Question:
$f : \mathbb Z\times\mathbb Z\to\mathbb Z\times\mathbb Z$ is defined by $f((x, y)) = (y, x)$. Write down whether $f^{-1}$ exists. If it does, write down $f^{-1}((3, 4))$. If it doesn’t, explain why not.
Answer: $f^{-1}((3, 4)) = (4, 3)$
So I can see that it meets the properties of being one-to-one and onto and that the composite of $f$ and $f^{-1}$ is equal to the composite of $f^{-1}$ and $f$.
But could I also argue that $f^{-1}(x,y) = (x,y)$ so that $f^{-1}(3,4) = (3,4)$?
In this case, the composite of $f$ and $f^{-1}$ would be $(4,3)$ and the composite of $f^{-1}$ and $f$ would also be $(4,3)$?
If $f(x,y)$ 1-1 and onto (bijective) an inverse exits.
Note that $(f\circ f)(x,y) = f(y,x) = (x,y)$
$f^{-1}(x,y) = f(x,y)$