Schedules

# Schedules

## Idea

Let $X$ be a topological space and $\mathcal{U}$ an open cover thereof. A (continuous) path $\gamma \colon I \to X$ can pass through many of the elements of $\mathcal{U}$ as it winds its way around $X$. We can decompose that path into segments such that each segment lies wholly inside one of the open sets in $\mathcal{U}$. The Schedule Theorem says that this can be done continuously over all paths in $X$.

This was proved by Dyer and Eilenberg and applied to the question of fibrations over numerable spaces.

## Schedules

The idea of a schedule is that it is a way of decomposing a length into pieces and then assigning a label to each piece. This clearly fits with the stated purpose of these things since we wish to decompose a path into pieces and assign an open set to each piece.

To make this precise, we start with a set of labels. Following Dyer and Eilenberg, let us write this as $A$. Lengths are positive real numbers and so we also need the set of such, Dyer and Eilenberg denote this by $T$; thus $T \coloneqq \mathbb{R}_{\ge 0}$.

###### Definition

The schedule monoid of $A$ is the free monoid on the set $A \times T$. It is written $S A$. Its elements are schedules in $A$.

A schedule in $A$ is thus a finite ordered list of pairs $(a,t)$ where $a \in A$ and $t \in T$.

There are two notions of length for a schedule. There is the word length which simply counts the number of pairs. Then there is the function $l \colon S A \to T$ defined by $l((a_1,t_1) \cdots (a_k,t_k)) = t_1 + \cdots + t_k$. There is also a right action of $T$ on $S A$ which simply multiplies all of the lengths: $((a_1,t_1) \cdots (a_k,t_k)) \cdot t = ((a_1,t_1 t) \cdots (a_k,t_k t))$. Then $l(s t) = l(s) t$.

###### Definition

A schedule is said to be reduced if all of its terms, $(a,t)$, have non-zero length, i.e. $t \gt 0$. The set of reduced schedules forms a submonoid of $S A$ which is written $R S A$.

The empty schedule is reduced.

There is a retraction map $\rho \colon S A \to R S A$ defined by removing all terms with zero length part.

The schedule monoid is given a topology so that the labels are discrete and the lengths topologised as usual. More concretely, given a word $a_1 a_2 \dots a_k$ of elements in $A$, the set of schedules of the form $(a_1,t_1) (a_2,t_2) \cdots (a_k,t_k)$ is in bijection with $T^k$ and we make that bijection a homeomorphism. Then $S A$ is topologised by taking the coproduct over the set of words in $A$. The reduced schedule monoid is topologised as the quotient of this.

## Paths

Let $X$ be a topological space. Let $P X$ denotes its Moore path space. Suppose that we have a family $\mathcal{U}$ of subsets of $X$ indexed by some set $A$. Then we consider a schedule in $A$ as giving an ordered list of these subsets together with the times to be spent in each. For a path in $X$, and a schedule of the appropriate length, then we can ask whether or not the path fits (or obeys) the schedule. We make that precise as follows.

###### Definition

Suppose that we have $\alpha \in P X$ and $s \in S A$, and suppose that $s = (a_1, t_1) \cdots (a_k,t_k)$. Then we say that $\alpha$ fits the schedule s, written $\alpha \Vert s$, if the following conditions hold:

1. $l(\alpha) = l(s)$
2. We can split $\alpha$ into subpaths according to the times $\{t_i\}$. Let $\alpha_i$ be the $i$th segment. Then $\alpha_i \in P U_{a_i}$.

Here, $l \colon P X \to T$ is the function that assigns to a Moore path its length. The schedule designates a decomposition of $[0,l]$ into subintervals with $t_i$ being the length of the $i$th subinterval. Then saying that $\alpha$ fits the schedule $s$ means that $\alpha$ spends the $i$th subinterval in the open set $U_{a_i}$.

## Schedule Theorem

We can now state the main theorem.

###### Theorem

Let $X$ be a topological space. Let $\mathcal{U}$ be a locally finite open covering of $X$ by numerable open sets with indexing set $A$. Then there is a covering $\mathcal{F}$ of $P X$ by closed sets and a family of continuous functions $f \colon F \to S A$, indexed by $F \in \mathcal{F}$ such that:

1. for each $\alpha \in P X$, there some finite subfamily $\{F_1, \dots, F_k\} \subseteq \mathcal{F}$ such that $\alpha$ is in the interior of $\bigcup F_j$,
2. for each $\alpha \in F$, $\alpha \Vert f_F(\alpha)$, and
3. for each $\alpha \in F \cap F'$, $\rho(f_F(\alpha)) = \rho(f_{F'}(\alpha))$

The first condition is purely about the covering. Dyer and Eilenberg use the term local covering for a covering by closed sets with this property.

###### Corollary

There exists a continuous function $h \colon P X \to R S A$ such that $\alpha \Vert h(\alpha)$ if $l(\alpha) \gt 0$ and $h(\alpha) = \Lambda$ if $l(\alpha) = 0$.

Here, $\Lambda \in R S A$ is the empty word.

## Globalisation Theorem

The original motivation for the notion of schedules was to prove the globalisation theorem for (Hurewicz) fibrations.

###### Theorem

Let $p \colon Y \to B$ be a continuous function. Suppose that $\mathcal{U}$ is a locally finite covering of $B$ by numerable open sets with the property that for each $U \in \mathcal{U}$ then the restriction $p_U \colon Y_U \to U$ is a fibration. Then $p$ is a fibration.

The link between the globalisation theorem and the schedule theorem is the characterisation of Hurewicz fibrations in terms of Hurewicz connections.

## Proof of the Schedule Theorem

Let $X$ be a topological space. Let $\mathcal{U}$ be a locally finite open covering of $X$ by numerable open sets and indexing set $A$.

Let us write $A^*$ for the free monoid on $A$. Then there is a function $A^* \times T \to S A$ which takes $(a_1 a_2 \cdots a_k, t)$ to the schedule $(a_1,t/k)(a_2,t/k)\cdots (a_k,t/k)$. We say that a path $\alpha \in P X$ evenly fits $s \in A^*$, and write this as $\alpha \Vert_e s$, if it fits the schedule corresponding to $(s,l(\alpha))$.

We need an initial technical result.

###### Lemma

There is a locally finite covering $\mathcal{W} = \{W_s \mid s \in A^*\}$ of $P X$ by numerable open sets such that for $\alpha \in W_s$ then $\alpha$ evenly fits the word $s$.

###### Remark

Let us explain why this is a reasonable result. Consider a path, $\alpha$, of length $l$. We pull back the cover $\mathcal{U}$ to a cover of $[0,l]$. Using compactness of $[0,l]$ we can replace the pull-back cover by a finite family of open subintervals of $[0,l]$ which cover $[0,l]$. Each subinterval is labelled by an element of $\mathcal{U}$ (though a label might be reused). As the family is finite, the intersections are finite and therefore have a minimum length. Choose $n$ big enough so that $l/n$ is less than this minimum length. Then consider the subdivision of $[0,l]$ given by $\{0,1/n,\dots,l/n\}$. Our conditions on $n$ guarantee that every intersection of subintervals contains at least one of these division points. We can therefore assign to each subinterval of the form $[k/n, (k+1)/n]$ one of the original family of subintervals that contains it. Then we can assign the corresponding element of $\mathcal{U}$. Thus $\alpha$ fits evenly the corresponding word.

Thus the sets $Y_s \coloneqq \{\alpha : \alpha \|_e s\}$ cover $P X$. That they are open follows from the fact that the condition for membership depends on certain compact sets lying in certain open sets and we use the compact-open topology on $P X$.

What is more complicated is reducing the family to a locally finite one.

As $\mathcal{W}$ is locally finite and its elements are numerable, we can choose a numeration that is also a partition of unity. That is, we can choose continuous functions $q_s \colon X \to [0,1]$ with the property that $q_s^{-1}((0,1]) = W_s$ and $\sum_s q_s = 1$.

Let $\mathcal{B}$ be the set of finite subsets of $A^* \setminus \Lambda$ (where $\Lambda$ is the empty word). For $b \in \mathcal{B}$ we define

\begin{aligned} D_b &\coloneqq \{\alpha \in P X \mid \sum_{s \in b} q_s(\alpha) = 1 \} \\ &=\{ \alpha \in P X \mid q_s(\alpha) = 0 \; \text{for all}\; s \notin b\} \end{aligned}

This is a covering of $P X$ by closed sets. As $\mathcal{W}$ is locally finite, for $\alpha \in P X$ there is some neighbourhood $V$ which meets only a finite number of the $\mathcal{W}$. These are indexed by elements of $A^*$, indeed of $A^* \setminus \Lambda$, and so the set of indices is an element, say $b$, of$\mathcal{B}$. Then for $s \notin b$, $q_s \mid V = 0$ and so for $\beta \in V$, $\sum_{s \in b} q_s(\beta) = 1$, whence $V \subseteq D_b$. Thus each $\alpha$ is contained in the interior of some $D_b$.

Now let us put a total ordering on $A^*$. This induces a total ordering on each $b \in \mathcal{B}$ and thus allows us to define the partial sums of the summation $\sum_{s \in b} q_s$. Write these as $Q_i$, with $Q_0$ as the zero function.

Fix $b \in \mathcal{B}$ and write it as $b = \{s_1,s_2,\dots,s_k\}$ in the inherited ordering. Let $e = (l_1,r_1,\dots,l_k,r_k)$ be a list of integers with the property that $1 \le l_i \le r_i \le \#s_i$ where $\#s_i$ is the word length of $s_i$. Define:

$D_{(b,e)} = \left\{ \alpha \in D_b \mid \frac{l_i -1}{\# s_i} \le Q_{i - 1}(\alpha) \le \frac{l_i}{\# s_i} \; \text{and} \; \frac{r_i - 1}{\# s_i} \le Q_i(\alpha) \le \frac{r_i}{\# s_i} \right\}.$

This is closed in $D_b$ and the collection $\{D_{(b,e)}\}$ is a finite cover of $D_b$. The family $\{D_{(b,e)}\}$ ranging over all $b \in \mathcal{B}$ and suitable $e$ is the family $\mathcal{F}$ that we are looking for. It has the required covering property since the interiors of the $D_b$ cover $P X$.

Define $f_{(b,e)} \colon D_{(b,e)} \to S A$ as follows:

$f_{(b,e)}(\alpha) = \sigma_1 \cdots \sigma_k l(\alpha)$

where $\sigma_i$ is the schedule with $\# \sigma_i = r_i - l_i + 1$ and $l(\sigma_i) = q_{s_i}(\alpha)$, and if $s_i = a_1 \cdots a_n$ then if $l_i \lt r_i$ we have

$\sigma_i = \left( a_{l_1}, \frac{l_i}{n} - Q_{i - 1}(\alpha)\right) \left(a_{l_i+1}, \frac{1}{n} \right) \cdots \left(a_{r_i - 1}, \frac{1}{n} \right) \left( a_{r_i}, Q_i(\alpha) - \frac{r_i - 1}{n} \right)$

otherwise, $\sigma_i = (a_{l_i}, Q_i(\alpha) - Q_{i-1}(\alpha))$.

This is continuous and for $\alpha \in D_{(b,e)}$ then $\alpha$ fits $f_{(b,e)}(\alpha)$. Moreover, for $\alpha \in D_{(b,e)} \cap D_{(b',e')}$ then $\rho f_{(b,e)}(\alpha) = \rho f_{(b',e')}(\alpha)$.

## References

Last revised on January 14, 2020 at 07:21:28. See the history of this page for a list of all contributions to it.