Proof $f,g : \mathbb{R} \to \mathbb{R}$ one-one $\implies f\circ g$ one-one 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.
 A: 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 $\mathbb R$.
A: 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 $f(g(a))=c$.

