If there is an surjection $f : \Bbb{Z}^+ \to a$, then there is an injection $f : a \to \Bbb{Z}^+$ I am trying to prove the following:

Suppose there is a surjection $f : \Bbb{Z}^+ \to a$ and $a \neq \varnothing$. How might we go about showing that there is an injection $h : a \to \Bbb{Z}^+$ without using the well-ordering principle (i.e. just relying on definitions of the relevant functions and maybe inverses)?

From the definition of a surjection, we know that for each $x \in a$ there exists some $n \in \Bbb{Z}^+$ such that $f(n) = x$. But now we need to show that (i) there exists a function $h : a \to \Bbb{Z}^+$ and that (ii) $h$ is injective, and it seems that we just can't get enough out of the definition of a surjection to prove either.
Do I need to make use of specific properties of $\Bbb{Z}^+$? I am at a loss here, and I wonder if anyone can help me out.
 A: Unfortunately, drawing the injection constructively fundamentally requires the ordering of $\mathbb{Z}^+$. More generally one could say that for a well-ordered set $S$ and any nonempty set $A$, there is a surjection $f : S \to A$ iff there exists an injection $g: A \to S$. But in $\mathsf{ZF}$ that's the best you can do: without choice there may be sets $A, B$ for which there exists a surjection $f : A \to B$ but no injection $g : B \to A$.
For completeness' sake, let's construct the injection in the well-ordered version. Let $(S, \leq)$ be a well-ordered set, let $A$ be a nonempty set, and suppose we have a surjective function $f : S \to A$. Then for each $a \in A$, the set $f^{-1}(\{a\})$ is nonempty. By well-ordering, the function $g : A \to S$ given by $g(a) = \min(f^{-1}(\{a\}))$ is therefore well-defined. We claim $g$ is injective.
Indeed, suppose $g(a_1) = g(a_2)$ for some $a_1, a_2 \in A$. Since clearly $g(a) \in f^{-1}(\{a\})$ for every $a \in A$, we simply apply $f$ and get
$$
a_1 = f(g(a_1)) = f(g(a_2)) = a_2.
$$
Hence $g$ is injective.
Addendum: By the way, when I say "the ordering of $\mathbb{Z}^+$" I don't mean to imply you need the standard ordering. Any well-ordering will do for the construction. I meant that you need to know that $\mathbb{Z}^+$ does have at least one well-ordering (and this would simply follow from its countability).
A: I am Assuming $A$ is a set.There can be three cases $A$ is countable infinite in that case there exist a bijection. $A$ is uncountable then answer is No Just take $A=\mathbb{R}$ Third $A$ is countable  finite obviously then there must exist two elements in $Z^{+}$ that goes to same element A(By Pegion Hole Principle) so no injection.
