📂Topological Data Analysis

Covering and Lifting in Algebraic Topology

Definitions ^{1 2}

Let $p : \widetilde{X} \to X$ be a continuous function between two topological spaces $\widetilde{X}, X$. Denote any index set as $\forall$, and let’s write the restriction function from $\widetilde{U}_{\alpha} \subset \widetilde{X}$ to $p$ simply as $p |_{\widetilde{U}_{\alpha}} : \widetilde{U}_{\alpha} \to U$.

• $I = [0,1]$ is the unit interval from $0$ to $1$.
• $\bigsqcup$ represents the union of disjoint sets.

Covering

1. An open set $U \subset X$ of $X$ is evenly covered by $p$ if, for all $\alpha \in \forall$ corresponding, all restriction functions $p |_{\widetilde{U}_{\alpha}}$ are homeomorphisms, and $$ \alpha_{1} \ne \alpha_{2} \implies \widetilde{U}_{\alpha_{1}} \cap \widetilde{U}_{\alpha_{2}} = \emptyset $$ is satisfied, that is, if $$ p^{-1} \left( U \right) = \bigsqcup_{\alpha \in \forall} \widetilde{U}_{\alpha} $$ applies for the disjoint open set $\widetilde{U}_{\alpha} \subset \widetilde{X}$ of $\widetilde{X}$.
2. If $p : \widetilde{X} \to X$ is a surjective function, and there exists an open neighborhood $U_{x} \subset X$ of $x$ that is evenly covered by $p$ for all $x \in X$, then $p : \widetilde{X} \to X$ is called a covering.
3. The domain $\widetilde{X}$ of covering $p$ is called covering space, and the codomain $X$ is called base space.

Lift

4. Let’s denote by $n \in \mathbb{N}$. If $f : I^{n} \to X$ and $\widetilde{f} : I^{n} \to \widetilde{X}$ satisfy the following, $\widetilde{f}$ is called the lift of $f$. $$ f = p \circ \widetilde{f} $$

Examples

Mathematical definitions are too complicated, let’s consider a simple example with $X = S^{1}$ and $\widetilde{X} = \mathbb{R}$. Frankly, the notions of covering and lift in the definitions generalize this example.

Intuitive Lift

Although written as $\widetilde{X} = \mathbb{R}$, this is depicted as a spiral embedded in $\mathbb{R}^{3}$, which is analogous to representing the spiral $h : \mathbb{R} \to \mathbb{R}^{3}$ as $$ s \mapsto \left( \cos 2 \pi s, \sin 2 \pi s , s \right) $$ Defining a path from $I = [0,1]$ to $\mathbb{R}$ as $$ \widetilde{\omega}_{n} (s) := ns $$ results in starting at $0$ and ending at $n$, winding around the spiral for $n \in \mathbb{Z}$ turns. Meanwhile, a sphere $S^{1}$ can be represented as a unit circle in $$ \omega_{n} (s) := \left( \cos 2 \pi n s , \sin 2 \pi n s \right) $$ -dimensional space, naturally making the projection $p : (x,y,z) \mapsto (x,y)$ a covering. Intuitively, $p$ appears as a projection that sends the unwound spiral onto a plane, whereas $\widetilde{\omega}_{n}$ lifts the repeatedly overlapped $\omega_{n}$ into three-dimensional space, making it aptly called a lift. Expressed in formula, it becomes $$ \omega_{n} = p \circ \widetilde{\omega}_{n} $$ Now looking back at the definition, so far we have one intuitive example of $I^{1}$, and there is no reason not to call them covering or lift if all listed conditions are met. In the context of algebraic topology, an immediate possibility to consider is the lift of $I^{2}$, i.e., the lift in homotopy $H : I^{2} \to X$.

Evenly Cover is Too Hard

The evenly cover written in the definitions might seem incredibly hard, but intuitively, it’s actually a simple concept.

The pre-image of $U \subset S^{1}$ is represented as the union of disjoint sets on the spiral, each being homeomorphic with little pieces $U$. However, in this example, indices are conveniently assigned to correspond to integers $k$, and the form is simple, but realistically, how peculiar the index set $\forall$ could be is unpredictable. Therefore, despite most mathematicians preferring easy and straightforward definitions, there’s no compromise on the description of evenly cover.