Proof of the Lifting Theorem in Algebraic Topology
Theorem 1 2
Definitions of Covering and Lifting: Let’s denote the unit interval as $I = [0,1]$.
- An open set $U \subset X$ of $X$ is evenly covered by $p$ if for every $\alpha \in \forall$, all corresponding restricted functions $p |_{\widetilde{U}_{\alpha}}$ are homeomorphisms, and $$ \alpha_{1} \ne \alpha_{2} \implies \widetilde{U}_{\alpha_{1}} \cap \widetilde{U}_{\alpha_{2}} = \emptyset $$ holds, meaning there exist disjoint open sets $\widetilde{U}_{\alpha} \subset \widetilde{X}$ in $\widetilde{X}$ such that $$ p^{-1} \left( U \right) = \bigsqcup_{\alpha \in \forall} \widetilde{U}_{\alpha} $$ is satisfied.
- If $p : \widetilde{X} \to X$ is a surjective function and for every $x \in X$, there exists an open neighborhood $U_{x} \subset X$ of $x$ that is evenly covered by $p$, then $p : \widetilde{X} \to X$ is called a covering.
- The domain $\widetilde{X}$ of the covering $p$ is called the covering space, and the codomain $X$ is called the base space.
- Let $n \in \mathbb{N}$. If $f : I^{n} \to X$ and $\widetilde{f} : I^{n} \to \widetilde{X}$ satisfy the following, $\widetilde{f}$ is called a lift of $f$. $$ f = p \circ \widetilde{f} $$
Let’s denote a covering with sphere $S^{1}$ as its codomain by $p : \mathbb{R} \to S^{1}$.
Path Lifting Theorem
A continuous function $f : I \to S^{1}$ has a lift $\widetilde{f} : I \to \mathbb{R}$. Specifically, for given $x_{0} \in S^{1}$ and $\widetilde{x}_{0} \in p^{-1} \left( x_{0} \right)$, there exists a unique $\widetilde{f}$ satisfying $\widetilde{f} \left( 0 \right) = \widetilde{x}_{0}$.
Homotopy Lifting Theorem
A continuous function $F : I^{2} \to S^{1}$ has a lift $\widetilde{F} : I^{2} \to \mathbb{R}$. Specifically, for given $x_{0} \in S^{1}$ and $\widetilde{x}_{0} \in p^{-1} \left( x_{0} \right)$, there exists a unique $\widetilde{F}$ satisfying $\widetilde{F} \left( 0 , 0 \right) = \widetilde{x}_{0}$.
Explanation
Lifting Theorems are often mentioned as auxiliary theorems for studying the properties of the unit circle $S^{1}$. Formally, the distinction between path lifting and homotopy lifting is not significant.
The question most mathematicians should be curious about is whether a generalization for $f: I^{m} \to X$, when $X \ne S^{1}$, is possible. It’s a fact that lifting theorems can be discussed even for continuous functions $f: Y \times I^{m} \to X$ over compact spaces $Y$. However, such extensions are said to be overly complicated and practically useless for direct study.
Proof
Strategy: We’ll only prove the path lifting theorem. Essentially, the proof of the homotopy lifting theorem is similar to that of the path lifting theorem. Like how the path lifting theorem is proved by subdividing the compact space $I$ into finite intervals, the homotopy lifting theorem involves subdividing the compact space $I^{2}$ into finite parts and repeating the same discussion.
Part 1. Setting
- If $p : \widetilde{X} \to X$ is a surjective function and for every $x \in X$, there exists an open neighborhood $U_{x} \subset X$ of $x$ that is evenly covered by $p$, then $p : \widetilde{X} \to X$ is called a covering.
Since $p : \mathbb{R} \to S^{1}$ is a covering, for every $x \in S^{1}$, there exists a neighborhood $U_{x} \subset S^{1}$ of $x$ that is evenly covered by $p$.
Since $I = [0,1]$ is compact, there exists a finite set of points $\left\{ a_{k} \right\}_{k=0}^{n} \subset I$ satisfying
$$
0 = a_{0} < a_{1} < \cdots < a_{n-1} < a_{n} = 1
$$
and whose intervals $\left[ a_{k-1} , a_{k} \right] \subset I$ have images in $S^{1}$, especially satisfying the inclusion relation for some open set $U \subset S^{1}$ as follows:
$$
f \left( \left[ a_{k-1} , a_{k} \right] \right) \subset U \subset S^{1}
$$
If we denote the disjoint preimages of $U$ under covering $p$ by $\widetilde{U}_{t} := p^{-1} \left( U_{t} \right)$, each of them is homeomorphic to $U$ for every $t \in \mathbb{Z}$.
Part 2. Inductive Construction
Let’s specifically fix $x_{0} \in S^{1}$ instead of any $x \in S^{1}$, and denote one of the preimages under $p$ as $\widetilde{x}_{0} := p^{-1} \left( x_{0} \right) \in \mathbb{R}$. According to the initial setting, there exists a bijection between these elements and $\mathbb{Z}$, and it doesn’t matter which one is chosen.
We aim to define the lifting $\widetilde{f}_{k}$ inductively for $\left[ 0, a_{k} \right]$, satisfying $\widetilde{f}_{k} (0) = \widetilde{x}_{0}$, instead of the entire $I$ at once, to eventually find $\widetilde{f}$.
- For $k = 0$, we simply set it as $\widetilde{f}_{0} (0) = \widetilde{x}_{0}$, with no other choice.
- Assume a unique continuous function $\widetilde{f}_{k} : \left[ 0 , a_{k} \right] \to \mathbb{R}$ is defined for $k \ne 0$.
- For some unique $\widetilde{U} \in \left\{ \widetilde{U}_{t} \right\}_{t \in \mathbb{Z}}$, $\widetilde{f} \left( a_{k} \right) \in \widetilde{U}$ holds.
- Since $\widetilde{f}_{k}$ is continuous and interval $\left[ a_{k} , a_{k+1} \right]$ is path-connected, the extension function $\widetilde{f}_{k+1}$, however defined, must map $\left[ a_{k} , a_{k+1} \right]$ strictly inside $\widetilde{U}$.
- Since $p$ is a covering, a homeomorphism $p | \widetilde{U}_{t} : \widetilde{U}_{t} \to U$ exists for all $t \in \mathbb{Z}$, leading to a unique function $\rho_{k} : \left[ a_{k} , a_{k+1} \right] \to \widetilde{U}$ satisfying $$ p \circ \rho_{k} = f | \left[ a_{k} , a_{k+1} \right] $$ The existence of such function $\rho_{k}$ is based on the existence of a homeomorphism as a restriction of $p$—in other words, due to its injectivity, ensuring $\rho_{k} \left( a_{k} \right) = \widetilde{f}_{k} \left( a_{k} \right)$ and the continuity of $\rho_{k}$.
Gluing Lemma: For a topological space $X,Y$ and two closed sets $A,B \subset X$ satisfying $A \cup B = X$, and two continuous functions $f : A \to Y$ and $g : B \to Y$ agree on all $x \in A \cap B$ as $f(x) = g(x)$, then the function $h$ defined as follows is continuous. $$ h(x) : = \begin{cases} f(x), & x \in A \\ g(x), & x \in B \end{cases} $$
- By the Gluing Lemma, a unique continuous function $\widetilde{f}_{k+1} : \left[ 0 , a_{k+1} \right] \to \mathbb{R}$ can be defined as follows: $$ \widetilde{f}_{k+1} := \begin{cases} \widetilde{f}_{k} (s) & , \text{if } s \in \left[ 0, a_{k} \right] \\ \rho_{k} (s) & , \text{if } s \in \left[ a_{k} , a_{k+1} \right] \end{cases} $$
According to mathematical induction, there specifically exists a lifting to the spiral winding around $S^{1}$ as many times as the wheel of $t \in \mathbb{Z}$. Here, $k = 0, 1, \cdots , n$ is not an index moving up and down in $\mathbb{R}$ but an index that slices $S^{1}$ into finite parts while rotating.
Part 3. Homotopy Lifting Theorem
- Let’s denote the set of integers from $0$ to $n-1$ simply as $0:n$.
Just as we could choose $0 = a_{0} < \cdots < a_{n} = 1$ based on the compactness of $I$, $I^{2}$ is also compact, allowing for the existence of finite natural numbers $n , m \in \mathbb{N}$ that subdivide the square into a grid. $$ \begin{align*} 0 = a_{0} < \cdots < a_{n} = 1 \\ 0 = b_{0} < \cdots < b_{m} = 1 \end{align*} $$ Defining each small square for $i = 0:n$ and $j = 0:m$ as $$ R_{i,j} := \left[ a_{i-1}, a_{i} \right] \times \left[ b_{j-1} , b_{j} \right] \subset I^{2} $$ yields a sequence of small rectangles as follows: $$ R_{0,0} , R_{0,1} , \cdots , R_{0,m} , R_{1,0} \cdots, R_{n,m} $$ Repeating the discussion from the path lifting theorem proves the homotopy lifting theorem.
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