Construct a Bijective Function $A \times B \rightarrow B \times A$ Suppose $A$ and $B$ are sets. Prove that there exists a bijective function $A \times B \rightarrow B \times A$.
Since this chapter precedes the one concerning infinite sets, I'm assuming that these sets are finite (if there is a way to include infinite sets, I would like to know how to go about the problem). 
Assuming that $A$ and $B$ are finite, let $|A| = m$ and $|B| = n$ for some $m, n \in \mathbb{N}$. 
It is here that I run into issues, for their are three relations that $m$ and $n$ have according to the Trichotomy law. The easiest case is where $m = n$, then a procedure to constructing a bijective function is straightforward, but I won't include it here, because I feel like there is another approach that may bypass these cases. What is a way to go about this problem? Could I just ignore the three cases, construct a function $\psi$, and show as a consequence of how $\psi$ is built, that $|A \times B| = |B \times A|$ thereby proving that $\psi$ is bijective?
 A: In this case it is much easier to construct such a bijection than using the abstract machinery of cardinality to show only the existence.  Consider the functions
\begin{align*}
 \varphi \colon A \times B \to B \times A, \quad (a,b) \mapsto (b,a) \\
 \psi \colon B \times A \to A \times B, \quad (b,a) \mapsto (a,b).
\end{align*}
Then $\varphi \circ \psi = \mathrm{id}_{B \times A}$ and $\psi \circ \varphi = \mathrm{id}_{A \times B}$. So $\varphi$ is bijective with $\varphi^{-1} = \psi$.
A: 
"Could I just ignore the three cases, construct a function ψ, and show as a consequence of how ψ is built, that |A×B|=|B×A||A×B|=|B×A| thereby proving that ψ is bijective? " 

Eventually yes, well,... it depends on how constructivist you want your mathematics to be. It is always a useful exercise to make a bijection anyways:
Consider $f(x,y)=(y,x)$ as our function $f | A \times B \rightarrow B \times A $. To prove that $f$ is a bijection, show that $f$ is injective and surjective. 
To see that $f$ is injective, suppose we have $(x_1,y_1), (x_2,y_2) \in A \times B$ and $f(x_1,y_1)=f(x_2,y_2)$. The equation $f(x_1,y_1)=f(x_2,y_2)$ gives us $(y_1,x_1)=(y_2,x_2)$ which immediately tells us $y_1=y_2$ and $x_1=x_2$. Thus $(x_1,y_1)= (x_2,y_2) $.
To see that $f$ is surjective, for any $(x,y) \in B \times A$. We see that $(x,y)$ is hit by $f(y,x)$ for $ (y,x) \in A \times B$.
