\chapter{Covering projections} A few chapters ago we talked about what a fundamental group was, but we didn't actually show how to compute any of them except for the most trivial case of a simply connected space. In this chapter we'll introduce the notion of a \emph{covering projection}, which will let us see how some of these groups can be found. \section{Even coverings and covering projections} \prototype{$\RR$ covers $S^1$.} What we want now is a notion where a big space $E$, a ``covering space'', can be projected down onto a base space $B$ in a nice way. Here is the notion of ``nice'': \begin{definition} Let $p : E \to B$ be a continuous function. Let $U$ be an open set of $B$. We call $U$ \vocab{evenly covered} (by $p$) if $p\pre(U)$ is a disjoint union of open sets (possibly infinite) such that $p$ restricted to any of these sets is a homeomorphism. \end{definition} Picture: \begin{center} \includegraphics[width=4cm]{media/even-covering.png} \\ \scriptsize Image from \cite{img:even_covering} \end{center} All we're saying is that $U$ is evenly covered if its pre-image is a bunch of copies of it. (Actually, a little more: each of the pancakes is homeomorphic to $U$, but we also require that $p$ is the homeomorphism.) \begin{definition} A \vocab{covering projection} $p : E \to B$ is a surjective continuous map such that every base point $b \in B$ has an open neighborhood $U \ni b$ which is evenly covered by $p$. \end{definition} \begin{exercise} [On requiring surjectivity of $p$] Let $p \colon E \to B$ be satisfying this definition, except that $p$ need not be surjective. Show that the image of $p$ is a connected component of $B$. Thus if $B$ is connected and $E$ is nonempty, then $p \colon E \to B$ is already surjective. For this reason, some authors omit the surjectivity hypothesis as usually $B$ is path-connected. \end{exercise} Here is the most stupid example of a covering projection. \begin{example}[Tautological covering projection] Let's take $n$ disconnected copies of any space $B$: formally, $E = B \times \{1, \dots, n\}$ with the discrete topology on $\{1, \dots, n\}$. Then there exists a tautological covering projection $E \to B$ by $(x,m) \mapsto x$; we just project all $n$ copies. This is a covering projection because \emph{every} open set in $B$ is evenly covered. \end{example} This is not really that interesting because $B \times [n]$ is not path-connected. A much more interesting example is that of $\RR$ and $S^1$. \begin{example}[Covering projection of $S^1$] Take $p : \RR \to S^1$ by $\theta \mapsto e^{2\pi i \theta}$. This is essentially wrapping the real line into a single helix and projecting it down. \end{example} \missingfigure{helix} We claim this is a covering projection. Indeed, consider the point $1 \in S^1$ (where we view $S^1$ as the unit circle in the complex plane). We can draw a small open neighborhood of it whose pre-image is a bunch of copies in $\RR$. \begin{center} \begin{asy} size(12cm); real[] t = {-2,-1,0,1,2}; xaxis(-3.5,3.5, graph.LeftTicks(Ticks=t), Arrows); pen bloo = blue+1.5; dotfactor *= 2; pair A,B; for (real x = -2; x <= 2; ++x) { A = (x-0.2, 0); B = (x+0.2, 0); draw(A--B, bloo); opendot(A, blue); opendot(B, blue); } MP("\mathbb R", (3,0), dir(90)); add(shift( (0,3) ) * CC()); path darrow = (0,2.5)--(0,1.5); MP("p", midpoint(darrow), dir(0)); draw(darrow, EndArrow); real r = 1.4; draw(scale(r)*unitcircle); MP("S^1", r*dir(45), dir(45)); A = r*dir(-20); B = r*dir(20); draw(arc(origin, A, B), bloo); opendot(A, blue); opendot(B, blue); dot("$1$", r*dir(0), dir(0)); \end{asy} \end{center} Note that not all open neighborhoods work this time: notably, $U = S^1$ does not work because the pre-image would be the entire $\RR$. \begin{example}[Covering of $S^1$ by itself] The map $S^1 \to S^1$ by $z \mapsto z^{3}$ is also a covering projection. Can you see why? \end{example} \begin{example} [Covering projections of $\CC \setminus \{0\}$] For those comfortable with complex arithmetic, \begin{enumerate}[(a)] \ii The exponential map $\exp : \CC \to \CC \setminus \{0\}$ is a covering projection. \ii For each $n$, the $n$th power map $-^n: \CC \setminus \{0\} \to \CC \setminus \{0\}$ is a covering projection. \end{enumerate} \end{example} \section{Lifting theorem} \prototype{$\RR$ covers $S^1$.} Now here's the key idea: we are going to try to interpret loops in $B$ as paths in $\RR$. This is often much simpler. For example, we had no idea how to compute the fundamental group of $S^1$, but the fundamental group of $\RR$ is just the trivial group. So if we can interpret loops in $S^1$ as paths in $\RR$, that might (and indeed it does!) make computing $\pi_1(S^1)$ tractable. \begin{definition} Let $\gamma : [0,1] \to B$ be a path and $p : E \to B$ a covering projection. A \vocab{lifting} of $\gamma$ is a path $\tilde\gamma : [0,1] \to E$ such that $p \circ \tilde\gamma = \gamma$. \end{definition} Picture: \begin{center} \begin{tikzcd} & E \ar[d, "p"] \\ {[0,1]} \ar[r, "\gamma"'] \ar[ru, "\tilde{\gamma}"] & B \end{tikzcd} \end{center} \begin{example}[Typical example of lifting] Take $p : \RR \to S^1 \subseteq \CC$ by $\theta \mapsto e^{2 \pi i \theta}$ (so $S^1$ is considered again as the unit circle). Consider the path $\gamma$ in $S^1$ which starts at $1 \in \CC$ and wraps around $S^1$ once, counterclockwise, ending at $1$ again. In symbols, $\gamma : [0,1] \to S^1$ by $t \mapsto e^{2\pi i t}$. Then one lifting $\tilde\gamma$ is the path which walks from $0$ to $1$. In fact, \emph{for any integer $n$}, walking from $n$ to $n+1$ works. \begin{center} \begin{asy} size(6cm); real[] t = {-1,0,1,2}; xaxis(-2,3, graph.LeftTicks(Ticks=t), Arrows); MP("\mathbb R", (2.5,0), dir(90)); path gt = (0,0.3)--(1,0.3); draw(gt, blue, EndArrow); label("$\tilde\gamma$", midpoint(gt), dir(90), blue); add(shift( (0,3) ) * CC()); path darrow = (0,2.5)--(0,1.5); MP("p", midpoint(darrow), dir(0)); draw(darrow, EndArrow); real r = 1.2; draw(scale(r)*unitcircle); MP("S^1", r*dir(45), dir(45)); dot("$1$", r*dir(0), dir(0)); path g = dir(20)..dir(100)..dir(180)..dir(260)..dir(340); draw(g, red, EndArrow); label("$\gamma$", midpoint(g), -dir(midpoint(g)), red); MP("p(0) = 1", (2.5,0.5)); MP("p(1) = 1", (2.5,0)); \end{asy} \end{center} Similarly, the counterclockwise path from $1 \in S^1$ to $-1 \in S^1$ has a lifting: for some integer $n$, the path from $n$ to $n+\half$. \label{example:lifting_circle} \end{example} The above is the primary example of a lifting. It seems like we have the following structure: given a path $\gamma$ in $B$ starting at $b_0$, we start at any point in the fiber $p\pre(b_0)$. (In our prototypical example, $B = S^1$, $b_0 = 1 \in \CC$ and that's why we start at any integer $n$.) After that we just trace along the path in $B$, and we get a corresponding path in $E$. \begin{ques} Take a path $\gamma$ in $S^1$ with $\gamma(0) = 1 \in \CC$. Convince yourself that once we select an integer $n \in \RR$, then there is exactly one lifting starting at $n$. \end{ques} It turns out this is true more generally. \begin{theorem}[Lifting paths] Suppose $\gamma : [0,1] \to B$ is a path with $\gamma(0) = b_0$, and $ p : (E,e_0) \to (B,b_0) $ is a covering projection. Then there exists a \emph{unique} lifting $\tilde\gamma : [0,1] \to E$ such that $\tilde\gamma(0) = e_0$. \end{theorem} \begin{proof} For every point $b \in B$, consider an evenly covered open neighborhood $U_b$ in $B$. Then the family of open sets \[ \left\{ \gamma\pre(U_b) \mid b \in B \right\} \] is an open cover of $[0,1]$. As $[0,1]$ is compact we can take a finite subcover. Thus we can chop $[0,1]$ into finitely many disjoint closed intervals $[0,1] = I_1 \sqcup I_2 \sqcup \dots \sqcup I_N$ in that order, such that for every $I_k$, $\gamma\im(I_k)$ is contained in some $U_b$. We'll construct $\tilde\gamma$ interval by interval now, starting at $I_1$. Initially, place a robot at $e_0 \in E$ and a mouse at $b_0 \in B$. For each interval $I_k$, the mouse moves around according to however $\gamma$ behaves on $I_k$. But the whole time it's in some evenly covered $U_k$; the fact that $p$ is a covering projection tells us that there are several copies of $U_k$ living in $E$. Exactly one of them, say $V_k$, contains our robot. So the robot just mimics the mouse until it gets to the end of $I_k$. Then the mouse is in some new evenly covered $U_{k+1}$, and we can repeat. \end{proof} The theorem can be generalized to a diagram \begin{center} \begin{tikzcd} & (E, e_0) \ar[d, "p"] \\ (Y, y_0) \ar[ru, "\tilde{f}"] \ar[r, "f"'] & (B, b_0) \end{tikzcd} \end{center} where $Y$ is some general path-connected space, as follows. \begin{theorem}[General lifting criterion] \label{thm:lifting} Let $f: (Y,y_0) \to (B, b_0)$ be continuous and consider a covering projection $p : (E, e_0) \to (B, b_0)$. (As usual, $Y$, $B$, $E$ are path-connected.) Then a lifting $\tilde f$ with $\tilde f(y_0) = e_0$ exists if and only if \[ f_\ast\im(\pi_1(Y, y_0)) \subseteq p_\ast\im(\pi_1(E, e_0)), \] i.e.\ the image of $\pi_1(Y, y_0)$ under $f$ is contained in the image of $\pi_1(E, e_0)$ under $p$ (both viewed as subgroups of $\pi_1(B, b_0)$). If this lifting exists, it is unique. \end{theorem} As $p_\ast$ is injective, we actually have $p_\ast\im(\pi_1(E, e_0)) \cong \pi_1(E, e_0)$. But in this case we are interested in the actual elements, not just the isomorphism classes of the groups. \begin{ques} What happens if we put $Y= [0,1]$? \end{ques} \begin{remark}[Lifting homotopies] Here's another cool special case: Recall that a homotopy can be encoded as a continuous function $[0,1] \times [0,1] \to X$. But $[0,1] \times [0,1]$ is also simply connected. Hence given a homotopy $\gamma_1 \simeq \gamma_2$ in the base space $B$, we can lift it to get a homotopy $\tilde\gamma_1 \simeq \tilde\gamma_2$ in $E$. \end{remark} Another nice application of this result is \Cref{ch:complex_log}. \section{Lifting correspondence} \prototype{$(\RR,0)$ covers $(S^1,1)$.} Let's return to the task of computing fundamental groups. Consider a covering projection $p : (E, e_0) \to (B, b_0)$. A loop $\gamma$ can be lifted uniquely to $\tilde\gamma$ in $E$ which starts at $e_0$ and ends at some point $e$ in the fiber $p\pre(b_0)$. You can easily check that this $e \in E$ does not change if we pick a different path $\gamma'$ homotopic to $\tilde\gamma$. \begin{ques} Look at the picture in \Cref{example:lifting_circle}. Put one finger at $1 \in S^1$, and one finger on $0 \in \RR$. Trace a loop homotopic to $\gamma$ in $S^1$ (meaning, you can go backwards and forwards but you must end with exactly one full counterclockwise rotation) and follow along with the other finger in $\RR$. Convince yourself that you have to end at the point $1 \in \RR$. \end{ques} Thus every homotopy class of a loop at $b_0$ (i.e.\ an element of $\pi_1(B, b_0)$) can be associated with some $e$ in the fiber of $b_0$. The below proposition summarizes this and more. \begin{proposition} Let $p : (E,e_0) \to (B,b_0)$ be a covering projection. Then we have a function of sets \[ \Phi : \pi_1(B, b_0) \to p\pre(b_0) \] by $[\gamma] \mapsto \tilde\gamma(1)$, where $\tilde\gamma$ is the unique lifting starting at $e_0$. Furthermore, \begin{itemize} \ii If $E$ is path-connected, then $\Phi$ is surjective. \ii If $E$ is simply connected, then $\Phi$ is injective. \end{itemize} \end{proposition} \begin{ques} Prove that $E$ path-connected implies $\Phi$ is surjective. (This is really offensively easy.) \end{ques} \begin{proof} To prove the proposition, we've done everything except show that $E$ simply connected implies $\Phi$ injective. To do this suppose that $\gamma_1$ and $\gamma_2$ are loops such that $\Phi([\gamma_1]) = \Phi([\gamma_2])$. Applying lifting, we get paths $\tilde\gamma_1$ and $\tilde\gamma_2$ both starting at some point $e_0 \in E$ and ending at some point $e_1 \in E$. Since $E$ is simply connected that means they are \emph{homotopic}, and we can write a homotopy $F : [0,1] \times [0,1] \to E$ which unites them. But then consider the composition of maps \[ [0,1] \times [0,1] \taking{F} E \taking{p} B. \] You can check this is a homotopy from $\gamma_1$ to $\gamma_2$. Hence $[\gamma_1] = [\gamma_2]$, done. \end{proof} This motivates: \begin{definition} A \vocab{universal cover} of a space $B$ is a covering projection $p : E \to B$ where $E$ is simply connected (and in particular path-connected). \end{definition} \begin{abuse} When $p$ is understood, we sometimes just say $E$ is the universal cover. \end{abuse} \begin{example}[Fundamental group of $S^1$] Let's return to our standard $p : \RR \to S^1$. Since $\RR$ is simply connected, this is a universal cover of $S^1$. And indeed, the fiber of any point in $S^1$ is a copy of the integers: naturally in bijection with loops in $S^1$. You can show (and it's intuitively obvious) that the bijection \[ \Phi : \pi_1(S^1) \leftrightarrow \ZZ \] is in fact a group homomorphism if we equip $\ZZ$ with its additive group structure $\ZZ$. Since it's a bijection, this leads us to conclude $\pi_1(S^1) \cong \ZZ$. \end{example} \section{Regular coverings} \prototype{$\RR \to S^1$ comes from $n \cdot x = n + x$} Here's another way to generate some coverings. Let $X$ be a topological space and $G$ a group acting on its points. Thus for every $g$, we get a map $X \to X$ by \[ x \mapsto g \cdot x. \] We require that this map is continuous\footnote{% Another way of phrasing this: the action, interpreted as a map $G \times X \to X$, should be continuous, where $G$ on the left-hand side is interpreted as a set with the discrete topology.} for every $g \in G$, and that the stabilizer of each point in $X$ is trivial. Then we can consider a quotient space $X/G$ defined by fusing any points in the same orbit of this action. Thus the points of $X/G$ are identified with the orbits of the action. Then we get a natural ``projection'' \[ X \to X/G \] by simply sending every point to the orbit it lives in. \begin{definition} Such a projection is called \vocab{regular}. (Terrible, I know.) \end{definition} \begin{example}[$\RR \to S^1$ is regular] Let $G = \ZZ$, $X = \RR$ and define the group action of $G$ on $X$ by \[ n \cdot x = n + x \] You can then think of $X/G$ as ``real numbers modulo $1$'', with $[0,1)$ a complete set of representatives and $0 \sim 1$. % chktex 9 \begin{center} \begin{asy} size(9cm); dotfactor *= 2; pair A = MP("0", (-5.1,0), 1.4*dir(90)); pair B = MP("1", (-3,0), 1.4*dir(90)); draw(A--B); Drawing("\frac13", (-4.4,0), 1.4*dir(90)); Drawing("\frac23", (-3.7,0), 1.4*dir(90)); MP("\mathbb R / G", (-4,-0.6), dir(-90)); dot(A); opendot(B); draw(unitcircle); draw( (-2.4,0)--(-1.6,0), EndArrow); dot("$0=1$", dir(0), dir(0)); dot("$\frac13$", dir(120), dir(120)); dot("$\frac23$", dir(240), dir(240)); label("$S^1$", origin, origin); \end{asy} \end{center} So we can identify $X/G$ with $S^1$ and the associated regular projection is just our usual $\exp : \theta \mapsto e^{2i\pi \theta}$. \end{example} \begin{example}[The torus] Let $G = \ZZ \times \ZZ$ and $X = \RR^2$, and define the group action of $G$ on $X$ by $(m,n) \cdot (x,y) = (m+x, n+y)$. As $[0,1)^2$ is a complete set of representatives, % chktex 9 you can think of it as a unit square with the edges identified. We obtain the torus $S^1 \times S^1$ and a covering projection $\RR^2 \to S^1 \times S^1$. \end{example} \begin{example}[$\mathbb {RP}^2$] Let $G = \Zc 2 = \left$ and let $X = S^2$ be the surface of the sphere, viewed as a subset of $\RR^3$. We'll let $G$ act on $X$ by sending $T \cdot \vec x = - \vec x$; hence the orbits are pairs of opposite points (e.g.\ North and South pole). Let's draw a picture of a space. All the orbits have size two: every point below the equator gets fused with a point above the equator. As for the points on the equator, we can take half of them; the other half gets fused with the corresponding antipodes. Now if we flatten everything, you can think of the result as a disk with half its boundary: this is $\RP^2$ from before. The resulting space has a name: \emph{real projective $2$-space}, denoted $\mathbb{RP}^2$. \begin{center} \begin{asy} size(3cm); dotfactor *= 2; draw(dir(-90)..dir(0)..dir(90)); draw(dir(90)..dir(180)..dir(-90), dashed); fill(unitcircle, yellow+opacity(0.2)); dot(dir(90)); opendot(dir(-90)); label("$\mathbb{RP}^2$", origin, origin); \end{asy} \end{center} This gives us a covering projection $S^2 \to \mathbb{RP}^2$ (note that the pre-image of a sufficiently small patch is just two copies of it on $S^2$.) \end{example} \begin{example} [Fundamental group of $\mathbb{RP}^2$] As above, we saw that there was a covering projection $S^2 \to \mathbb{RP}^2$. Moreover the fiber of any point has size two. Since $S^2$ is simply connected, we have a natural bijection $\pi_1(\mathbb{RP}^2)$ to a set of size two; that is, \[ \left\lvert \pi_1(\mathbb{RP}^2) \right\rvert = 2. \] This can only occur if $\pi_1(\mathbb{RP}^2) \cong \Zc 2$, as there is only one group of order two! \end{example} \begin{ques} Show each of the continuous maps $x \mapsto g \cdot x$ is in fact a homeomorphism. (Name its continuous inverse). \end{ques} % WOW I thought this was always a covering projection gg \section{The algebra of fundamental groups} \prototype{$S^1$, with fundamental group $\ZZ$.} Next up, we're going to turn functions between spaces into homomorphisms of fundamental groups. Let $X$ and $Y$ be topological spaces and $f : (X, x_0) \to (Y, y_0)$. Recall that we defined a group homomorphism \[ f_\sharp : \pi_1(X, x_0) \to \pi_1(Y_0, y_0) \quad\text{by}\quad [\gamma] \mapsto [f \circ \gamma]. \] % which gave us a functor $\catname{Top}_\ast \to \catname{Grp}$. More importantly, we have: \begin{proposition} Let $p : (E,e_0) \to (B,b_0)$ be a covering projection of path-connected spaces. Then the homomorphism $p_\sharp : \pi_1(E, e_0) \to \pi_1(B, b_0)$ is \emph{injective}. Hence $p_\sharp \im(\pi_1(E, e_0))$ is an isomorphic copy of $\pi_1(E, e_0)$ as a subgroup of $\pi_1(B, b_0)$. \end{proposition} \begin{proof} We'll show $\ker p_\sharp$ is trivial. It suffices to show if $\gamma$ is a nulhomotopic loop in $B$ then its lift is nulhomotopic. By definition, there's a homotopy $F : [0,1] \times [0,1] \to B$ taking $\gamma$ to the constant loop $1_B$. We can lift it to a homotopy $\tilde F : [0,1] \times [0,1] \to E$ that establishes $\tilde\gamma \simeq \tilde 1_B$. But $1_E$ is a lift of $1_B$ (duh) and lifts are unique. \end{proof} \begin{example}[Subgroups of $\ZZ$] Let's look at the space $S^1$ with fundamental group $\ZZ$. The group $\ZZ$ has two types of subgroups: \begin{itemize} \ii The trivial subgroup. This corresponds to the canonical projection $\RR \to S^1$, since $\pi_1(\RR)$ is the trivial group ($\RR$ is simply connected) and hence its image in $\ZZ$ is the trivial group. \ii $n\ZZ$ for $n \ge 1$. This is given by the covering projection $S^1 \to S^1$ by $z \mapsto z^n$. The image of a loop in the covering $S^1$ is a ``multiple of $n$'' in the base $S^1$. \end{itemize} \end{example} It turns out that these are the \emph{only} covering projections of $S^n$ by path-connected spaces: there's one for each subgroup of $\ZZ$. (We don't care about disconnected spaces because, again, a covering projection via disconnected spaces is just a bunch of unrelated ``good'' coverings.) For this statement to make sense I need to tell you what it means for two covering projections to be equivalent. \begin{definition} Fix a space $B$. Given two covering projections $p_1 : E_1 \to B$ and $p_2 : E_2 \to B$ a \vocab{map of covering projections} is a continuous function $f : E_1 \to E_2$ such that $p_2 \circ f = p_1$. \begin{center} \begin{tikzcd} E_1 \ar[r, "f"] \ar[rd, "p_1"'] & E_2 \ar[d, "p_2"] \\ & B \end{tikzcd} \end{center} Then two covering projections $p_1$ and $p_2$ are isomorphic if there are $f : E_1 \to E_2$ and $g : E_2 \to E_1$ such that $f \circ g = \id_{E_1}$ and $g \circ f = \id_{E_2}$. \end{definition} \begin{remark} [For category theorists] The set of covering projections forms a category in this way. \end{remark} It's an absolute miracle that this is true more generally: the greatest triumph of covering spaces is the following result. Suppose a space $X$ satisfies some nice conditions, like: \begin{definition} A space $X$ is called \vocab{locally connected} if for each point $x \in X$ and open neighborhood $V$ of it, there is a connected open set $U$ with $x \in U \subseteq V$. \end{definition} \begin{definition} A space $X$ is \vocab{semi-locally simply connected} if for every point $x \in X$ there is an open neighborhood $U$ such that all loops in $U$ are nulhomotopic. (But the contraction need not take place in $U$.) \end{definition} \begin{example}[These conditions are weak] Pretty much every space I've shown you has these two properties. In other words, they are rather mild conditions, and you can think of them as just saying ``the space is not too pathological''. \end{example} Then we get: \begin{theorem}[Group theory via covering spaces] Suppose $B$ is a locally connected, semi-locally simply connected space. Then: \begin{itemize} \ii Every subgroup $H \subseteq \pi_1(B)$ corresponds to exactly one covering projection $p : E \to B$ with $E$ path-connected (up to isomorphism). (Specifically, $H$ is the image of $\pi_1(E)$ in $\pi_1(B)$ through $p_\sharp$.) \ii Moreover, the \emph{normal} subgroups of $\pi_1(B)$ correspond exactly to the regular covering projections. \end{itemize} \end{theorem} Hence it's possible to understand the group theory of $\pi_1(B)$ completely in terms of the covering projections. Moreover, this is how the ``universal cover'' gets its name: it is the one corresponding to the trivial subgroup of $\pi_1(B)$. Actually, you can show that it really is universal in the sense that if $p : E \to B$ is another covering projection, then $E$ is in turn covered by the universal space. More generally, if $H_1 \subseteq H_2 \subseteq G$ are subgroups, then the space corresponding to $H_2$ can be covered by the space corresponding to $H_1$. % According to \cite{ref:covering_all_we_know}, this statement and % its extension to group actions are ``pretty much all there is to know % about covering projections''. \section{\problemhead} \todo{problems}