So we have $f$ that is onto, and $g$ which is onto and one-to-one. Is $f \circ g$ onto and one-to-one?
My attempt at to solve this problem was to state the following scenarios and come to a conclusion. My proof method is below:
We know $f$ is onto so every element in its domain is mapped to one or more codomain making it onto.
$g$ is one-to-one and onto, so every element in $g$'s domain is mapped to a unique element in its codomain.
We know that $g$ goes to the domain of $f$. $f$ can map to the same element of the codomain; knowing that $f$ does this I can conclude that $f \circ g$ is not one-to-one and onto because of the fact that $f$ is onto.
Ex: Set $A$ contains $a$ and $b$, set $B$ contains $c$ and $d$, set $C$ contains $e$. $g$'s domain is $A$ codomain is $B$, and $f$'s domain is $B$ and codomain is $C$.
$$\text{}(f \circ g) = e \\ g(a) = c \\ f(g(a)) = e \\ \text{but} \\ g(b) =d \\ f(g(b)) = e \\ \therefore \text{$(f \circ g)$ is not one-to-one a because the domain maps to multiple things in the codomain.}$$
My questions are: is this correct, and is this a valid way of doing this proof?
Best Answer
No proof is necessarily needed since a counter-example will suffice and you have provided one, good job! Another counter-example in a similar spirit would be;
Take $f:\mathbb{R}\to\mathbb{R}^{\geq 0}$, $x\mapsto x^{2}$ and $g:\mathbb{R}\to\mathbb{R}$, $x\mapsto x$.
Then $g$ maps one-to-one and onto $\mathbb{R}$ and $f$ maps onto $\mathbb{R^{\geq 0}}$.
However, $(f\circ g)=f:\mathbb{R}\to\mathbb{R^{\geq 0}}$, $x\mapsto x^{2}$ is not one-to-one.