📂Topological Data Analysis

Monodromy Theorem Proof

Theorem ¹

Let’s say there are two paths $f_{0} \simeq f_{1}$ that are equivalent starting at point $1 := (1,0)$ on the sphere $S^{1}$. If each of their lifts satisfies $\widetilde{f}_{0}, \widetilde{f}_{1}$ equals $\widetilde{f}_{0} (0) = \widetilde{f}_{1} (0)$, then it follows that $\widetilde{f}_{0} (1) = \widetilde{f}_{1} (1)$.

Description

$1 = (1,0)$

Strictly speaking, $1 = (1,0)$ should be distinguished, but for convenience, if we simply state $1$ as being from the unit circle $S^{1}$, then there’s nothing else to mention besides $(1,0)$. In fact, if one imagines a vertical line spread out on a plane, it’s not that far off from such notation. $$ \mathbb{R} \ni 1 = (1,0) \in \mathbb{R}^{2} $$

Monodromy?

What the statement of the theorem itself says is simply that the lifts of homotopic paths have the same endpoint when they have the same starting point. However, it is ambiguous what this actually implies.

First of all, note that the Monodromy Theorem introduced here applies specifically to $X = S^{1}$. According to the condition of the theorem, when the starting point of the lift is $0$, the endpoints are the same, and being at the same endpoint within $\widetilde{X} = \mathbb{R}$ means that the exact number of turns made around $\widetilde{f}_{0} (0) = \widetilde{f}_{1} (0)$ are identical. This becomes a crucial lemma when demonstrating that the fundamental group $\pi_{1} \left( S^{1}, 1 \right)$ of $S^{1}$ is isomorphic to the integer group $\mathbb{Z}$.

On the other hand, because it leads to a conclusion about how many turns were made, the term Monodromy appears. Although the endpoints of the lifts are continuous, the number of rotations is an integer, thus revealing a Singularity. Instead of being curious about what monodromy means, just accept this as what’s referred to as monodromy.

Proof

Homotopy Lifting Property: A continuous function $F : I^{2} \to S^{1}$ has a lift $\widetilde{F} : I^{2} \to \mathbb{R}$. In particular, given $x_{0} \in S^{1}$ and $\widetilde{x}_{0} \in p^{-1} \left( x_{0} \right)$, $\widetilde{F} \left( 0 , 0 \right) = \widetilde{x}_{0}$ that is $\widetilde{F}$ exists uniquely.

Let the homotopy of $f_{0}$ and $f_{1}$ be $F$. According to the assumption and the Homotopy Lifting Property, a unique lift $\widetilde{F} : I^{2} \to \mathbb{R}$ that is $\widetilde{F} (0,0) = \widetilde{f}_{0} (0) = \widetilde{f}_{1} (0)$ exists. By the definition of homotopic $F(t,s)$, $$ \begin{align*} F (t,0) =& f_{0} (t) \\ F (t,1) =& f_{1} (t) \end{align*} \implies \begin{align*} \widetilde{F} (t,0) =& \widetilde{f}_{0} (t) \\ \widetilde{F} (t,1) =& \widetilde{f}_{1} (t) \end{align*} $$ and since $F(1,t) = f_{0}(1) = f_{1}(1)$, therefore, $\widetilde{F} (1,t)$ is a path from $\widetilde{f}_{0} (1)$ to $\widetilde{f}_{1} (1)$. However, $$ \widetilde{F} (1,t) \in p^{-1} \left( f_{0} (1) \right) \simeq \mathbb{Z} $$ hence, the continuous function $\widetilde{F} (1,t)$, that is, the path $\widetilde{F} (1,t)$, can only be a constant function. Consequently, $\widetilde{f}_{0} (1) = \widetilde{f}_{1} (1)$.

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