Help me with these Mathematical Proofs.

[Scrapster]

Problem:

Prove the following statements about composition of functions.

1) The composition of two injections is an injection.

2) The composition of two surjections is a surjection.

3) The composition of two bijections is a bijection.

4) If f: A -> B and g: B -> C are bijections, then (g o f)^(-1) =
f^(-1) o g^(-1). (Hint: Use associativity of composition to prove that the function f^(-1) o g^(-1) must be the inverse of the function g o f.)

How do I prove these? We just went over bi/sur/in'jections today in class.

Any ideas?

 

[hendon]

I think you can start from the definitions of in/surjections and prove 1 and 2... 3 follows from 1 and 2.

For one-one (can't remember if it's in or surjection)
If f,g are one-one, it means
f(x1)= f(x2) <=> x1 = x2 and
g(x1) = g(x2) <=> x1 = x2
then (f o g)(x1) = (f o g)(x2) => ...... => x1 = x2
then prove the other direction (i.e. <= )

The other case (no. 1 or 2) should be similar...

For 4:
(g o f) o [ f^-1 o g^-1 ]
= g o f o f^-1 o g^-1 (from associativity of composition)
simplify this to the identity function (x)
and then you proved that the term in the square brackets is the inverse of (g o f)

hope i'm clear...

 

[Scrapster]

Pretty clear, except for the identity function (x). More specific?

 

[hendon]

Identity function just means the function I(x) = x
ie f o f^-1 = I