Interval covering relation ($J⟶K$)

For closed intervals $J$ and $K$, write $J⟶K$ under $F$ when $F(J)\supseteq K$. The intermediate value theorem implies that there is a closed subinterval $L\subseteq J$ with $F(L)=K$.

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

The covering words $IJJJJ$ and $IJIJJ$ both occur under the interval coverings in period three implies all periods. Their points cannot be the boundary three-cycle. Each has least period five, because five is prime and each itinerary visits both intervals. Their respective one and two visits to the interior of $I$ distinguish the orbits even after cyclic changes of starting point.

Given finitely many intervals $J_1,\dots ,J_r$, their directed covering graph has an directed edge $J_i\to J_j$ whenever $F(J_i)\supseteq J_j$.

Every closed walkin a directed covering graph of an interval map has a point $x\in J_0$ with $F^j(x)\in J_j$ and $F^n(x)=x$. This follows by successively pulling $J_0$ back through the covering relations and applying the fixed-point property of a closed interval. If the itinerary has least period $n$ and avoids shared endpoints, $x$ lies on an $n$-cycle.

If $A$ is the adjacency matrix of a directed covering graph, then $\mathrm{tr}(A^n)$ counts its pointed closed walks of length $n$. For a prime number $p$, subtracting the $\mathrm{tr}(A)$ constant walks and identifying the $p$ cyclic choices of starting point givesprimitive closed itineraries of length $p$.

For cycle points ordered $x_3<x_1<x_0<x_2<x_4$, successive gap vertices have arrows $0\to 3$, $1\to 1,2$, $2\to 0$, $3\to 0,1$. Their transition-matrix traces through length four are $1,3,1,7$, forcing periods one, two and four but not three. The affine interpolation through ordered values $(4,3,1,0,2)$ has no three-cycle: its only length-three closed interval itinerary is $111$, on which the map is $5-2x$ and its third iterate has only its fixed point. Two distinct length-four return words at vertex one give a horseshoe for the fourth iterate.

A five-cycle whose points are in orbit order along the real line gives four interval vertices with arrows $0\to 1\to 2\to 3$ and $3\to 0,1,2,3$. The transition-matrix traces for lengths one through four are $1,3,7,15$. Primitive closed words force at least one fixed point, one two-cycle, two three-cycles and three four-cycles. The second iterate has a two-branch horseshoe for an interval map on vertices two and three. The map itself need not have a one-step horseshoe: the unimodal affine example $F(x)=x+1$ for $x\le 3$, $F(x)=16-4x$ for $x\ge 3$ has precisely this five-cycle and has no two disjoint self-covering branches.