Solutions

a) Define $f(z)=x$ for $z=g(x)$. The function $f$ is valid, and we prove this by cases. Let $x,y\in\operatorname{Dom}(g)$. If $x=y$, then we have

$$g(x)=g(y)\implies f(g(x))=f(g(y)).$$

On the other hand, if $x\ne y$, then we have

$$g(x)\ne g(y)\implies f(g(x))=x\ne y=f(g(y)).$$

Hence, there are no two pairs in $f$ where the first component is the same.

b) Let $b\in\operatorname{Range}(f)$. There exists $a$ such that $f(a)=b$. Define in this way $g(b)=a$ for all $b\in\operatorname{Range}(f)$. It then follows that

$$f(g(b))=f(a)=b$$

for all $b\in\operatorname{Range}(f)$.