I am trying to prove the following:
Let $A$, $B$, $C$, and $D $ be sets. Let $ g:A\to C $ and let $h:B\to D$. Then, we can define a function $$\phi(x)=\begin{cases} g(x), & \text{if } x\in A \\ h(x), & \text{if } x\in B \end{cases}$$ Then if $g$ is a bijection from $ A $ to $ C$, and $h$ is a bijection from $ B $ to $ D$, $A\cap B = \emptyset$, and $ C\cap D = \emptyset$, then $ \phi $ is a bijection from $ X $ to $ Y $, and for all $ y \in Y $ we have $$ \phi ^{-1}(y)=\begin{cases} g^{-1}(y), & \text{if } y\in C \\ h^{-1}(y), & \text{if } y\in D \end{cases}$$
I am mainly stuck on showing that $\phi$ has an inverse. Any help is appreciated!
You don't need to show in a vacuum that $\phi$ has an inverse -- since the problem explicitly tells you what the inverse will be, you can simply go ahead and define $$ \psi(y)=\begin{cases} g^{-1}(y), & \text{if } y\in C \\ h^{-1}(y), & \text{if } y\in D \end{cases}$$ and then show directly element for element that $\psi$ satisfies the condition of being an inverse of $\phi$.
Once you have shown that $\phi$ has an inverse, this will tell you for free that it is a bijection.