# composite functions and their injectivity and surjectivity

• Jan 8th 2010, 09:57 AM
osudude
composite functions and their injectivity and surjectivity
Here is the question:

let f : A $\displaystyle \rightarrow$ B and g : B $\displaystyle \rightarrow$ C be functions.

i) show that if g $\displaystyle \circ$ f is injective, then f is injective

ii) show that if g $\displaystyle \circ$ f is surjective, then g is surjective.

Here is what I have so far:

$\displaystyle g \circ f : A \rightarrow C.$ Now assume that $\displaystyle g \circ f$ is injective. then $\displaystyle g \circ f$ is one to one and $\displaystyle g(f(x_1)) = g(f(x_2)) \Rightarrow f(x_1) = f(x_2)$

So from here I know the goal is to show that $\displaystyle f(x_1)=f(x_2) \Rightarrow x_1=x_2$. But how?? my guess is that inverse functions are involved?? maybe not.

As far as part b goes, I'm thinking that it is a similar process, almost reversed. but I'm still somewhat confused on this part. thank you for the help!
• Jan 8th 2010, 10:19 AM
Plato
Realize that if $\displaystyle t=s$ then $\displaystyle g(t)=g(s)$ must be true.

So suppose that $\displaystyle f(p)=f(q)$. It must be true that $\displaystyle g(f(p))=g(f(q))$.
From the given what follows?
• Jan 8th 2010, 10:54 AM
artvandalay11
if the composition is injective, then $\displaystyle (g\circ f)(x_1)=(g\circ f)(x_2)\Rightarrow x_1=x_2$, in your first line your implication actually says g is injective, not g composed with f