how to prove Euler formula with disconnected graph with k components? (8 points) Let G be a graph with an $\mathbb{R_{2}}$-embedding having f faces. Euler’s formula tells us that if G is connected, then $\lvert V \lvert − \lvert E \lvert + f = 2$. What is $\lvert V \lvert − \lvert E \lvert + f$$ if G has k connected components? Prove that your answer always works!
How should I approach this? I see all the connected components share 1 common external face, although their vertex and edge sets are distinct. 
That implies that $\sum_{i}^{k}\lvert V_{i}\lvert -\sum_{i}^{k}\lvert E_{i}\lvert$ +$\sum_{i}^{k}\lvert f_{i}\lvert - k=2$
and I can induct on k to k+1 through induction?

I think the formula should be
$|V|-|E|+f=k+1$
If you consider one component at a time, for component $1\leq i\leq k$, with $|V_i|$ vertices and $|E_i|$ edges and $f_i$ faces, you can simply write
$|V_i|-|E_i|+f_i=2$
Add them all, for all $i$, to get
$\sum_{i}|V_i|-\sum_{i}|E_i|+\sum_{i}f_i=|V|-|E|+\sum_{i}f_i=2\times k$
As you have mentioned, there is only one face that is common between all of them and it has been counted $k$ times. So write the last equation as
$|V|-|E|+\sum_{i}f_i=|V|-|E|+f+(k-1)=2\times k$
Finally
$|V|-|E|+f=k+1$
