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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.

question
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  • 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 $f$. – Mesmerized student Oct 02 '15 at 10:34
  • 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 $x$. – Git Gud Oct 02 '15 at 10:34
  • 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. – Git Gud Oct 02 '15 at 10:42
  • @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? – Zoran Loncarevic Oct 02 '15 at 10:45
  • @ZoranLoncarevic Actually it wasn't you. If you go to the edit historyyou'll see someone did it before you. In any case the answerers posted their answers before any edit was made, so they also made the same mistake of reading '$\mathbb R$'. Should you change '$\mathbb R$' to $R$? I think that's for the best unless the OP opposes it. – Git Gud Oct 02 '15 at 10:48
  • Why is this tagged (relations)? – Martin Sleziak Oct 02 '15 at 13:26
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    See here and here. – Martin Sleziak Oct 02 '15 at 13:33

3 Answers3

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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.

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

R.N
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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 $h(g(f(x)))$

Börge
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