Let $f:A \to B$ and $g:C\to D$. Prove that $f \times g$ is a function from $A \times C$ to $B \times D$. 
Let $f:A \to B$ and $g:C \to D$. Define
$$f \times g = \{((a,c),(b,d)) : (a,b) \in f\text{ and }(c,d) \in g\}.$$ Prove that $f \times g$ is a function from $A \times C$ to $B \times D$.

My proof (so far):
Let $(a,c) \in A \times C$, then $a \in A$ and $c \in C$.
Let $(b,d) \in B \times D$, $b \in B$ and $d \in D$.
If $p \in A \times C$ then $p=(a,c)$.
Define $(f \times g)(p)$ as $(f \times g)(p) = (f \times g)(a,c) = (f(a),g(c)) = (b,d)$, however $b \in B$ and $d \in D$ and $(b,d)\in B \times D$.
This shows $f \times g$ is a function from $A \times C$ to $B \times D$ as $(f \times g) : A \times C\to B \times D$ then $(a,b) \to (f(a),g(c))$.
I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f \times g$.
 A: You're missing something very important: the definition of a function. The set $f \times g$ is defined as a relation, specifically a binary relation between the sets $A \times C$ and $B \times D$. Functions are a specific type of relation. Specifically,

A relation $f \subseteq X \times Y$ between $X$ and $Y$ is a function if, for all $x \in X$, there exists a unique $y \in Y$ such that $(x, y) \in f$.

You need to use this property in your proof, as $f$ and $g$ are both assumed to have this property, as well as prove this property for $f \times g$.
In your proof, you skirt around this a little by equivalently defining $f \times g$ in terms of function notation, specifically defining $(f \times g)(p)$ for all $p \in A \times C$. As a proof technique, it's right on the border between acceptable and unacceptable. It buries the logic simply by a change of notation. By using function notation, you're implicitly assuming $f \times g$ is a function. You could instead define a new function $h : A \times C \to B \times D$ by saying $h(a, c) := (f(a), g(c))$, but then you should prove that the function $h$ is equal to the relation $f \times g$.
Either way, something is missing from your approach. You haven't assumed anything false or difficult to show, but some steps need to be added in order to rectify the proof. That said, I would highly recommend proving it from the definition above, because I strongly suspect you're not comfortable enough with it yet.
So, let me get you started here. Suppose $(a, b) \in A \times B$. Then $a \in A$ and $b \in B$. Since $f : A \to C$ is a function, there must exist some $c \in C$ such that $(a, c) \in f$. Can you finish the rest?
A: What you need to check is:

*

*Is $f\times g$ a subset of $(A\times C)\times(B\times D)$?

*Is there, for every $(a,c)\in A\times C,$ some $(b,d) \in (B\times D)$ such that $((a,c), (b,d)) \in f\times g$?

*Is the pair $(b,d)$ in the previous point unique for each $(a,c)\in A\times C$?

