1) If $F: A \rightarrow B$ has a left inverse then $F$ is injective.
2) If $F: A \rightarrow B$ has a right inverse then $F$ is surjective.
I've already proved those statements in an iff implication. Now I just want to check if it's possible to prove those claims by contradiction... Here's what I've got:
Let $G: B \rightarrow A$ be the left inverse of $F$ in (1) and right inverse of $F$ in (2)
1) Since $G$ is a left inverse, we have that: $$ G(F(a)) = a \text{ is valid }\forall a \in A $$ Now suppose that $F$ is not injective, therefore $\exists a_1,a_1 \in A$ such that $a_1 \neq a_2$ and $F(a_1) = F(a_2) = b' \in B$, therefore: $$ G(F(a_1)) = a_1 \text{ and } G(F(a_2)) = a_2 $$ but since $F(a_1) = F(a_2) = b'$ it follows that $$ G(b') = a_1 \text{ and } G(b') = a_2 $$ Since $G$ is a function, that is clearly a contradiction, therefore $F$ is injective.
2) Since $G$ is a right inverse, we have that: $$ F(G(b)) = b \text{ is valid } \forall b \in B $$ Now suppose $F$ is not surjective, therefore $\exists b' \in B$ where $b' \notin \text{Im}(F)$. Now see that $$ F(G(b'))=b' $$ hence $b' \in \text{Im}(F)$, which is a contradiction. Conclusion: $F$ is surjective.
Here is something that I'd like to mention before you post your comment:
As I've said, I've already proved those statements in an easier and more immediate way... The reason for this post is to check a different approach to the proof, using proof by contradiction. For you, what I've done is correct? If not, could you explain why and how would you do to prove it by contradiction?
Thanks!