Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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:… BTW a related question:… – Martin Sleziak Jan 8 '12 at 21:04
up vote 2 down vote accepted

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 $g \neq j ∘ f$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.