Covering Maps and Evenly Covered Neighborhoods

Evenly Covered Neighborhoods

A covering map describes a map that, near each point of its target, looks like several separate copies of the same neighborhood. The word “separate” matters: each copy must map homeomorphically onto the entire neighborhood, not merely onto some smaller open subset. This distinction makes covering maps more restrictive than local homeomorphisms.

Let $(E,\mathcal{T}_E)$ and $(B,\mathcal{T}_B)$ be topological spaces. A map $p:E\to B$ is surjective if every point of $B$ is the image of a point of $E$. A homeomorphism is a bijective continuous map whose inverse is continuous.

Definition: An open set $U\subseteq B$ is evenly covered by $p$ if there is an index set $J$ and pairwise disjoint open sets $V_j\subseteq E$, for $j\in J$, such that $$ p^{-1}(U)=\bigcup_{j\in J}V_j $$ and, for every $j\in J$, the restriction $p|_{V_j}:V_j\to U$ is a homeomorphism. The sets $V_j$ are called the sheets over $U$. A covering map is a surjective continuous map $p:E\to B$ such that every $b\in B$ has an open neighborhood $U$ evenly covered by $p$.

In particular, every point over $U$ belongs to exactly one sheet, and each sheet contains exactly one point over each $b\in U$. The neighborhood $U$ need not be connected. The number of sheets can vary from one neighborhood to another, so the definition does not assert that the same number of sheets works throughout the target.

Covering Maps Are Local Homeomorphisms

A map $f:E\to B$ is a local homeomorphism if it is continuous and, for every $e\in E$, there is an open neighborhood $W$ of $e$ such that $f(W)$ is open in $B$ and $f|_W:W\to f(W)$ is a homeomorphism. The image $f(W)$ in this definition can depend on $e$, and it need not contain a neighborhood of any other point in the same fiber.

Proposition: Every covering map is a local homeomorphism. Moreover, if $U$ is evenly covered, then every $b\in U$ has as many preimages as there are sheets over $U$.

Proof. Let $p:E\to B$ be a covering map and take any $e\in E$. By the covering condition, there is an evenly covered open neighborhood $U$ of $p(e)$. There is a sheet $V_j$ containing $e$, since $p^{-1}(U)$ is the union of the sheets. The restriction $p|_{V_j}:V_j\to U$ is a homeomorphism. Thus the open neighborhood $V_j$ of $e$ is mapped homeomorphically onto the open set $U$, which proves that $p$ is a local homeomorphism. For the fiber count, each homeomorphism $p|_{V_j}$ gives exactly one point of $V_j$ over every $b\in U$. The sheets are disjoint and their union is $p^{-1}(U)$, so the fiber over every $b\in U$ has exactly one point from each sheet and no other points. $\square$

The converse does not follow from the definition. A local homeomorphism gives a suitable open piece around each point of $E$; a covering map requires all pieces over one target neighborhood to map onto that whole neighborhood. The next examples make the distinction precise.

Worked Example: A Two-Sheet Covering of an Interval

Let $B=(0,1)$ with its standard topology, let $F=\{L,R\}$ with the discrete topology, and give $E=B\times F$ the product topology. Define $p:E\to B$ by $p(x,s)=x$.

The map is continuous because for every open $O\subseteq B$, its preimage is $O\times F$, which is open in the product topology. It is surjective: for each $x\in(0,1)$, $p(x,L)=x$. Fix $x\in B$, and choose any open interval $U\subseteq(0,1)$ containing $x$. Its preimage is $$ p^{-1}(U)=(U\times\{L\})\cup(U\times\{R\}). $$ These two sets are disjoint and open in $E$, since $\{L\}$ and $\{R\}$ are open in the discrete space $F$. On each one, $p$ is a homeomorphism onto $U$; the inverse maps are $y\mapsto(y,L)$ and $y\mapsto(y,R)$. Thus $U$ is evenly covered, and $p$ is a covering map.

Check: for every $y\in U$, the fiber is $p^{-1}(\{y\})=\{(y,L),(y,R)\}$, with one point in each sheet. This verifies the two-sheet count directly.

Worked Example: A Covering Over a Disconnected Space

Give $B=\{u,v\}$ the discrete topology and $E=\{a,b,c\}$ the discrete topology. Define $p:E\to B$ by $p(a)=u$, $p(b)=u$, and $p(c)=v$. The map is continuous because every subset of $E$ is open, so the preimage of every open subset of $B$ is open. It is surjective because both $u$ and $v$ are images.

For $u$, take the open neighborhood $\{u\}$. Its preimage is $\{a,b\}$, the disjoint union of the open sheets $\{a\}$ and $\{b\}$. The map from either singleton sheet to $\{u\}$ is a homeomorphism. For $v$, the neighborhood $\{v\}$ has preimage $\{c\}$, and $\{c\}\to\{v\}$ is a homeomorphism. Therefore every point of $B$ has an evenly covered neighborhood and $p$ is a covering map.

Check: the fibers have different sizes: $p^{-1}(\{u\})=\{a,b\}$, whereas $p^{-1}(\{v\})=\{c\}$. There is no contradiction with the proposition: the evenly covered neighborhoods used here are different, and the two points lie in different components of the discrete target.

Worked Example: A Surjective Local Homeomorphism That Is Not a Covering Map

Let $B=(0,1)$ with its standard topology. Let $E$ be the disjoint union of two spaces: a copy $E_1=(0,1)$ and a copy $E_2=(0,\tfrac12)$, each with its standard topology, and give $E$ the disjoint union topology. Define $p:E\to B$ to be the identity on $E_1$ and the inclusion $x\mapsto x$ on $E_2$.

The map is surjective because its restriction to $E_1$ is the identity on $(0,1)$. It is continuous: for each open $O\subseteq B$, the preimage consists of the open subset $O\cap E_1$ in the first component together with the open subset $O\cap(0,\tfrac12)$ in the second. Each point of $E_1$ has an open interval neighborhood within $E_1$ mapped homeomorphically onto an open interval in $B$. Each point $x\in E_2$ also has such a neighborhood: choose $r>0$ with $r<x$ and $r<\tfrac12-x$; then $(x-r,x+r)\subseteq E_2$ is mapped homeomorphically onto the open interval $(x-r,x+r)\subseteq B$. Thus $p$ is a local homeomorphism.

However, $p$ is not a covering map. The fiber over $\tfrac12$ has one point, from $E_1$, while the fiber over $\tfrac14$ has two points, one from each component. Every open neighborhood $U$ of $\tfrac12$ in $B$ contains an interval $(\tfrac12-r,\tfrac12+r)$ for some $r>0$ small enough that this interval lies in $B$. In particular, $U$ contains $y=\tfrac12-\tfrac r2$ and $z=\tfrac12+\tfrac r2$. Since $0<y<\tfrac12<z<1$, the fiber over $y$ has two points and the fiber over $z$ has one. If $U$ were evenly covered, the proposition’s fiber-count conclusion would give the same number of preimages for every point of $U$. That is impossible. So $\tfrac12$ has no evenly covered neighborhood.

Check: at $y=\tfrac14$, both components contribute, so the fiber has two points; at $z=\tfrac34$, only $E_1$ contributes, so the fiber has one. These values lie on opposite sides of $\tfrac12$, as required by the neighborhood argument.

A Common Mistake: Confusing the Two Conditions

Common pitfall. A tempting but incorrect argument is: “Every point has a neighborhood mapped homeomorphically onto an open set, so the map is a covering map.” That verifies only the local homeomorphism condition. In the preceding example, a neighborhood around a point in $E_2$ maps onto an interval contained in $(0,\tfrac12)$, not onto a single neighborhood that works evenly for points near $\tfrac12$. The fiber calculation exposes the failure: any neighborhood of $\tfrac12$ contains points with one preimage and points with two.

Conversely, an evenly covered neighborhood supplies the full covering condition: its preimage must be exhausted by disjoint sheets, and every sheet must map onto the same $U$. Checking only that a few chosen pieces map homeomorphically onto open subsets is not enough; one must also check that those pieces cover the entire preimage $p^{-1}(U)$.

Condition	What it requires	What it does not require
Local homeomorphism	Each point of the domain has a neighborhood mapped homeomorphically onto an open subset of the target.	One target neighborhood evenly covered by all points above it.
Covering map	Surjectivity and, around each target point, a neighborhood whose full preimage is a disjoint union of sheets mapping homeomorphically onto it.	A fixed number of sheets across different components of the target.

Check Your Understanding

For each question, use the definitions to justify the conclusion.

In the two-sheet interval example, why is $U\times\{L\}$ open in the product topology?
In the finite discrete example, identify the sheets over $\{u\}$ and verify that each restriction is a homeomorphism.
For the surjective local homeomorphism example, compute the fibers over $\tfrac13$ and $\tfrac23$. What does their difference imply about an evenly covered neighborhood of $\tfrac12$?
Does a covering map have to use finitely many sheets over each evenly covered neighborhood? Answer from the definition.
Explain why the identity map $\operatorname{id}_X:X\to X$ is a covering map for any topological space $X$, including $X=\varnothing$.

Covering Maps and Evenly Covered Neighborhoods

What You'll Learn