Prove that f is surjective The problem:
Prove or refute the following:

If $f,g,h: \mathbb{R} \to \mathbb{R}$ and $f \circ g \circ h$ is surjective then $f$ is surjective.

My solution:
(The definition of surjective: iff $∀y ∈ T ,∃x ∈ S \implies f(x) = y$)
Let $f\colon A \to B$, $g\colon B \to C$ and $h: C \to D$.
Lets say $b ∈ A$, $a ∈  B$. We know by definition that $f(g(h(a)) = b$
Therefore $f(a) = b$, $g(a) = b$, $h(a)=b$, so $f$ is surjective.
I am kind of confused (as you can see from my solution)
Please help, am I at least on the right track or completely wrong.
 A: Let $y\in\mathbb{R}$. As $f\circ g\circ h$ is surjective, there exists $x\in\mathbb{R}$ such that $(f\circ g\circ h)(x) = y$.
So, if we let $x' = (g\circ h)(x)$, we have $f(x') = y$,
which proves that $f$ is surjective.
A: $\mathbb{R}=(f\circ g\circ h) (\mathbb{R})\subseteq{f(\mathbb{R}})\subseteq \mathbb{R}$, so $f(\mathbb{R})=\mathbb{R}$.
even
$f\colon E \to F$, $g\colon D \to E$ and $h: C \to D$, then since $\mathbb{F}=(f\circ g\circ h) (\mathbb{C})\subseteq{f(E})\subseteq \mathbb{F}$, so $f(\mathbb{E})=F$.
A: I don't know how you got to 

Therefore f(a) = b , g(a) = b , h(a) =b , so f is surjective.

by the way you don't need A,B,C and D because you know by definition of f,g,h that they go from R to R..... so A,B,C,D would simply be R
As you said we know per definition $\forall y \in R\; \exists x\in R : f(g(h(x)))=y$ Then simply substitude $g(h(x))$ by $x'$ now you have $\forall y \in R\; \exists x'\in R: g(x')=y$ . But thats excactly the definition of surjective so you'r finished....
PS: be carful with the notation $f\circ g \circ h $ because it's not clearly defined wether it means $f(g(h(x)))$ or $h(g(f(x)))$ 
