If $f:\Bbb R\to \Bbb R$ and $f\circ f$ has an unique fixed point, prove $f$ has an unique fixed point too.
I've tried contradiction but only proved that $f(x)$ has at most one fixed point:
Suppose for contradiction that $f(x)$ has two different fixed point $a$ and $b$, thus we have $f(f(a))=f(a)=a$ and $f(f(b))=f(b)=b$, a contradiction.
Any help will be appreciated!
Suppose that $a$ is a fixed point of $f\circ f\implies f\circ f(a)=a\implies f\circ f\circ f(a)=f(a)\implies f(a)$ is a fixed point of $f\circ f\implies f(a)=a$ since $f\circ f$ has a unique fixed point.
So $a$ is a fixed point of $f$.
Now you apply what you have done.
You shoud use the fact that if x is a fixed point of f then f(x) is a fixed point of f. Acoording to this, if a and b are fixed point of f we have that f(f(a))=f(a)=a and f(f(b))=f(b)=b. Since f(f(x)) have a unique fixed point we conclude that a=b.
There is a fixed point to f due to if x is a fixed point of f(f(x)) then f(f(f(x)))=f(x). So we have f(x)=x by the uniqueness of x
