I am considering the following statement:
If $f$ is a function, then $\operatorname{card} \operatorname{ran} f \leq \operatorname{card} \operatorname{dom} f$.
This can be proved using the axiom of choice. Since $f^{-1}$ is a relation, the axiom of choice implies that there is a function $g \subseteq f^{-1}$ with $\operatorname{dom} g = \operatorname{dom} f^{-1} (= \operatorname{ran} f)$. We have $\operatorname{ran} g \subseteq \operatorname{ran} f^{-1} = \operatorname{dom} f$. Also, $g(x) = g(y)$ implies $x = f(g(x)) = f(g(y)) = y$, so $g$ is one-to-one. Thus, there is a one-to-one function from $\operatorname{ran} f$ into $\operatorname{dom} f$.
What I am wondering about is this: does the above statement imply the axiom of choice, so that they are equivalent?
