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 $x = y$

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.

  • 1
    $\begingroup$ Welcome to MSE! Please use the mathjax basic tutorial and quick reference guide and enhance your question $\endgroup$
    – Laufen
    Feb 24 '21 at 1:45
  • $\begingroup$ If $f$ is injective, is $$f: A \to f(A)$$ bijective? $\endgroup$
    – macton
    Feb 24 '21 at 4:32
  • $\begingroup$ By the way, is $F3$ supposed to be a particular symbol? $\endgroup$
    – fleablood
    Feb 24 '21 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.


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