Local Structure of Covering Maps

Evenly Covered Neighborhoods as Local Coordinates

A covering map can have many points over a single point of its target. Its defining feature is that, near any target point, those preimages separate into disjoint regions, and each region maps homeomorphically onto the same neighborhood. These regions are called sheets. This local description gives more than a way to recognize a covering: it shows exactly how nearby points in the target have corresponding preimages.

We use the following definition. A homeomorphism is a continuous bijection whose inverse is continuous. A subset has the subspace topology inherited from its ambient space.

Definition: Let $X$ and $Y$ be topological spaces, with topologies understood. A map $p:X\to Y$ is a covering map if it is surjective and, for every $y\in Y$, there is an open neighborhood $U$ of $y$ such that $$ p^{-1}(U)=\bigsqcup_{\alpha\in J}V_\alpha, $$ where each $V_\alpha$ is open in $X$, the sets $V_\alpha$ are pairwise disjoint, and each restriction $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism. Such a $U$ is an evenly covered neighborhood of $y$; its sets $V_\alpha$ are the sheets over $U$. A map is a local homeomorphism if every point of its domain has an open neighborhood that the map sends homeomorphically onto an open subset of its target.

The index set $J$ may be finite or infinite. The disjoint-union requirement means that every point of $p^{-1}(U)$ belongs to exactly one sheet. The homeomorphism condition then says that each sheet contains exactly one preimage of each point of $U$.

Covering Maps Are Local Homeomorphisms

Theorem: Every covering map $p:X\to Y$ is a local homeomorphism.

Proof. Let $x\in X$, and set $y=p(x)$. Choose an evenly covered open neighborhood $U$ of $y$. Since $x\in p^{-1}(U)$, it belongs to one of the sheets, say $V_\alpha$. By the definition of evenly covered, $V_\alpha$ is open in $X$, and $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism. The set $U$ is open in $Y$. Thus $V_\alpha$ is an open neighborhood of $x$ mapped homeomorphically onto an open subset of $Y$, as required. $\square$

This conclusion is local: at each point of the domain, one can use a sheet as a neighborhood on which the map is a homeomorphism. It does not say that the entire map is one-to-one or a homeomorphism. Different sheets may map onto the same neighborhood.

Fibers Are Discrete

For $y\in Y$, the fiber over $y$ is $p^{-1}(\{y\})=\{x\in X:p(x)=y\}$, given the subspace topology from $X$. A topological space is discrete if every one of its subsets is open; equivalently, each singleton is open in that space.

Theorem: If $p:X\to Y$ is a covering map, then every fiber $p^{-1}(\{y\})$ is discrete in its subspace topology.

Proof. Fix $y\in Y$ and $x\in p^{-1}(\{y\})$. Choose an evenly covered neighborhood $U$ of $y$, and let $V_\alpha$ be the sheet containing $x$. Because $p|_{V_\alpha}$ is a bijection onto $U$, the only point of $V_\alpha$ mapping to $y$ is $x$. Hence $$ V_\alpha\cap p^{-1}(\{y\})=\{x\}. $$ The set $V_\alpha$ is open in $X$, so this equality shows that $\{x\}$ is open in the subspace $p^{-1}(\{y\})$. This holds for every point $x$ of the fiber. Every subset of the fiber is a union of such open singletons, so the fiber is discrete. $\square$

Discreteness is about the topology on the fiber, not about how many points it contains: a discrete fiber may be finite or infinite. The theorem also does not claim that a fiber is closed in $X$; that would require additional assumptions about the target topology.

Worked Example: A Two-Sheet Cover of the Real Line

Let $X=\mathbb{R}\times\{1,2\}$, where $\mathbb{R}$ has its standard topology and $\{1,2\}$ has the discrete topology. Give $X$ the product topology, and define $p:X\to\mathbb{R}$ by $p(x,i)=x$. We check that this is a covering map and compute its fibers.

For any open set $U\subseteq\mathbb{R}$, the product-topology basis gives open sets $U\times\{1\}$ and $U\times\{2\}$ in $X$. They are disjoint and $$ p^{-1}(U)=(U\times\{1\})\sqcup(U\times\{2\}). $$ On either set $U\times\{i\}$, the restriction of $p$ maps $(x,i)$ to $x$. It is a bijection onto $U$, and its inverse $x\mapsto(x,i)$ is continuous: the preimage of a basic open set $W\times\{j\}$ under this inverse is $W$ if $j=i$, and is empty if $j\ne i$. Thus both restrictions are homeomorphisms onto $U$.

Taking $U=\mathbb{R}$ verifies the covering condition, including surjectivity. For each $a\in\mathbb{R}$, $$ p^{-1}(\{a\})=\{(a,1),(a,2)\}. $$ The fiber has two points. Each is isolated in the fiber, since the open set $\mathbb{R}\times\{1\}$ meets it in $\{(a,1)\}$, and $\mathbb{R}\times\{2\}$ meets it in $\{(a,2)\}$.

Check: for every $a\in\mathbb{R}$, both points in the displayed fiber map to $a$, and there are no others because $p(x,i)=a$ forces $x=a$. Thus the fiber calculation agrees with the two sheets over $\mathbb{R}$.

Matching Points in Nearby Fibers

The sheets over one evenly covered neighborhood provide a local way to compare fibers. Suppose $U$ is evenly covered, with sheets $(V_\alpha)_{\alpha\in J}$, and take $y,z\in U$. Each sheet $V_\alpha$ contains exactly one point over $y$ and exactly one point over $z$. Matching those two points for each $\alpha$ gives a bijection between the fibers.

Theorem: If $U\subseteq Y$ is an evenly covered neighborhood for $p:X\to Y$, then for every $y,z\in U$ there is a bijection $p^{-1}(\{y\})\to p^{-1}(\{z\})$ that matches points lying in the same sheet. In particular, all fibers over points of $U$ have the same cardinality.

Proof. Write $p^{-1}(U)=\bigsqcup_{\alpha\in J}V_\alpha$ as in the definition. For each $\alpha$, the homeomorphism $p|_{V_\alpha}:V_\alpha\to U$ has exactly one inverse image of $y$ and exactly one inverse image of $z$. Define a map from the fiber over $y$ to the fiber over $z$ by sending the point in $V_\alpha$ over $y$ to the point in $V_\alpha$ over $z$. Each point of either fiber lies in exactly one sheet, so this rule is defined for every point and gives exactly one image. Reversing $y$ and $z$ defines an inverse map. It is therefore a bijection, and the fibers have equal cardinality. $\square$

There is also a useful local description in terms of inverse maps. For each sheet $V_\alpha$, let $s_\alpha:U\to V_\alpha$ be the inverse of $p|_{V_\alpha}$. Each $s_\alpha$ is continuous, and $$ p(s_\alpha(y))=y \qquad (y\in U). $$ Thus each sheet supplies a continuous choice of one preimage for every point in $U$. These choices are separate: the covering definition does not select one sheet as preferred.

Worked Example: The Circle Covering

Give $\mathbb{R}$ its standard topology and $S^1=\{z\in\mathbb{C}:|z|=1\}$ the subspace topology from $\mathbb{C}$. Consider $p:\mathbb{R}\to S^1$ defined by $p(t)=e^{2\pi i t}$. We exhibit evenly covered neighborhoods and describe the local inverse branches.

Fix $t_0\in\mathbb{R}$ and choose $0<\delta<\tfrac12$. Set $U=p((t_0-\delta,t_0+\delta))$. The map $p$ is one-to-one on this interval: if $p(s)=p(t)$, then $s-t\in\mathbb{Z}$; for $s,t$ in this interval, $|s-t|<2\delta<1$, so the only possible integer difference is $0$. The interval parametrizes an open arc of the circle, and this parametrization is a homeomorphism onto $U$.

Since $p(t+k)=p(t)$ for every integer $k$, the full preimage of $U$ is $$ p^{-1}(U)=\bigsqcup_{k\in\mathbb{Z}}(t_0-\delta+k,t_0+\delta+k). $$ To see that these are exactly the preimage points, if $p(s)\in U$, then $p(s)=p(t)$ for some $t\in(t_0-\delta,t_0+\delta)$. Hence $s-t\in\mathbb{Z}$, so $s$ belongs to one of the displayed intervals. Distinct intervals are disjoint because their lengths are $2\delta<1$. Translation by $k$ identifies each interval with the original one, and $p(t+k)=p(t)$; therefore every restriction to a displayed interval is a homeomorphism onto $U$.

The fiber over $p(t_0)$ is $\{t_0+k:k\in\mathbb{Z}\}$. Indeed, $p(t)=p(t_0)$ holds exactly when $t-t_0\in\mathbb{Z}$. The branch over the $k$-th interval sends $p(t)\in U$, with $t\in(t_0-\delta,t_0+\delta)$, to $t+k$. In this way the same target point has one preimage in each sheet.

Check: applying $p$ to the branch value gives $p(t+k)=p(t)$. The intervals are disjoint, so these branch values are distinct, and the preimage calculation shows that every preimage has been included.

A Discrete Fiber Is Not Enough

The discreteness theorem gives a necessary property of covering maps, not a test that works in reverse. In fact, the nearby-fiber theorem gives a stronger necessary condition: over one evenly covered neighborhood, all fibers have the same cardinality.

Worked Example: Why the Squaring Map Is Not a Covering

Give $\mathbb{C}$ its standard topology and consider $f:\mathbb{C}\to\mathbb{C}$, $f(z)=z^2$. This map is surjective, since every complex number has a square root. Its fiber over $0$ is $\{0\}$. For $w\ne0$, the two square roots are distinct: if $z^2=w$, then $(-z)^2=w$, and $z=-z$ would imply $z=0$, contradicting $w\ne0$. Thus the fiber over $w\ne0$ has two points.

If $f$ were a covering map, there would be an evenly covered open neighborhood $U$ of $0$. Since $U$ is open and contains $0$, it contains a nonzero point $w$. The nearby-fiber theorem would imply that the fibers over $0$ and $w$ have the same cardinality. But the first has one point and the second has two. This contradiction proves that $f$ is not a covering map.

The direct local-homeomorphism test also fails at $0$. Every open neighborhood of $0$ contains a disk $B(0,r)$ for some $r>0$. Choose $z$ with $0<|z|<r$. Then $z$ and $-z$ are distinct points in that neighborhood, but $f(z)=f(-z)$, so the restriction cannot be one-to-one.

Common pitfall: the fiber $\{0\}$ is discrete, but that fact does not imply that $f$ is a covering map. Discrete fibers are necessary; they do not ensure evenly covered neighborhoods. Here the change from one preimage over $0$ to two over every nonzero point in a neighborhood rules out a covering.

What the Local Description Provides

Evenly covered neighborhoods let us study a covering map one sheet at a time. On each sheet, the map has a continuous inverse, so local questions about the domain can be translated into questions about the target. At the same time, the sheets keep distinct preimages separate. The matching of fibers is valid within a single evenly covered neighborhood; it does not by itself provide a preferred matching over the entire target.

Keep these conclusions distinct:

A covering map is a local homeomorphism because every point of its domain lies in a sheet.
Each fiber is discrete in its subspace topology because each point can be isolated by its sheet.
Fibers over points in one evenly covered neighborhood are in bijection, by matching points within each sheet.
Discrete fibers alone do not imply that a map is a covering map.

Check Your Understanding

Use the definitions and local arguments above to answer these questions.

In an evenly covered neighborhood, why does each sheet contain exactly one point over each point of the neighborhood?
Explain why the proof that fibers are discrete uses the subspace topology on the fiber.
For the map $p(x,i)=x$ from $\mathbb{R}\times\{1,2\}$ to $\mathbb{R}$, identify the two local inverse branches over an open set $U\subseteq\mathbb{R}$.
For $p(t)=e^{2\pi i t}$, why must the interval used to parametrize an arc have length less than $1$?
Which necessary covering-map property shows that $z\mapsto z^2$ cannot be a covering near $0$?