Showing if f and g are bijective then so is g o f

1. Oct 3, 2013

bonfire09

1. The problem statement, all variables and given/known data
If $f$ and $g$ are bijective functions and $f:A→B$ and $g:B→C$ then $g \circ f$ is bijective.

2. Relevant equations

3. The attempt at a solution
Showing $g \circ f$ is one to one
Suppose that $g\circ f(a_1)=g\circ f(a_2)=> g(f(a_1))=g((f(a_2))$ Since $g$ is one to one then $f(a_1)=f(a_2)=b$. But since f is bijective there exists $a_1$ and $a_2$ in $A$ such that $f(a_1)=b$ and $f(a_2)=b$. Since f is one to one then $a_1=a_2$
Showing $g \circ f$ is onto
Since $f$ is onto there exists a $a\in A$ such that $f(a)=b$ where $b\in B$. Then $g(b)=c$ for a $c\in C$ since g is onto. Thus $g(b)=g(f(a))=c$ implies that $g \circ f$ is onto.

Would this be right?

Last edited: Oct 3, 2013
2. Oct 3, 2013

haruspex

Not a good start. You need to start with supposing $g \circ f (a_1) = g \circ f (a_2)$

3. Oct 3, 2013

bonfire09

I fixed it now. Is the onto part correct?

4. Oct 3, 2013

haruspex

Not really. You should start with some arbitrary c in C then prove there exists an a in A such that etc.

5. Oct 3, 2013

bonfire09

Oh I was thinking that the range of f is in the domain of g so I thought I could do that. I'm curious why is the way I did it was wrong?

6. Oct 4, 2013

haruspex

You started with a b in B and found from that an a in A and a c in C. But the second of those did not require g to be onto. g is defined on B, so there would have to be a c in C. So you did not show that for any c in C there is an a in A.

7. Oct 4, 2013

bonfire09

I fixed the onto part.Here it is. Since $g$ is onto then for every $c\in C$ there exists a unique $b\in B$ such that $g(b)=c$. Since $f$ is onto there exists a $a\in A$ such that $f(a)=b$. Now we see $c=g(b)=g(f(a))=g\circ f(a)$. Thus $g \circ f$ is onto

8. Oct 5, 2013

haruspex

Good. (But it is not necessary that b is unique for this part of the proof.)