Relation based on function composition I have got the following problem.
I am sure that I miss something but can not get what I am missing.
Let $\mathcal{F}=\{f | f:\mathbb{R}\to\mathbb{R}\}$ and define a relation $\mathbf{R}$ on $\mathcal{F}$ as follows:
$\mathbf{R}=\{(f,g)\in\mathcal{F}\times\mathcal{F}|\exists h \in \mathcal{F}(f=h \circ g) \}$
Let $f(x)=x^2+1$, $g(x)=x^3+1$ and $h(x)=x^4+1$.
Prove that $hRf$ , but it is not the case that $gRf$.
Prove the first claim is simple, let $c(x)=(x-1)^2+1$ then 
$c(f(x))=c(x^2+1)=x^4+1=h(x)$ so $hRf$. 
My problem is that from my point of view we can also prove that $gRf$.
If we take into account that there is a function $b(x)=(|x-1|)^{3/2}+1$.
Domane of this function is $\mathbb{R}$ and because $(x^2+1)-1$ is always positive we have got $(x^2)^{3/2} + 1 = x^3 + 1$.
Could someone kindly explain what I miss?
 A: Well thank you for @ ryanblack tip, I have got why could not be such function.
Range of $f(x)$ is $\mathbb{R}+$, but range of $g(x)$  is $\mathbb{R}$.
So let x be some negative value,  -a, then $f(x)=f(a)=f(-a)=a^2+1$
But $g(a)\neq g(-a)$ which means that function $b(c)$ if such one exists should have two outputs for the same value, in our case for $a^2+1$ it should have outpus $-a^3+1$ and $a^3+1$ Which contradicts function defintion.
A: Suppose there was a function $\alpha:\mathbb{R}\rightarrow \mathbb{R}$ with $\alpha \circ f = g$. Then $0 =g(-1) = \alpha \circ f (-1) = \alpha (2) = \alpha \circ f (1) = g(1) = 2$, which is a contradiction.
More generally, any composition $\alpha \circ f $ won't be injective (because $f$ is not).
A: In a more general setting we can define a relation $R$ on functions $X\to Y$ by stating that: $$fRg\iff\forall a,b\in X\; [g(a)=g(b)\implies f(a)=f(b)]$$ In that case we say that function $f$ "respects" function $g$. It is evident that this will be the case if $f=h\circ g$ for some function $h:Y\to Y$. Also the converse of this is true. We can start with $y\mapsto f(y)$  as the prescription of a function $h:\text{im }g\to Y$. The fact that $f$ respects $g$ implies that this is a well defined function, and secondly without troubles the domain of $h$ can be expanded to become $Y$. Just pick some element of $x_0\in X$ and define $h(y)=x_0$ for every $y\in Y\setminus\text{im }g$. Then we will have $f=h\circ g$ for our constructed $h:Y\to Y$ and we are ready.

Projection on your case tells us that: $$f\mathbf Rg\iff f\text{ respects }g$$
Now observe that $h$ respects $f$ since:
$$a^2+1=b^2+1\implies a^2=b^2\implies a^4=(a^2)^2=(b^2)^2=b^4\implies a^4+1=b^4+1$$
But from $a^2+1=b^2+1$ it cannot be concluded that $a^3+1=b^3+1$ and we conclude that $g$ does not respect $f$.
