[SOLVED] Cantor-Schroder-Bernstein

Consider the injections $\displaystyle f : \mathbb{N} \rightarrow \mathbb{N} \sqcup \mathbb{N} : n \rightarrow (0, n)$ and $\displaystyle g : \mathbb{N} \sqcup \mathbb{N} \rightarrow \mathbb{N}$ given by

$\displaystyle (i, n) \rightarrow 3n + i$, where $\displaystyle i = 0$ or $\displaystyle i = 1$.

(a) From the proof of Cantor-Schroder-Bernstein compute F(n) directly for n =

0, 1, 2, 3, 4 and find $\displaystyle F^{-1} (0,4)$.

Is this just simply finding g o f? Like for n=2 it would be f(2) = 2 then g(0,2) = 6?

Or is it... Find x such that (g o f)(x) = 0,1,2,3,4?

OR! Is it something to do with the fact that... if i=0 then F(n) can be either 0,3,6, ... (3n + 0 basically). And if i=1 then F(n) can be either 1,4,7, ... (3n+1)..?

Since F isn't actually defined I'm not sure what it is.

AND. Since the CSB theorem is involved, there will be bijection A -> B so it makes me think that some values of n will equal 0,3,6, ... or 1,4,7 ... and the rest will have to be defined as something else...