Let $f:A\rightarrow B\;$be a bijective mapping, $x$ an element of $A$ and $y$ an element of $B$. Prove that there exists a bijective mapping $g:A\rightarrow B \;$ so that $g(x)=y$.

So the above is the full question and I am frustrated. Could anyone hint me and correct the thoughts below if they are incorrect?

1. If $f:A\rightarrow B\;$ be a bijective mapping, $x$ an element of $A$ and $y$ an element of $B$ then $\exists\, g:A\rightarrow B \;$ such that $g(x)=y$.
2. Prove by contrapositive: If there isn't a $g:A\rightarrow B \;$ such that $g(x)=y \;\forall x \in A \; \forall y \in B$, then there doesn't exist a bijective mapping $f:A\rightarrow B$.

Is proof by contrapositive correct?

If not, may I have some directions?

Hints preferred.