Cardinality: $\left|\Bbb N^{\{1,2\}}\right| = \left|\Bbb N^{\{1,2,3\}}\right|$ Left $F$ be the set of functions from $\{1,2\}$ to $\Bbb N$ and $G$ be the set of functions from $\{1,2,3\}$ to $\Bbb N$, prove that $\left|F\right| = \left|G\right|$.
Seems intuitive to me that the sets are cardinal equivalent as $\aleph_0^2 = \aleph_0^3$.
I've been trying to write a proof that is not based on cardinal number arithmetic. My approach was to show that $\left|F\right|\le\left|G\right|$ and $\left|G\right|\le\left|F\right|$ then according to Schröder–Bernstein theorm $\Rightarrow \left|F\right|=\left|G\right| \Rightarrow F\sim G$ .
It is clear that $\{1,2\}\subset\{1,2,3\}$ therefore $F\subseteq G$ therefore $\left|F\right|\le\left|G\right|$.
Another explanation is that for every $f\in F$ we can match $g\in G$ such as:
$$g(x) = 
\begin{cases}
f(x) + 1 & x\in\{1,2\} \\
0  & x\in \{3\}
\end{cases}$$
I need help to prove the second part.

HINT: For $f\in\Bbb N^{\{1,2,3\}}$ define $\hat f\in\Bbb N^{\{1,2\}}$ by setting $\hat f(1)=f(1)$ and $\hat f(2)=2^{f(2)}\cdot3^{f(3)}$. Show that the map $f\mapsto\hat f$ is injective.
