# Right Inverse for Surjective Function

Prove that if $f:X\to Y$ is a surjective function between sets, then there must exist a function $g:Y\rightarrow X$ such that $f\circ g=1_Y$.

I know that the identity function is onto, and if $f$ has right inverse, then $f$ must be onto; although, I haven't seen a proof of this. A good hint would be nice.

• Hint: axiom of choice. – Berci Jul 10 '13 at 21:51

Suppose that $f\colon A\to B$ is surjective, then for every $b\in B$ the set $F_b=\{a\in A\mid f(a)=b\}$ is non-empty. Therefore, using the axiom of choice, there is some $g$ which selects an element from $F_b$, that is $g(F_b)\in F_b$.
Now show that $g$ is actually a function from $B$ into $A$, and that $g$ is injective.
What would an inverse look like? For every element $y$ of $Y$, it would send $y$ to an element of $X$ in the preimage of $Y$. You can just make up such a map by choosing $g(y)$ to be any point in $f^{-1}(y).$
• $\Large{\ldots\rm choosing \ldots}$ – Pedro Tamaroff Jul 10 '13 at 23:15