When Is a Covering Map a Homeomorphism?

A covering map is locally a disjoint collection of copies of the base. The natural question is when that local collection consists of exactly one copy everywhere. Under the usual convention that covering maps are surjective, the answer is simple: a covering map is a homeomorphism precisely when every fiber has one point.

This also gives a useful criterion for deciding whether two covering spaces are homeomorphic over the same base.

The basic definitions

A map $p:E\to B$ is a covering map if it is surjective and every point $b\in B$ has an open neighborhood $U$ such that

$$p^{-1}(U)=\coprod_{\alpha\in A} V_\alpha,$$

where each $V_\alpha$ is open in $E$ and the restriction

$$p\big\vert_{V_\alpha}:V_\alpha\to U$$

is a homeomorphism. The sets $V_\alpha$ are called sheets over $U$.

The fiber over $b$ is

$$p^{-1}(b)=\{e\in E:p(e)=b\}.$$

Each sheet contains exactly one point lying over $b$. Thus, for a covering map, the number of points in a fiber is the number of sheets over any sufficiently small evenly covered neighborhood.

Criterion for a covering map

Let $p:E\to B$ be a covering map. Then

$$p\text{ is a homeomorphism}\quad\Longleftrightarrow\quad |p^{-1}(b)|=1\text{ for every }b\in B.$$

The forward implication is immediate: a homeomorphism is bijective, so every fiber has exactly one point.

For the converse, assume every fiber has exactly one point. Since a covering map is surjective, $p$ is already bijective. It remains to prove that $p^{-1}$ is continuous.

Take $b\in B$ and choose an evenly covered open neighborhood $U$ of $b$. Write

$$p^{-1}(U)=\coprod_{\alpha\in A}V_\alpha,$$

with every restriction $p\big\vert_{V_\alpha}:V_\alpha\to U$ a homeomorphism. If there were two distinct sheets, then a point of $U$ would have at least two preimages, contradicting the assumption that every fiber has one point. Therefore there is exactly one sheet, say $V$, and

$$p^{-1}(U)=V.$$

Consequently, $p\big\vert_V:V\to U$ is a homeomorphism. On $U$, the global inverse $p^{-1}$ agrees with the continuous inverse of this restriction. Since every point of $B$ has such a neighborhood $U$, the global inverse is continuous. Hence $p$ is a homeomorphism.

Worked examples

The exponential covering of the circle

Consider

$$p:\mathbb{R}\to S^1,\qquad p(t)=e^{2\pi i t}.$$

This is a covering map. For example, a sufficiently short arc on $S^1$ has infinitely many disjoint inverse-image intervals in $\mathbb{R}$, one around each lift of a point in the arc; for an arc around $1$, these intervals are near the integers.

The fiber over $1\in S^1$ is

$$p^{-1}(1)=\mathbb{Z}.$$

Since the fibers are not singletons, $p$ is not a homeomorphism. In fact, it is not even injective: $p(t)=p(t+n)$ for every integer $n$.

A one-sheet covering

Let $B$ be any topological space and define

$$p:B\to B,\qquad p(b)=b.$$

Every fiber contains exactly one point, so this is a homeomorphism and a one-sheet covering map.

More generally, if $E$ is homeomorphic to $B$ and $p:E\to B$ is that homeomorphism, then $p$ is a one-sheet covering.

A finite covering of the circle

For a positive integer $n$, define

$$p:S^1\to S^1,\qquad p(z)=z^n.$$

Every point of $S^1$ has exactly $n$ preimages. Therefore this is an $n$-sheeted covering. It is a homeomorphism only when $n=1$.

For $n=2$, for instance, both $1$ and $-1$ map to $1$, so the map cannot be injective.

Maps between two covering spaces

Suppose

p:E\to B\qquad\text{and}\qquad q:F\to B$$ are covering maps. A map $h:E\to F$ is a map **over $B$** if

q\circ h=p.

This means that $h$ sends each point of the fiber $p^{-1}(b)$ into the fiber $q^{-1}(b)$. A useful theorem is: > If $h:E\to F$ is continuous, satisfies $q\circ h=p$, and restricts to a bijection > $$h:p^{-1}(b)\to q^{-1}(b)$$ > for every $b\in B$, then $h$ is a homeomorphism. To see why, choose an evenly covered neighborhood $U$ of a point $b$ that is evenly covered for both $p$ and $q$; intersecting evenly covered neighborhoods gives such a $U$. Write

p^{-1}(U)=\coprod_\alpha V_\alpha,\qquad q^{-1}(U)=\coprod_\beta W_\beta.

Fiberwise bijectivity does not necessarily send an entire sheet to an entire sheet when $U$ is disconnected, so argue as follows. For a sheet $V=V_\alpha$ and a sheet $W=W_\beta$, set

S_{V,W}=p\bigl(V\cap h^{-1}(W)\bigr)\subseteq U.

Because $p\big\vert_V:V\to U$ is a homeomorphism and $h$ is continuous, $S_{V,W}$ is open in $U$. Moreover,

h(V)\cap W=(q\big\vert_W)^{-1}(S_{V,W}).

Thus $h(V)$, which is the union of these intersections over all sheets $W$, is open in $F$. Fiberwise bijectivity makes $h$ globally bijective, and on $h(V)$ its inverse is

h^{-1}\big\vert_{h(V)}=(p\big\vert_V)^{-1}\circ q.

This is continuous, so $h$ is a homeomorphism. This criterion is stronger than merely asking whether $E$ and $F$ are abstractly homeomorphic. Two total spaces might be homeomorphic while their projections to $B$ are different. The equation $q\circ h=p$ is what makes the homeomorphism one of covering spaces over the specified base. ## Common mistakes ### Confusing a covering map with a homeomorphism A covering map is locally a homeomorphism, not necessarily globally one. Several sheets can fit over each neighborhood, producing multiple points in a fiber. ### Checking only one fiber For a general covering space, one fiber having one point does not automatically establish that the map is a homeomorphism unless additional connectedness or constant-sheet-number hypotheses are available. The safe criterion is to check every fiber. If the base is connected, the number of points in the fibers of a covering is constant. In that common setting, checking one fiber is enough: one point in one fiber implies a one-sheet covering everywhere. ### Ignoring surjectivity conventions Some authors define a covering map to be surjective; others allow a map onto a union of components or state surjectivity separately. If surjectivity is not assumed, a bijective map onto its image can be a homeomorphism onto that image without being a homeomorphism $E\to B$. State the convention before applying the criterion.