A Universe of Sorts

§ Covering spaces: Intuition

Consider the map $p(z) = z^2 : \mathbb C^\times \rightarrow \mathbb C^\times$ . This is a 2-to-1 map. We can try to define an inverse regardless.
We do define a "square root" if we want. Cut out a half-line $[0, -infty)$ called $B$ for branch cuts. We get two functions on $q_+, q_-: \mathbb C - B \rightarrow \mathbb C^\times$ , such that $p(q_+(z)) = z$ . Here, we have $q_- = - q_+$ .
The point of taking the branch cut is to preserve simply connectedness. $\mathbb C^\times$ is not simply connected, while $\mathbb C/B$ is simply connected! (This seems so crucial, why has no one told me this before?!)
Eg 2: exponential. Pick $exp: \mathbb C \rightarrow \mathbb C^\times$ . This is surjective, and infinite to 1. $e^{z + 2 \pi n} = e^{iz}$ .
Again, on $\mathbb C / B$ , we have $q_n \equiv \log + 2 \pi i n$ , such that $exp(q_n(z)) = z$ .
A covering map is, roughly speaking, something like the above. It's a map that's n-to-1, which has n local inverse defined on simply connected subsets of the target.
So if we have $p: Y \rightarrow X$ , we have $q: U \rightarrow Y$ (for $U \subseteq X$ ) such that $p(q(z)) = z, \forall z \in U$ .

§ Covering spaces: Definition

A subset $U \subset X$ is a called as an elementary neighbourhood if there is a discrete set $F$ and a homeomorphism $h: p^{-1}(U) \rightarrow U \times F$ such that $p|_{p^{-1}(U)}(y) = fst(h)$ or $p|_{p^{-1}(U)} = pr_1 \circ h$ .
Alternative definition A subset $U \subset X$ is called as evenly covered/elementary nbhd if $p^{-1}(U) = \sqcup \alpha V_\alpha$ where the $V_\alpha$ are disjoint and open, and $p|_{V_\alpha} : V_\alpha \rightarrow U$ is a homeomorphism for all $\alpha$ .
An elementary neighbourhood is the region where we have the local inverses (the complement of a branch cut).
We get for each $i \in F$ , a map $q_i : U \rightarrow U \times F; q_i(x) = (x, i)$ and then along $h^{-1}$ sending $h^{-1}(x, i) \in p^{-1}(U)$ .
We say $p$ is a covering map if $X$ is covered by elementary neighbourhoods.
We say $V \subseteq Y$ is an elementary sheet if it is path connected and $p(V)$ is an elementary neighbourhood.
So, consider $p(x) = e^{ix}: \mathbb R \rightarrow S^1$ . If we cut the space at $(0, 0)$ , then we will have elementary neighbourhood $S^1 - \{(0, 0)\}$ and elementary sheets $(2 \pi k, 2 \pi+1)$ .
The point is that the inverse projection $p^{-1}$ takes $U$ to some object of the form $U \times F$ : a local product! So even though the global covering space $\mathbb R$ does not look like a product of circles, it locally does. So it's some sort of fiber bundle?

Slogan: Covering space is locally disjoint copies of the original space.

§ Path lifting and Monodromy

Monodromy is behaviour that's induced in the covering space, on moving in a loop in a base.
Etymology: Mono --- single, drome --- running. So running in a single loop / running around a single time.
Holonomy is a type of monodromy that occurs due to parallel transport in a loop, to detect curvature
Loop on the base is an element of $\pi_1(X)$ .
Pick some point $x \in X$ . Consider $F \equiv \pi^{-1}(x)$ ( $F$ for fiber).
Now move in a small loop on the base, $\gamma$ . The local movement will cause movement of the elements of the fiber.
Since $\gamma(1) = \gamma(0)$ , the elements of the fiber at the end of the movement are equal to the original set $F$ .
So moving in a loop induces a permutation of the elements of the fiber $F$ .
Every element of $\pi_1(X)$ induces a permutation of elements of the fiber $F$ .
This lets us detect non-triviality of $\pi_1(X)$ . The action of $\pi_1(X)$ on the fiber lets us "detect" what $\pi_1(X)$ is.
We will define what is means to "move the fiber along the path".

§ Path lifting lemma

Theorem :Suppose $p: y \rightarrow X$ is a covering map. Let $\delta: [0, 1] \rightarrow X$ be a path such that $\delta(0) = x$ , and let $y \in p^{-1}(x)$ [ $y$ is in the fiber of $x$ ]. Then there is a unique path $\gamma: [0,1] \rightarrow Y$ which "lifts" $\delta$ . That is, $\delta(p(y)) = \gamma(y)$ , such that $\gamma(0) = Y$ .

Slogan: Paths can be lifted. Given how to begin the lift, can be extended all the way.

Let $N$ be a collection of elementary neighbourhoods of $X$ .
$\{ \delta^{-1}(U) : U \in N \}$ is an open cover (in the compactness sense) of $[0, 1]$ .
By compactness, find a finite subcover. Divide interval into subintervals $0 = t_0 < t_1 < \dots t_n = 1$ such that $\delta|k = \delta|_{[t_k, t_{k+1}]}$ lands in $U_k$ , an elementary neighbourhood.
Build $\gamma$ by induction on $k$ .
We know that $\gamma(0)$ should be $y$ .
Since we have an elementary neighbourhood, it means that there are a elementary sheets living over $U_0$ , indexed by some discrete set $F$ . $y$ lives in one of thse sheets. We have local inverses $q_m$ . One of them lands on the sheet of $y$ , call it $q$ . So we get a map $q: U_0 \rightarrow Y$ such that $q(x) = y$ .
Define $\gamma(0) \equiv q(\delta(0)) = q(x) = y$ .
Extend $\gamma$ upto $t_1$ .
Continue all the way upto $t_k$ .
To get $\gamma$ from $(t_k, t_{k+1}$ , there exists a $q_k: U_k \rightarrow Y$ such that $q_k(\delta(t_k)) = \gamma(t_k)$ . Define $\gamma(t_k \leq t \leq t_{k_1}) \equiv q_k(\delta(t_k))$ .
This is continuous because $\delta$ continuous by definition, $q_k$ continuous by neighbourhood, $\gamma$ is pieced together such that endpoints fit, and is thus continuous.
Can check this is a lift! We get $p \circ \gamma = p \circ q_k \circ \delta_k$ . Since $q_k$ is a local inverse of $p$ , we get $p \circ \gamma = \delta_k$ in the region.

§ 7.03: Path lifting: uniqueness

If we have a space $X$ and a covering space $Y$ , for a path $gamma$ that starts at $x$ , we can find a path $\gamma'$ which starts at $y \in p^{-1}(x)$ and projects down to $\gamma$ : $\gamma(t) = p(\gamma'(t))$ . We want to show that this path lift is unique

§ Lemma

Let $p: Y \rightarrow X$ be a covering space. Let $T$ be a connected space Let $F: T \rightarrow X$ be a continuous map (for us, $T \simeq [0, 1]$ ). Let $F_1, F_2: T \rightarrow Y$ be lifts of $F$ ( $p \circ F_1 = F$ , $p \circ F_2 = F$ ). We will show that $F_1 = F_2$ iff the lifts are equal for some $t \in$ T.

Slogan: Lifts of paths are unique: if they agree at one point, they agree at all points!

We just need to show that if $F_1$ and $F_2$ agree somewhere in $Y$ , they agree everywhere. It is clear that if they agree everywhere, they must agree somewhere.
To show this, pick the set $S$ where $F_1, F_2$ agree in $Y$ : $S \equiv \{ t \in T : F_1(t) = F_2(t) \}$ .
We will show that $S$ is open and closed. Since $T$ is connected, $S$ must be either the full space or the empty set. Since $S$ is assumed to be non-empty, $S = T$ and the two functions agree everywhere.
(Intuition: if both $S$ and $S^c$ are open, then we can build a function that colors $T = S \cup S^c$ in two colors continuously; ie, we can partition it continuously; ie the spaces must be disconnected. Since $T$ is connected, we cannot allow that to happen, hence $S = \emptyset$ or $S = T$ .)
Let $t \ in T$ . Let $U$ be an evenly covered neighbourhood/elementary neighbourhood of $F(t)$ downstairs (in $X$ ). Then we have $p^{-1}(U) = \sqcup_\alpha V_\alpha$ such that $p|_V{\alpha}: V_\alpha \rightarrow U$ is a local homeomorphism.
Since $F_1, F_2$ are continuous, we will have opens $V_1, V_2$ in $V_\alpha$ , which contain $F_1(t), F_2(t)$ upstairs (mirrroring $U$ containing $F(t)$ downstairs).
The pre-images of $V_1$ , $V_2$ along $F_1, F_2$ give us open sets $t \in T_1, T_2 \subseteq T$ .
Define $T* = T_1 \cap T_2$ . If $F_1(t) \neq F_2(t)$ , then $V_1 \neq V_2$ and thus $F_1 \neq F_2$ on all of $T*$ . So, $S^c = T*$ is open.
If $F_1(t) = F_2(t)$ , then $V_1 = V_2$ and thus $F_1 = F_2$ on $T*$ (since $p \circ F_1 = F = p \circ F_2$ , and $p$ is injective within $U$ , ie within $V_1, V_2$ ). So $S$ is open.
Hence we are done, as $S$ is non-empty and clopen and is thus equal to $T$ . Thus, the two functions agree on all of $T$ .

§ Homotopy lifting, Monodromy

Given a loop $\gamma$ in $X$ based at $x$ , the monodromy around $\gamma$ is a permutation $\sigma_\gamma : p^{-1}(x) \rightarrow p^{-1}(x)$ , where $\sigma_{\gamma}(y) \equiv \gamma^y(1)$ where $\gamma^y$ is the unique lift of $\gamma$ staring at $y$ . We have that $\sigma_{\gamma} \in Perm(p^{-1}(x))$ .
Claim: if $\gamma_1 \simeq \gamma_2$ then $\sigma_{\gamma_1} = \sigma_{\gamma_2}$ .
We need a tool: homotopy lifting lemma.

Slogan: permutation of monodromy depends only on homotopy type

§ Homotopy lifting lemma/property of covering spaces

Suppose $p: Y \rightarrow X$ is a covering map and $\gamma_s$ is a homotopy of paths rel. endpoints ( $\gamma_s(0)$ and $\gamma_s(1)$ are independent of $s$ / endpoints are fixed throughout the homotopy). Then there exists for each lift $\gamma'_0 : [0, 1] \rightarrow Y$ of $\gamma_0:[0,1] \rightarrow X$ (ie, $p \circ \gamma'_0 = gamma_0$ ), a completion of the lifted homotopy $\gamma'_s: [0, 1] \rightarrow Y$ (ie, $p \circ gamma'_s = gamma_s$ ). Moreover, this lifted homotopy is rel endpoints: ie, the endpoints of $gamma'$ are independent of $s$ .

Slogan: homotopy lifted at 0 can be lifted for all time

Let $H: [0, 1] \times [0, 1] \rightarrow X$ be the homotopy in $X$ such that $H(s, t) = \gamma_s(t)$ . Subdivide the square into rectangles $R_{ij}$ such that $H(R_{ij})$ is contained in $U_{ij}$ for some elementary neighbourhood $U_{ij}$ . We build $H': [0, 1] \times [0, 1] \rightarrow Y$ by building local inverses $q_{ij} : U_{ij} \rightarrow Y$ such that $p \circ q_{ij} = R_{ij}$ . We then set $H'|_{R_{ij}} = q_{ij} \circ H$ .

§ Covering Spaces