Composite function

**nikie1o2** · February 15th 2010, 08:18 AM

Let A,B and C be sets and let f: A->B and g: B->c be functions. the composite function denoted by g "o" f is a function from A to C defined by
g "o" f(x)= g(f(x)) for every x in A. Prove that if g "o" f is one to one, the f is one to one. Then prove that if g "o" f is onto, then g onto...

**Drexel28** · February 15th 2010, 07:17 PM

Originally Posted by nikie1o2

Let A,B and C be sets and let f: A->B and g: B->c be functions. the composite function denoted by g "o" f is a function from A to C defined by
g "o" f(x)= g(f(x)) for every x in A. Prove that if g "o" f is one to one, the f is one to one. Then prove that if g "o" f is onto, then g onto...

Suppose $gf:A\mapsto C$ is one to one but $f$ is not. Then there exists some $x,y\in A$ such that $f(x)=f(y)$ but $x\ne y$ and so $g(f(x))=g(f(y))$ and $x\ne y$ . Contradiction.

Try the other one yourself.

**nikie1o2** · February 16th 2010, 05:32 PM

Originally Posted by Drexel28

Suppose $gf:A\mapsto C$ is one to one but $f$ is not. Then there exists some $x,y\in A$ such that $f(x)=f(y)$ but $x\ne y$ and so $g(f(x))=g(f(y))$ and $x\ne y$ . Contradiction.

Try the other one yourself.

In your hypothesis, what does "gf" mean, the composite? And it's a contradiction because f is one to one ??

Do i approach the onto problem using a proof by contradiction also?

**Drexel28** · February 16th 2010, 05:35 PM

Originally Posted by Drexel28

Suppose $gf:A\mapsto C$ is one to one but $f$ is not. Then there exists some $x,y\in A$ such that $f(x)=f(y)$ but $x\ne y$ and so $g(f(x))=g(f(y))$ and $x\ne y$ . Contradiction.

Try the other one yourself.

Originally Posted by nikie1o2

In your hypothesis, what does "gf" mean, the composite? And it's a contradiction because f is one to one ??

Do i approach the onto problem using a proof by contradiction also?

Yes, it's the composite function. And no, it's a contradiction because assuming that $f$ is non-injective implies that $gf$ is non-injective.

For the other case it may be easier to approach it this way.

Let $c\in C$ then there exists some $x\in A$ such that $g(f(x))=c$ (since $gf$ is surjective)...but $f(x)\in B$ ...so

Thread: Composite function

LinkBack

Thread Tools

Display