Proof $f,g : \mathbb{R} \to \mathbb{R}$ one-one $\implies f\circ g$ one-one

Question

I thought this question would be interesting enough for someone to have asked this here, but it turns out that I can't find the answer to this anywhere on this forum:

Let $f,g : \mathbb{R} \to \mathbb{R}$ be two functions that are one-one. Proof (or disprove) that $f\circ g$, the composition of the two functions, is also one-one.

I tried playing around with some one-one functions $f$ and $g$ and the statement seems to hold true, but i have no idea where to start for a general proof.

By $fg$, do you mean $fg(x) = f(x)\cdot g(x)$ (in which case @DavidMitra provided a counterexample) or do you mean $f\circ g$? — 5xum, Feb 17 '14 at 12:34
Suppose $f(g(x_1)) = f(g(x_2))$ (I suppose with $f , g$ you mean the composition of the two functions, not the product). Then ... and hence ... so $x_1 = x_2$. — Daniel Fischer, Feb 17 '14 at 12:35
@5xum I meant the composition function, thanks. (Can't find the circle though) — Yiyuan Lee, Feb 17 '14 at 12:37
You should mention the domains and ranges of $f$ and $g$, as it may influence the answer to the question (as noted by dkuper). — SomeStrangeUser, Feb 17 '14 at 12:42

score 2 · Accepted Answer · answered Feb 17 '14 at 12:37

2

It depends what you call one-one. If it is being injective, then it is true: $f\circ g(x)=fg(y)$ implies $g(x)=g(y)$ (because $f$ injective) implies $x=y$ (because $g$ injective).

On the other hand, if it is being a bijection, then it is wrong, because you can have $g:A\to B$ and $f:B'\to C$ where $B\subset B'$, that are both bijective, but $f\circ g$ only reaches a subset of $C$. For instance take $g$ the identity on $\mathbb N$ and $f$ the identity on $\mathbb R$.

answered Feb 17 '14 at 12:37

Denis

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My apologies, I didn't state the domain and range of $f,g$ in my initial question. – Yiyuan Lee Feb 17 '14 at 12:46
"One-one" almost always refers to being injective. If it's quoted from a textbook i would be willing to stake pretty much anything betting that it does mean injective. – Arthur Feb 17 '14 at 12:47

score 1 · Answer 2 · answered Feb 17 '14 at 12:48

Just go by definition. You have a bijective function $g:A\rightarrow B$ and a bijective function $f: B\rightarrow C$. This means:

If $f(x) = f(y)$ for some $x,y\in C$, then $x=y$. If $g(x) = g(y)$ for some $x,y\in B$, then $x=y$.
For every $b\in B$, there exists some $a\in A$ that $b=g(a)$. For every $c\in C$, there exists some $b\in B$ that $c=f(b)$

Now prove that $f\circ g$ is bijective.

$f$ is injective: Take any two $x,y\in C$ for which $f(g(x))=f(g(y))$. Prove that $x=y$.
$f$ is surjective: Take some $c\in C$. Find such a $a\in A$ that $f(g(a))=c$.

Proof $f,g : \mathbb{R} \to \mathbb{R}$ one-one $\implies f\circ g$ one-one

2 Answers2