Prove that there exists $\gamma \in (a,b)$ such that $f \circ f(\gamma)=\gamma$

Question

If $a,b,c \in R$ such that $a<b<c$ and $f:R \to R$ is a continuous function satisfying $f(a)=b,f(b)=c,f(c)=a$, then there exists $\gamma \in (a,b)$ such that $(f \circ f)(\gamma)=\gamma$

My try:
We have $f(a)=b, f(b)=c$ $\implies$ $f \circ f(a)=c$.

But $a=f(c)$, so we get $f \circ f \circ f(c)=c$

Similarly $f \circ f \circ f(b)=b$ and $f \circ f \circ f(a)=a$.

Now since $a<b<c$ we have $f \circ f \circ f(a)<f \circ f \circ f(b)<f \circ f \circ f(c)$

So this means $f \circ f \circ f(x)$ is strictly increasing in $(a,b)\cup[b,c)$ and so it is One-One.

Now since $f \circ f \circ f$ is One-One, that means $f \circ f$ is also One one which means $f$ is One-One.

But how to show that there exists $\gamma \in (a,b)$ such that $f \circ f(\gamma)=\gamma$

Answer 1

I am afraid that your idea does not work. $$ f(f(f(a))) < f(f(f(b))) < f(f(f(c))) $$ for $a < b < c$ does not imply that $f \circ f \circ f$ is strictly increasing on $[a, b]$. Also the conclusion that $f$ is one-to-one cannot be correct because $f(a) < f(b)$ and $f(b) > f(c)$.

Instead use that $$ f(f(a)) = c > a \\ f(f(b)) = a < b $$ implies that $f(f(x)) - x$ has a zero in $(a, b)$, due to the intermediate value theorem.