Logic: on the Axiom of Choice

Question

Let $X,Y,Z$ be infinite sets and $f:X \rightarrow Y$ be a surjective function. How can I prove that if $|Y| \le |Z|$ and for every $y \in Y$ is $|f^{-1}(y)| \le |Z|$, the following inequality holds: $|X| \le |Z|$ ?

Answer 1

If you are allowed to use cardinal arithmetic, then the following calculate is immediate:

$$|X|=\left|\bigcup_{y\in Y}f^{-1}(y)\right|=\sum_{y\in Y}|f^{-1}(y)|=|Y|\cdot\sup\{|f^{-1}(y)|\mid y\in Y\}\leq|Z|\cdot|Z|=|Z|$$

If you want to actually find an injection from $X$ to $Z$, you might want to use the fact that you have a bijection between $Z$ and $Y\times Z$. Now for every $y\in Y$ choose an injection from $f^{-1}(y)$ into $Z$ and finish the proof in the obvious way.

Answer 2

Hint: Using $X \cong \displaystyle\coprod_{y \in Y} f^{-1}(y)$, can you embed it in something that you know (by AC) has cardinality $|Z|$? You may want to think about the cardinality of $\displaystyle \coprod_{s \in S} T$ for arbitrary sets $S$ and $T$.