Let $B$ and $C$ be non-empty sets. A map $\varphi: B\to C$ is called countable-to-one if for every $c\in C$, $\varphi^{-1}(\{c\})$ is countable. Prove that if such a map exists then $|B|\le \aleph_0|C|$ .
My approach:
Need to prove that $B\preceq \mathbb{N}\times C$, i.e. that there exists an injection from $B$ to $\mathbb{N}\times C$.
Since $\varphi^{-1}(\{c\})$ is countable, there exists an injective function $f:\varphi^{-1}(\{c\})\to \mathbb{N}$. Let $F$ be the set of all fibers $\phi^{-1}(\{c\})$, for all $c\in C$. Let $g_c:\phi^{-1}(\{c\}) \to \mathbb{N}\times C$ be the function defined as $g(b)=(f(b),\phi(b))$, where $b$ is some element in $\varphi^{-1}(\{c\})$, unless $\varphi^{-1}(\{c\})$ is empty. Then $g_c$ is injective. Let $g:=\bigcup\limits_{c\in C} g_c$, then $g$ is the infinite union of injective functions. By the axiom of choice, $g$ is injective.
This implies that $|B|\le \aleph_0|C|$.
Please let me know if this proof is correct or not very much so.