Proving the Bijection Property of Composed Functions

[teme92]

Homework Statement

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

Homework Equations

f_og = f(g(x))

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.

 

[pasmith]

Homework Statement

Let f: A→B and g: B→C both be bijections. Prove that the composed function f_og is also a bijection. Write (f_og)² in terms of f^-1 and g^-1.

Homework Equations

f_og = f(g(x))

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)).

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.

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.

 

[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 f_og not g_of?

Injectivity: If x,y \inA and g(f(x))=g(f(y)) then f(x)=f(y) and x=y

Surjectivity: If c \inC 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?

 

[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.

 

[teme92]

Ok thanks for the help :)