Show that a function which has a unique right inverse is necessarily bijective

Question

I try to prove a problem

Show that a function which has a unique right inverse is necessarily bijective.

I proved it is surjective, but I can not prove it is injective to finish the proof. I need help. Thank all!

Answer 1

If $f:X\to Y$ is a function then function $g:Y\to X$ will serve as a right-inverse for $f$ if and only if $$\forall y\in Y \left[g(y)\in f^{-1}(\{y\})\right]$$

(Note that $g(y)\in f^{-1}(\{y\}$ and $g(f(y))=y$ are equivalent statements.)

So condition for existence of a right-inverse is that the sets $f^{-1}(\{y\})$ are not empty (i.e. $f$ must be surjective) and condition for a unique right-inverse is that moreover these sets are singletons (so $f$ must be injective).

Answer 2

Assume the function $f$ is not injective. Then there are two different values $x \neq y$ such that $f(x) = f(y)$. How can you construct two different right inverses in this case, contradicting the uniqueness?