Is there at least a continuous mapping from an arbitrary interval $[a,b]$ to $[0,1]$ with a periodic points of period $3$? $$\Large-\textbf{Problem}-$$
Is there at least a continuous mapping from an arbitrary interval $[a,b]$ to $[0,1]$ with a periodic point of period $3$?
$$\Large-\textbf{Thoughts and Ideas}-$$
Let
$$T_{\lambda}(\theta) = \theta + \dfrac{2\pi p}{3}$$
where $(p,3) = 1$, $\lambda = p/3$ and $p$ is an integer.  Denote this by $T:S^1\rightarrow S^1$, where $S^1$ is the set of values modulo $2\pi$ radians.  I was thinking a bit about the translation of the circle with periodic points of period $3$.  However, I can't find a continuous mapping $f:[a,b] \rightarrow [0,1]$ from here.  So I find another approach.
For a mapping to have periodic point with period $3$, it needs not to be a homeomorphism; that is because a homeomorphism can have no periodic points with prime period greater than $2$.  Then, I need to find a continuous mapping that is either 1-1 or onto or (possibly) not both.  This type of approach, to me, is fine, but it doesn't help me a lot to find such mapping.
Any suggestions or comments?
 A: You can rescale $\sin(x)$ or $\cos(x)$ . You know $\sin(x)$ has period $2\pi$. So, rescale the argument $x$ so that either $\sin(x)$ or $\cos(x)$ can hit every value three times in the interval $[a,b]$. This gives a continuous mapping from an arbitrary interval $[a,b] \to [0,1]$ with a periodic point of period $3$.
A: Let's construct a function $f$ from $[0,1]$ to $[0,1]$ such that some point $x_0$ is periodic with period $3$.  That is, $f(x_0) = x_1$, $f(x_1) = x_2$, $f(x_2) = x_0$, for some distinct 
$x_0$, $x_1$ and $x_2$ in $[0,1]$.  In fact, we can take any three distinct points.  For example, try $x_0 = 0$, $x_1 = 1$, $x_2 = 1/2$, and interpolate linearly:
$$ f(x) = \cases{ 1 - 2 x & for $0 \le x \le 1/2$\cr
                 -1/2 + x & for $1/2 \le x \le 1$\cr}$$     
If you want $f: [a,b] \to [0,1]$ with $[a,b]$ arbitrary, you'll have trouble if $[0,1]$ and $[a,b]$ don't overlap, because $f \circ f$ won't be defined.
If $[0,1] \cap [a,b] = [c,d]$ is an interval of positive length, then you could 
define $f$ similar to the above on $[c,d]$ and extend to $[a,b]$ so that the
values stay in $[c,d]$.
