Prove the following:

If $\displaystyle f : A \rightarrow B $ and $\displaystyle g : C \rightarrow D $, then $\displaystyle f \cap g : A \cap C \rightarrow B \cap D$.

Here's my thoughts/attempt:

Proof:

Let A, B, C, and D be sets. Assume $\displaystyle f : A \rightarrow B $ and $\displaystyle g : C \rightarrow D $. Let $\displaystyle a \in A $. Since f is a function from A to B, there is some $\displaystyle y \in B $ such that $\displaystyle (a, y) \in f $. Let $\displaystyle b \in B $ be such an element, that is, let $\displaystyle b \in B $ such that $\displaystyle (a,b) \in f $. Let $\displaystyle c \in C $. Since g is a function from C to D, there is some $\displaystyle z \in D $ such that $\displaystyle (c, z) \in g $. Let $\displaystyle d \in D $ be such an element, that is, let $\displaystyle d \in D $ such that $\displaystyle (c,d) \in g $.

This is all I have so far.

Would I have to break it into cases where $\displaystyle a = c $ and $\displaystyle a \not= c $? If $\displaystyle a = c $, $\displaystyle A \cap C $ contains an element, but if $\displaystyle a \not= c $, $\displaystyle A \cap C $ is empty since a and c were arbitrary. The same argument holds for $\displaystyle B \cap D $. So, taking these things into account, $\displaystyle f \cap g $ is either a function from the set containing a to the set containing b, or its a function from the empty set to the empty set.

Does this make any sense, is it necessary, and how should I write it in my proof?

Thanks in advance.