Let $f$ be a bijective function from $A$ to $B$. Let $x \in A$ and $y \in B$ . Prove that there exists a bijection $g$ defined from $A$ to $B$ such that $g(x)=y$ .
Here is my solution but i don't know if it is correct. Assume that $f(x) \neq y$ , because if they were equal taking $g$ the same bijection as $f$ would work.
Since $f$ is a bijection we have $|A|=|B|$ and there exists $x_1 \in A$ and $y_1 \in B $ such that $f(x_1)=y$ and $f(x)=y_1$ Now taking $g(a)=f(a)$ if $a \neq x_1$ or$ x$ , and $g(x)=y $ and $g(x_1)=y_1$ completes the proof.