Proving that the cross product of two injective functions is bijective.

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:

Injectivity:

• 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 $$x = y$$

Surjectivity: ???

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.

• Welcome to MSE! Please use the mathjax basic tutorial and quick reference guide and enhance your question Feb 24 at 1:45
• If $f$ is injective, is $$f: A \to f(A)$$ bijective? Feb 24 at 4:32
• By the way, is $F3$ supposed to be a particular symbol? Feb 24 at 6:43

Suppose $$f: A \to X$$ is injective. And $$g:B\to Y$$ is injective.

Note if we define $$H:A\times B \to X\times Y$$ via $$H(a,b) = (f(a),g(b))$$ then $$H$$ is might NOT be surjective. But that was not the question that was asked.

We aren't defining $$H:A\times B \to X \times Y$$; we are defining $$F3:A\times B \to f(A)\times f(B)$$. And $$f(A)\times f(B)$$ is not the same codomain as $$X \times Y$$ would be. We do have $$f(A)\times f(B)\subset X\times Y$$ but it could be that that $$f(A)\times f(B) \subsetneq X\times Y$$, a proper subset.

If $$f$$ is not surjective then there will be some $$x \in X$$ where there is no $$a\in A$$ so that $$f(a) =x$$. But then $$x \in X$$ but $$x$$ is NOT in $$f(A)$$. And $$f(A)\subsetneq X$$.

So if $$f$$ or $$g$$ are not both surjective then $$f(A)\times f(B) \subsetneq X\times Y$$.

Now... the proof that $$F3: A\times B \to f(A)\times f(B)$$ is downright trivial!

If $$(x,y) \in f(A)\times f(B)$$ then $$x \in f(A)$$ and $$y \in f(B)$$.

But if $$x \in f(A) \subset X$$ then there is an $$\alpha \in A$$ so that $$f(\alpha) = x$$. And if $$y \in f(B)$$ then there is a $$\beta \in A$$ so that $$f(\beta) = y$$.

And so ..... $$F3(\alpha, \beta) = (f(\alpha), f(\beta)) = (x,y)$$.

So $$F3$$ is surjective.