this with the inequality and the original equation, we see that $4|F|\le 2|E|$ (and hence $2|F|\le|E|$: the number of edges of a planar bipartite graph is at least twice the number of faces).
In the example above $|F|=3$ and $|E|=6$, so in this case the inequality is actually an equality: the number of edges is exactly twice the number of faces.
We define faces of a plane graph $G$ to be the connected regions of $\mathbb R^2 - G$.
*-----* *-----*
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* *
By this definition, the graph above has $3$ faces:
- Two "normal" faces of length $3$, whose boundaries are $3$-cycles;
- One "weird" face of length $6$ whose boundary consists of two $3$-cycles.
The "weird" face is not necessarily the unbounded face; if we draw one triangle inside the other, then the region between the two triangles is the "weird" face. Here is what this looks like:
*---------*
\ *---* /
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*
In general, though we'd like the boundary of a face to be a cycle in the graph, this can fail in two ways:
- When the graph is not connected, the boundary of a face can have multiple pieces, as we see here.
- Each piece may be a closed walk, not necessarily a cycle, when the graph has bridges.
Note, however, that Euler's formula
$$
(\text{# vertices}) - (\text{# edges}) + (\text{# faces}) = 2
$$
only applies to connected graphs! For the $6$-vertex example we're looking at, it would tell us that there are $2$ faces, which is nonsense. The general rule is that
$$
(\text{# vertices}) - (\text{# edges}) + (\text{# faces}) = (\text{# components}) + 1.
$$
Best Answer
No, it is not necessarily possible.
Suppose there is an odd face $F$, such that each adjacent face is a square, so you cannot subdivide edges of $F$. Then you are asking to add inside $F$ an arbitrary planar graph $H$, such that $F + H$ is bipartite. In $F + H$, the outer face is odd, while any inner face is even. Thus its dual graph has exactly one odd-degree vertex, which is impossible.