4
$\begingroup$

I apologize in advance if this question is too basic to warrant a post. I just ran into the following question:

Let $f: A \to B$. If $A$ and $B$ are finite sets with the same number of elements, then $f: A \to B$ is bijective if and only if $f$ is injective if and only if $f$ is surjective.

The use of two "if and only if" statements has confused me; I am not exactly sure what I am trying to prove. I understand the function concepts being presented in the question, but I'm just not sure what I need to show. Therefore, I ask not for hints toward the solution of the question, but only the meaning of the question.

Thanks!

$\endgroup$
  • $\begingroup$ +1 for your nice question. May I ask why did you remove this ? I voted it to be undelete again and think it deserves to be alive again here at the site. Keep it on. Thanks. $\endgroup$ – mrs Jul 9 '13 at 15:35
8
$\begingroup$

The question is precisely equivalent to the following question:

Let $f:A\to B$, where $A$ and $B$ are finite sets with the same number of elements. Show that the following three statements are equivalent:

  1. $f:A\to B$ is bijective.
  2. $f:A\to B$ is injective.
  3. $f:A\to B$ is surjective.

In other words, given the hypotheses, each of the three statements implies and is implied by each of the others.

$\endgroup$
4
$\begingroup$

This sentence is the same as saying that the three statements are equivalent (after all, "if and only if" is the same as logical equivalence). Usually you'd see it written like this:

The following are equivalent (or, TFAE for short): (1) $f$ is bijective, (2) $f$ is injective, (3) $f$ is surjective.

A nice way to prove these is circularly; show that (1) implies (2), that (2) implies (3), and that (3) implies (1). Of course, you could rearrange the statements if certain implications are easier to prove than others.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.