Show $f \cup g$ is a function $\iff f|C = g|C$ where $\text{dom}(f) = A, \text{dom}(g) = B \text{ and } \emptyset \ne C = A \cap B$ Assume $(x, y) \in f \cup g.$ Then $(x, y) \in f$ or $(x, y) \in g.$ Assume $(x, y) \in g.$ By definition, $(x, y) \in f|C \implies (x, y) \in f \text{ and } x \in C$. Since $(x, y) \in g$ and $x \in C$, we have that $(x, y) \in g|C.$ The other case is similar.
If $(x, y) \in f \cup g$, then $(x, y) \in f$ or $(x, y) \in g$ and so $x \in A$ or $x \in B$ meaning $x \in A \cup B$ implying $A \cup B$ is the domain of $f \cup g.$  Now suppose $(x, y) \in f|C = g|C.$ Then $x \in C = A \cap B$ meaning $x \in A$ and $x 
\in B.$ Now $x \in A \implies x \in A$ or $x \in B$ meaning $x \in A \cup B.$ Similarly, if $x \in B$, then $x \in A \cup B.$ So $x$ in the domain of $f \cup g.$ Now suppose $f \cup g(x_1) \ne f\cup g(x_2)$ with $x_1 = x_2.$ Then $x_1$ has two different images under $f \cup g$ which contradicts the definition of function and so $x_1 \ne x_2.$ But then $f \cup g$ is a function by definition. 

Just want to verify if the proof above works. For example, I didn't use the fact that $f \cup g$ is a function when showing $f|C = g|C$ unless it was implicit in what I wrote (without me knowing it). And in the second paragraph there's probably a lot of redundancy and maybe some wrong conclusions. If the proof is incorrect, how can I fix and make it better? Thanks.
 A: In both parts of the proof you really should say clearly what you’re proving.
It really is necessary to use the hypothesis that $f\cup g$ is a function to get the conclusion that $f\upharpoonright C=g\upharpoonright C$, and your argument in the first paragraph really does have a gap for just that reason. Assuming that $\langle x,y\rangle\in f\upharpoonright C$ and inferring that $\langle x,y\rangle\in f$ and $x\in C$ is fine, but this does not immediately imply that $\langle x,y\rangle\in g$: $\langle x,y\rangle$ in $g$ iff $g(x)=y$, and it’s only because we’re assuming that $f\cup g$ is a function that we can guarantee that $g(x)=y$. For suppose instead that $g(x)=z\ne y$; then $\langle x,y\rangle,\langle x,z\rangle\in f\cup g$ with $y\ne z$, contradicting the assumption that $f\cup g$ is a function. Thus, $g(x)=y$, and we can continue as you did, observing that $\langle x,y\rangle\in g\upharpoonright C$.
For the second part we’re assuming that $f\upharpoonright C=g\upharpoonright C$ and trying to show that $f\cup g$ is a function. To do this we need only show that it’s a set of ordered pairs that satisfies the function condition, namely, that if $\langle x,y\rangle,\langle x,z\rangle\in f\cup g$, then $y=z$. Suppose, then, that $\langle x,y\rangle,\langle x,z\rangle\in f\cup g$. Clearly $x\in\operatorname{dom}(f\cup g)=\operatorname{dom}f\cup\operatorname{dom}g=A\cup B$. If $x\in A\setminus C=A\setminus B$, then $x\notin\operatorname{dom}g$, so $\langle x,y\rangle,\langle x,z\rangle\in f$, and therefore $y=z$, since $f$ is a function. A similar argument takes care of the case $x\in B\setminus A$. The only remaining possibility is that $x\in C$, in which case 
$$\langle x,y\rangle,\langle x,z\rangle\in(f\cup g)\upharpoonright C=(f\upharpoonright C)\cup(g\upharpoonright C)\overset{*}=f\upharpoonright C\subseteq f\;,$$
where the starred equality follows from the hypothesis that $f\upharpoonright C=g\upharpoonright C$. And since $f$ is a function, we conclude that $y=z$. Thus, in all cases $y=z$, and $f\cup g$ is indeed a function.
Note that your notation $f\cup g(x_1)$ is potentially ambiguous or worse: it could have your intended meaning, but it is more naturally be read as the union of $f$ and $g(x_1)$. In this case the latter doesn’t actually make sense, but in general it could, and your intended meaning really should be written $(f\cup g)(x_1)$.
