I've been asked to prove the following property:
If $f$ and $g$ are injective then $F3$ is bijective, where $F3$ is given by $A×B→f(A)×g(B), (x,y)⟼(f(x),g(y))$
Here's what I have so far...
To prove that this is bijective, we must show that it is both injective and surjective:
- We must prove that if $(f(x), g(x)) = (f(y), g(y))$, then $x = y$
- Suppose that $(f(x), g(x)) = (f(y), g(y))$
- By the equality of ordered pairs, we know that $f(x) = f(y)$ and that $g(x) = g(y) $
- $f$ is injective according to the hypothesis and $f(x) = f(y)$, implying that $x = y$
I'm not too sure how to prove the surjective nature of the cross product and am therefore stuck. I was thinking of finding the inverse in order to show surjectivity, but am not too show how to proceed.