Question

Prove: a function is bijective iff it is surjective and injective.

I'm really inexperienced with proofs. Please help.

Answer 1

take two sets x and y , choose the elements of x such as they are surjective and injective and y should then be biijective

Answer 2

I understand why it's true. I just can't prove it formally

Answer 3

I don't believe there is anything to prove. I believe what you are trying to prove is the definition of bijective. We don't prove definitions. If not, you should post the definitions, so we can go from there.

Answer 4

\[f:A \rightarrow B \] is injective iff \[\forall a, a' \in A, f(a) = f(a') => a = a'\]

\[f:A \rightarrow B \] is surjective iff \[\forall b \in B, \exists a \in A : f(a) = b\]

\[f:A \rightarrow B \] is bijective iff \[\forall b \in B, \exists! a \in A : f(a) = b\]

Answer 5

A good argument to this lemma would be some kind of proof by contradiction where a function cannot be bijective and not injective because then there doesn't exist exactly one element in A : f(a) = b. But how do I put that into symbols...

Answer 6

This is one direction of the proof.