I have an idea of where to go with this proof, but would like a second opinion as to wether I have actually made a logical argument.
$f(g(x)) \neq f(g(y))$ where $x, y \in \mathbb{R}$ and $x \neq y$.
Given that $g(x)$ is a $1$-$1$ function $g(x) \neq g(y)$. For simplicities sake we will say that $g(x) = a$ and $g(y) = b$, where $a, b \in \mathbb{R}$ and $a \neq b$
This leaves us with $f(a) \neq f(b)$. As we have established that $a \neq b$, and we know that $f(x)$ is a $1$-$1$ function this statement is true, hence proving that $f(g(x))$ is a $1$-$1$ function
I also have to establish whether it is true if $f(g(x))$ is $1$−$1$ must $f(x)$ and $g(x)$ be $1$−$1$ on their domains?
To me this is true as the $g(x)$ must be $1$-$1$ for $f(g(x))$ to be $1$-$1$, and if say $f(g(x)) = f(g(-x))$ (i.e. $f(x)$ is not $1$-$1$) then that would contradict the fact that $f(g(x))$ is $1$-$1$.
Any help is greatly appreciated!
Let $f : B \rightarrow C$ be bijective.
Let $g : A \rightarrow B$ be bijective.
Injectivity:
$$f(g(x)) = f(g(y)) \Rightarrow g(x) = g(y) \Rightarrow x = y$$
Surjectivity:
$$f(g(A)) = f(B) = C$$