Let $\ f_1:A \rightarrow B$ and $\ f_2:A \rightarrow B$. Prove or disprove $f_1 \cap f_2$ iff $f_1=f_2$. Here is the question I am working on (screenshot):



So, I haven't worked with function proofs very much (especially in the context of iff statements and with intersections). I am looking to see if I am on the right track for how to approach this proof, and to look for feedback in the holes of my proof knowledge. 
Strategy: I will prove this statement (not disprove).
Forward:
Prove directly. Let $(x,y) \in f_1 \cap f_2$. This is equivalent to saying that $x \in f_1$ and $x \in f_2$, given by the intersection. From here, I am uncertain where to go and how to show this indicates $A=B$.
Converse:
I was thinking that I should proceed by contrapositive. To do the contrapositive, my statement would be "If $A \neq B$, then $f_1 \cup f_2$ is not a function".
I am much more stuck with the converse direction, so a push in the right direction would be helpful.
Additional question: Can you prove either side of a biconditional statement by contradiction?
 A: As you said, we can think of $f_{1}$ and $f_{2}$ as subsets of $A \times B$, with $f_{1} = \{(x, f_{1}(x)) \in A \times B  \}$ and $f_{2} = \{(x, f_{2}(x)) \in A \times B  \}$.
Now, we need to prove $f_{1} \cap f_{2}$ is a function from $A$ to $B$ iff $f_{1} = f_{2}$.
Let's prove the forward direction first.  Suppose $f_{1} \cap f_{2}$ is a function from $A$ to $B$.  We want to show $f_{1} = f_{2}$.  Let's do this by contradiction.  So suppose $f_{1} \neq f_{2}$.  Then without loss of generality we can assume there is some $(x, f_{1}(x)) \in f_{1}$ but $(x, f_{1}(x)) \not \in f_{2}$.  Since $(x, f_{1}(x)) \not \in f_{2}$, then the function $f_{1} \cap f_{2}$ cannot be a function from $A$ to $B$ because for it to be one, it should be defined for every input $a \in A$, but for the particular $x$ we isolated above, $f_{1} \cap f_{2}$ is not defined on this $x$ (why?).
Now let's prove the backward direction.  Suppose $f_{1} = f_{2}$.  We want to show $f_{1} \cap f_{2}$ is a function from $A$ into $B$.  Well, let $a \in A$.  What is $(f_{1} \cap f_{2})(a)$?  Well, if we are thinking about $f_{1}$ and $f_{2}$ as the sets described above, then we consider their intersection.  Since $f_{1} = f_{2}$, for each $a$, $f_{1}(a) = f_{2}(a)$, so $(a, f_{1}(a)) = (a, f_{2}(a))$ for each $a$, which tells us that $f_{1} \cap f_{2}$ as a function takes $a$ to the value $f_{1}(a)$ (which equals $f_{2}(a)$), and this is true for every $a$.  It is up to you to now determine why this proves the statement.
