(ZF) Existence of injection and surjection I'm trying to prove the following statement related to the Zermelo–Fraenkel set theory, which looks rather basic yet I'm still unable to solve it.
Problem: Let $A$ be a nonempty well-ordered set. Prove that for every set $B$ the following two statements are equivalent:


*

*There exists an injective function from $A$ to $B$.

*There exists surjective function from $B$ to $A$.


Any comments, ideas and suggestions would be very much welcome. As usual, thank you in advance!
EDIT: Here is the a modification of the original:
Let $A$ be a nonempty well-ordered set. Prove that for every set $B$ such that there exists an injective function from $A$ to $B$ then there exists surjective function from $B$ to $A$ and vice versa. 
 A: Pedro is right. You need $B$ to be well-ordered.
It is possible that there is a set $A$ and an ordinal $\alpha$ such that $A$ can be mapped onto $\alpha$, and $\alpha$ cannot be mapped injectively into $A$. Some examples include:


*

*You can have $A=\Bbb R$ and $\alpha=\omega_1$ (the least uncountable ordinal).

*You can have $A\subseteq\Bbb R$ and $\alpha=\omega$.


On the other hand, if $B$ is well-ordered, and not $A$ (or if $A$ is mapped onto $B$, and $B$ is injected into $A$), then you can do the following:


*

*Prove that in general, if $X$ is non-empty, and $f\colon X\to Y$ is injective, then there is a surjective map $g\colon Y\to X$; by dealing with those elements of $Y$ which are in the image of $f$; and separately with those which are not in the image of $f$ (here you use the fact $X$ is non-empty).

*Use the fact that $A$ is well-ordered, so if $f\colon A\to B$ is surjective, you can pick canonically $a_b\in A$ such that $f(a_b)=b$ for each $b\in B$; and this guarantees that you defined an injection (why?)
