# Thread: composite functions and their injectivity and surjectivity

1. ## composite functions and their injectivity and surjectivity

Here is the question:

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

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

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

Here is what I have so far:

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

So from here I know the goal is to show that $f(x_1)=f(x_2) \Rightarrow x_1=x_2$. 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!

2. Realize that if $t=s$ then $g(t)=g(s)$ must be true.

So suppose that $f(p)=f(q)$. It must be true that $g(f(p))=g(f(q))$.
From the given what follows?

3. if the composition is injective, then $(g\circ f)(x_1)=(g\circ f)(x_2)\Rightarrow x_1=x_2$, in your first line your implication actually says g is injective, not g composed with f