# Hints needed on basic proof involving functions and relations.

Let $F = \{f\mid f\colon \mathbb R \to \mathbb R\}$, and define a relation $S$ on $F$ as follows: $S = \{(f,g) ∈ F \times F \mid \exists h \in F :f = h\circ g\}$. Let $f$, $g$ and $h$ be the functions from $\mathbb R$ to $\mathbb R$ defined by the formulas $f(x) = x^2 + 1$, $g(x) = x^3 + 1$, and $h(x) = x^4 + 1$. Prove that $h\,S\,f$, but that it is not the case that $g\, S\, f$.

By letting $j(x) = x^2 - 2x + 2$, then $h = j \circ f$, thus $h\, S\,f$.

I've been struggling to show that it is not the case that $g\,S\,f$. In specific, I need to show that $g$ is not equal to $j \circ f$, for arbitrary $j \colon \mathbb R \to \mathbb R$.

Any help would be appreciated.

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Maybe this could help: If $g=h\circ f$ and $f(x)=f(y)$ for some $x$, $y$, what can you say about values of $g$ at $x$ and $y$? – Martin Sleziak Jan 8 '12 at 19:52
Since it is only your second post here and the question sounds a little like a homework, I think I should mention this: How to ask a homework question? – Martin Sleziak Jan 8 '12 at 19:54
g(x)=h(f(x))=h(f(y))=g(y). If $x \neq y$, then this is a contradiction, since g is injective. Thus, g $\neq$ h∘f. – Freddie Jan 8 '12 at 20:31
Seems ok to me. You might as well post your comment as an answer; perhaps adding more details. It's ok to answer your own question: blog.stackoverflow.com/2011/07/… BTW a related question: math.stackexchange.com/questions/28123/… – Martin Sleziak Jan 8 '12 at 21:04

Let $j:\mathbb{R}\rightarrow\mathbb{R}$ be arbitrary. Let $a=1, b=-1$, so $j(f(a))=j(f(b))$, thus $j ∘ f$ is not injective. Since, $g$ is injective, then $g \neq j ∘ f$.