Show that $f$ also has an orbit of period $2$.

Question

Given $f : [\alpha,\beta] \to [\alpha,\beta]$ with an orbit of period four $\{a,b,c,d\}$ ($a<b<c<d$), and given also $f(a)=b$, $f(b)=c$, $f(c)=d$, $f(d)=a$, is there a way I can show that $f$ has an orbit of period $2$, and maybe even an orbit of period $1$? (I cannot use Sharkovskii's theorem for this.)

Answer 1

Suppose $f$ is continuous on $[\alpha,\beta]$, then so is $g(x) = x-f(f(x))$.

As $g(b) = b-d < 0 < c-a = g(c)$, there exists a point $e \in (b,c)$ such that $g(e)=0$, i.e., $e = f(f(e))$.