Prove $f \cup g : A \cup B \to C $ iff $ f|(A \cap B) = g|(A \cap B)$ Prove $f \cup g : A \cup B \to C $  iff   $ f|(A \cap B) = g|(A \cap B)$
Given $f : A \to C$ and $g: B \to C$ and $ f|(A \cap B)$ is restriction of $f$ to $A \cap B$
ATTEMPT
Let $x \in A \cap B$. So let $(x,y) \in f|(A \cap B)$. Also $x \in A \cup B$. So $ \exists z \in C$ such that $(x,z) \in f \cup g $
Case 1: $(x,z) \in g$
Also $(x,y) \in f$ and so $ {(x,y),(x,z)} \in f \cup z$ so $y=z$. So $(x,y) \in g$. So $f \subseteq g$
Is this correct way to go about this?

If $f:A\to C$ and $g:B\to C$ are both functions then $f\cup g$ is actually a set of ordered pairs $(x,y)$ with $x\in A\cup B$ and $y\in C$. You could call a subset of $(A\cup B)\times C$ or a relation. 
Now the question is: is this relation a function?
For this is must satisfy the rule that for every $x\in A\cup B$ there is exactly one $y\in C$ such that $(x,y)\in f\cup g$.
It is evident that this is the case for $x\in A-B$ and $x\in B-A$ since $f$ and $g$ are functions. It can only go wrong for $x\in A\cap B$ because there are pairs $(x,f(x))$ and $(x,g(x))$ both belong to $f\cup g$ somehow endangering the uniqueness of $y$. 
However the statement $f|A\cap B=g|A\cap B$ ensures that of $x\in A\cap B$ we have $f(x)=g(x)$ hence uniqueness.
Then our conclusion is that $f\cup g$ is indeed a function $A\cup B\to C$.
