1. ## Composite function

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...

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

Try the other one yourself.

3. Originally Posted by Drexel28
Suppose $\displaystyle gf:A\mapsto C$ is one to one but $\displaystyle f$ is not. Then there exists some $\displaystyle x,y\in A$ such that $\displaystyle f(x)=f(y)$ but $\displaystyle x\ne y$ and so $\displaystyle g(f(x))=g(f(y))$ and $\displaystyle x\ne y$. 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?

4. Originally Posted by Drexel28
Suppose $\displaystyle gf:A\mapsto C$ is one to one but $\displaystyle f$ is not. Then there exists some $\displaystyle x,y\in A$ such that $\displaystyle f(x)=f(y)$ but $\displaystyle x\ne y$ and so $\displaystyle g(f(x))=g(f(y))$ and $\displaystyle x\ne y$. 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 $\displaystyle f$ is non-injective implies that $\displaystyle gf$ is non-injective.

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

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