# Composition of functions

1. May 8, 2014

### teme92

1. The problem statement, all variables and given/known data

Let f: A→B and g: B→C both be bijections. Prove that the composed function fog is also a bijection. Write (fog)-1 in terms of f-1 and g-1.

2. Relevant equations

fog = f(g(x))

3. The attempt at a solution

I know what the composed function means and what bijective means (one-to-one). I can do these problems when I have values for f(x) and g(x) but I'm having trouble with this type of problem. I nod in the right direction would be much appreciated.

Last edited: May 8, 2014
2. May 8, 2014

### pasmith

If $f: A \to B$ and $g: B \to C$ then the composition of the two which makes sense is $g \circ f : A \to C : x \mapsto g(f(x))$.

A bijection is both an injection and a surjection. Have you shown that a composition of injections is an injection, and a composition of surjections is a surjection?

Alternatively, you could exhibit an inverse of $g \circ f$ in terms of the inverses of $g$ and $f$.

3. May 8, 2014

### teme92

First of all thanks, and I made a mistake in the question which I changed ie. the composed function was a square but now an inverse.

Secondly, the question asked for fog not gof?

Injectivity: If x,y $\in$A and g(f(x))=g(f(y)) then f(x)=f(y) and x=y

Surjectivity: If c $\in$C then g(b)=c and f(a)=b

Hence g(f(a))=g(b)=c

Is that sufficient to prove its a bijection?

And how do I write Write (fog)-1 in terms of f-1 and g-1?

4. May 8, 2014

### xiavatar

Yes, the logic and solution is fine. Just justify the parts where you assume a certain property of a function, so that the reader(grader) knows that you understand what your doing at each step. So instead of saying f(x)=f(y),x=y.
Instead say f(x)=f(y) implies x=y because the function f is bijective.

5. May 11, 2014

### teme92

Ok thanks for the help :)