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.

  • $\begingroup$ You missed the definitions of functions. Following your definition $f(g(x)):A \to B$, but $g:B\to C$, so $f(g())=f(C)$ but we know nothing about $C$ and $f$. $\endgroup$ – Mesmerized student Oct 2 '15 at 10:34
  • $\begingroup$ You seem to be a bit confused. You're given that $f,g,h$ are all functions from $R$ to $R$, so what's that about $A,B,C$ and $D$? You're hypothesis is that $\forall b\in R\exists a\in R(f(g(h(a)))=b)$ and you want to prove that $\forall y\in R\exists x\in R(f(x)=y)$. So take an arbitrary $y$. You need to find a suitable $x$. Use the hypothesis with $b:=y$ and find $x$. $\endgroup$ – Git Gud Oct 2 '15 at 10:34
  • $\begingroup$ I think people were too quick to read 'R' as '$\mathbb R$'. Specially since this is not only true in general, but also because the proof looks exactly the same, nothing about the real numbers is used. $\endgroup$ – Git Gud Oct 2 '15 at 10:42
  • $\begingroup$ @Git Gud I agree, and I was the culprit who changed $R$ to $\mathbb{R}$ in the text of the question. Should we change it back? $\endgroup$ – Zoran Loncarevic Oct 2 '15 at 10:45
  • 1
    $\begingroup$ See here and here. $\endgroup$ – Martin Sleziak Oct 2 '15 at 13:33

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.


$\mathbb{R}=(f\circ g\circ h) (\mathbb{R})\subseteq{f(\mathbb{R}})\subseteq \mathbb{R}$, 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 $f(\mathbb{E})=F$.


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 $h(g(f(x)))$

  • $\begingroup$ Thank you so much Börge, I finally understood it $\endgroup$ – question Oct 2 '15 at 10:43
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    $\begingroup$ Very important points. Plus the fact that you don't unnecessarily do it for the real numbers (+1). $\endgroup$ – Git Gud Oct 2 '15 at 10:45

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