nLab
schedule

Idea

Let $X$ be a topological space and $𝒰$ an open cover thereof. A (continuous) path $\gamma :I\to X$ can pass through many of the elements of $𝒰$ 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 $𝒰$. 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 $T≔{ℝ}_{\ge 0}$.

Definition

The schedule monoid of $A$ is the free monoid on the set $A\times T$. It is written $SA$. 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 $l:SA\to T$ 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 $SA$ 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 $l(st)=l(s)t$.

Definition

A schedule is said to be reduced if all of its terms, $(a,t)$, has non-zero length, i.e. $t>0$. The set of reduced schedules forms a submonoid of $SA$ which is written $RSA$.

The empty schedule is reduced.

There is a retraction map $\rho :SA\to RSA$ 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 $SA$ 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 $PX$ denotes its Moore path space. Suppose that we have a family $𝒰$ 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 $\alpha \in PX$ and $s\in SA$, and suppose that $s=(a_1,t_1)\cdots (a_k,t_k)$. Then we say that $\alpha$ fits the schedule s, written $\alpha \parallel s$, 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, $l:PX\to T$ 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 $𝒰$ be a locally finite open covering of $X$ by numerable open sets with indexing set $A$. Then there is a covering $ℱ$ of $PX$ by closed sets and a family of continuous functions $f:F\to SA$, indexed by $F\in ℱ$ such that:

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

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 $h:PX\to RSA$ such that $\alpha \parallel h(\alpha )$ if $l(\alpha )>0$ and $h(\alpha )=\Lambda$ if $l(\alpha )=0$.

Here, $\Lambda \in RSA$ 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 $p:Y\to B$ be a continuous function. Suppose that $𝒰$ is a locally finite covering of $B$ by numerable open sets with the property that for each $U\in 𝒰$ then the restriction $p_U:Y_U\to U$ 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 $𝒰$ 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 $A^{*}\times T\to SA$ 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 $\alpha \in PX$ evenly fits $s\in A^{*}$, and write this as $\alpha {\parallel }_{e}s$, if it fits the schedule corresponding to $(s,l(\alpha ))$.

We need an initial technical result.

Lemma

There is a locally finite covering $𝒲=\{W_s\mid s\in A^{*}\}$ of $PX$ 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 $𝒰$ 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 $𝒰$ (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 $𝒰$. Thus $\alpha$ fits evenly the corresponding word.

Thus the sets $Y_s≔\{\alpha :\alpha {\parallel }_{e}s\}$ cover $PX$. 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 $PX$.

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

As $𝒲$ 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 $q_s:X\to [0,1]$ with the property that $q_s^{-1}((0,1])=W_s$ and ${\sum }_s{q}_s=1$.

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

$$\begin{array}{rl}{D}_b& ≔\{\alpha \in PX\mid \sum _{s\in b}{q}_s(\alpha )=1\}\\ & =\{\alpha \in PX\mid q_s(\alpha )=0\text{for all}s\notin b\}\end{array}$$\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 $PX$ by closed sets. As $𝒲$ is locally finite, for $\alpha \in PX$ there is some neighbourhood $V$ which meets only a finite number of the $𝒲$. These are indexed by elements of $A^{*}$, indeed of $A^{*}\setminus \Lambda$, and so the set of indices is an element, say $b$, of$ℬ$. 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 $b\in ℬ$ 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 $b\in ℬ$ 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:

$$D_{(b,e)}=\{\alpha \in D_b\mid \frac{l_i-1}{\#s_i}\le Q_{i-1}(\alpha )\le \frac{l_i}{\#s_i}\text{and}\frac{r_i-1}{\#s_i}\le Q_i(\alpha )\le \frac{r_i}{\#s_i}\}.$$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 $b\in ℬ$ and suitable $e$ is the family $ℱ$ that we are looking for. It has the required covering property since the interiors of the $D_b$ cover $PX$.

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

$$f_{(b,e)}(\alpha )={\sigma }_1\cdots {\sigma }_{k}l(\alpha )$$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 $l_i<r_i$ we have

$${\sigma }_i=(a_{l_1},\frac{l_i}{n}-Q_{i-1}(\alpha ))(a_{l_i+1},\frac{1}{n})\cdots (a_{r_i-1},\frac{1}{n})(a_{r_i},Q_i(\alpha )-\frac{r_i-1}{n})$$\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 $\alpha \in D_{(b,e)}\cap D_{(b\prime ,e\prime )}$ then $\rho f_{(b,e)}(\alpha )=\rho f_{(b\prime ,e\prime )}(\alpha )$.

References

• Dyer, E. and E., Samuel. (1988). Globalizing fibrations by schedules. Fund. Math., 130, 125–136. MR0963792

• Dyer, Eilenberg, MR0963792