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Assuming $G$ is a planar graph with $k$ components, I need to determine an equation relating vertices, edges, faces and components. It is given that when $k=1$, $v-e+f=2$ (Euler's formula).

So from this I have gotten:

$v=e-f+2 \implies v=0-1+2=1$ (True)

$f=e-v+2 \implies f=0-1+2=1$ (True)

$e=v+f-2 \implies e=1+1-2=0$ (True)

From here I need to prove this. This is where I am getting confused. I can show multiple cases where these hold true, but I am having a hard time actually proving it.

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  • $\begingroup$ What does happen in the general case is that $v - e + f = 1 + k$ $\endgroup$ – Misguided Nov 26 '15 at 2:57
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Hints

  • If the separate components have $v_1, v_2, \ldots, v_k$ vertices, then the entire graph has $v_1 + v_2 + \cdots + v_k$ vertices.

  • If the separate components have $e_1, e_2, \ldots, e_k$ edges, then the entire graph has $e_1 + e_2 + \cdots + e_k$ edges.

  • If the separate components have $f_1, f_2, \ldots, f_k$ faces, then the entire graph has $f_1 + f_2 + \cdots + f_k$ faces.

  • For each $i$, $v_i - e_i + f_i = 2$.

  • What is $\sum_{i=1}^k (v_i - e_i + f_i)$?

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