1-1 Function Composition I am asked to consider f(x) and g(x) which are both 1-1 functions on their respective domains. I have been asked to show f ∘ g is a 1-1 function and then a followup statement of if f ∘ g is 1-1 does that mean f(x) and g(x) are both 1-1 on their domains.
The problem I am having with this question is how do I show the result? I understand when a function is 1-1 over a certain domain and I have tried composing my own functions but am unsure how I prove the result is true because I do not believe providing an example of two functions f(x) and g(x) where the first statement holds is really what I am supposed to do.
Sorry for posting again so soon, especially for such a similar and straightforward question.
Thank you.
 A: If $f$ and $g$ are one-to-one functions then $f\circ g$ is one-to-one.
From the assumption, $f(u)=f(v)\implies u=v$ and similarly for $g$. Thus
$$\begin{align}
(f\circ g)(u)=(f\circ g)(v) &\implies f(g(u))=f(g(v)) &\\
 &\implies g(u)=g(v) &\text{since $f$ is one-to-one} \\
 &\implies u=v &\text{since $g$ is one-to-one}
\end{align}$$
Therefore $f\circ g$ is one-to-one.
If $f\circ g$ is a one-to-one function then $g$ is one-to-one.
Let's assume that $g$ is not one-to-one. Then there exist $u$ and $v$ in the domain of $g$ such that $u\ne v$ and $g(u)=g(v)$. Then
$$(f\circ g)(u)=f(g(u))=f(g(v))=(f\circ g)(v)$$
Thus $f\circ g$ is not one-to-one, and the desired statement is the contrapositive of what we have just shown.
If $f\circ g$ is a one-to-one function then $f$ may or may not be one-to-one.
Let $f(x)=x^2$ and $g(x)=e^x$. The domains and co-domains are both the real numbers, $f$ is not one-to-one and $g$ is one-to-one. Then
$$(f\circ g)(x)=f(g(x))=\left(e^x\right)^2=e^{2x}$$
We see that $f\circ g$ is a one-to-one function, even though $f$ is not. This is so since the range of $g$ (the positive reals) does not include subsets where $f$ takes the same value for differing values of $x$. In other words, $f$ is one-to-one on the range of $g$ even though it is not on all real numbers.
Of course, we can easily find examples where all of $f$, $g$, and $f\circ g$ are one-to-one.
