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A branched covering is a map that is a covering space away from a branch set and has a specified local covering model at that set. For a smooth two-fold covering branched over a codimension-two submanifold, the local model is $(z,u)\mapsto (z^2,u)$. A two-fold branched cover of a knot is the basic three-dimensional example.

A covering homomorphism is a surjective homomorphism of topological groups whose underlying map is a covering space. Its discrete kernel is the fiber over the identity.

For a covering map with nonempty connected base, the degree of a covering map is the common cardinality of its fibres. Each evenly covered neighbourhood makes this number locally constant, so connectedness makes it constant. The total covering space need not be connected. On a path-connected base the path lifting theorem gives a fibre bijection by path lifting. A disconnected base may have different degrees on different connected components.

There are three connected two-sheeted covering spaces of the wedge of two circles, distinguished by the three nonzero homomorphisms from its fundamental group to $\mathbb{Z}/2\mathbb{Z}$. If exactly one circle exchanges sheets, the covering graph has two parallel joining edges and one loop at each vertex. If both circles exchange sheets, it has four parallel joining edges. These give two homeomorphism types of total spaces even though they give three inequivalent covers of the fixed base.

If a finite group $G$ is generated by $h$ elements, send the $a$-generators of the genus-$h$ surface group to those generators and its $b$-generators to the identity. The surface relation is satisfied, so this is a surjective group homomorphism. Its kernel defines a connected regular covering space, on which the deck transformation group $G$ acts freely. The lifted orientation and multiplicativity of the Euler characteristic give genus $g=1+|G|(h-1)$.

A connected $n$-sheeted covering space of a closed orientable surface of genus $g$ has genus $1+n(g-1)$. Lifting a finite cell decomposition gives $n$ cells above every base cell, so its Euler characteristic is $n(2-2g)$. Equating this with $2-2g^{\prime }$ gives the formula. For $g\ge 1$, every positive $n$ occurs: send one handle generator of the fundamental group to $1$ in $\mathbb{Z}/n$ and all remaining generators to zero, then use the classification of connected covering spaces for the kernel of that surjection.

For an $n$-sheeted covering space $f:X\to Y$, define the transfer of a singular simplex as the sum of its $n$ lifts. Restriction of these lifts to any face gives exactly the lifts of that face, so transfer is a chain map. Its induced map $\mathrm{tr}:H_i(Y;R)\to H_i(X;R)$ satisfies $f_{\ast }\mathrm{tr}=n\mathrm{id}$ for any commutative coefficient ring $R$. If $n$ is invertible in $R$, then $n^{-1}\mathrm{tr}$ is a right inverse of $f_{\ast }$. In particular a finite covering surjects onto the base's rational homology.

A homotopy into the base of a covering space has a unique continuous lift once its initial map is continuously lifted. Local inverse sheets construct the lift along each time interval, and compactness of that interval permits a finite subdivision locally around each point of the homotopy's parameter space. Uniqueness on overlapping intervals follows from path lifting theorem uniqueness. In a homotopy of paths with endpoints fixed, a chosen lifted endpoint remains fixed because it varies continuously in a discrete fiber.

The connected covering space corresponding to $\ker \alpha$ has deck transformation group $\mathbb{Z}$. Choosing a deck generator makes its integral homology a module over the Laurent polynomial ring $\mathbb{Z}[t,t^{-1}]$.

For $C_2\ast \mathbb{Z}\to S_3$ taking the two generators to a transposition and a $3$-cycle, the kernel's six-sheeted regular covering space consists of three spheres and two circles. Each sphere has two marked points, and each circle meets one point on each sphere. It retracts onto a rank-four graph by replacing each sphere by a meridian arc, but admits no deformation retraction onto any graph because its degree-two homology is ${\mathbb{Z}}^3$.

Lifting a based loop in a covering space from every point of its finite fibre gives a permutation of that fibre. Composition of lifted loops yields the monodromy action of a covering space. If a closed geodesic of length $\ell$ has a cycle of length $m$ in its monodromy permutation, the corresponding lift closes after $m$ traversals and has length $m\ell$. For a primitive base geodesic, these cycles describe the primitive lift components.

An oriented coordinate neighborhood of a smooth manifold gives two sheets distinguished by the sign of its orientation. On an overlap, the sheets agree where the two local orientation forms differ by a positive function. These open agreement sets define a two-sheeted smooth covering space, whose projection is locally a diffeomorphism.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

Use the standard integral cohomology ring $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, $|u|=2$. Any self-homeomorphism $h$ sends $u$ to $u$ or $-u$, and hence sends $u^2$ to $u^2$. Its Lefschetz number is therefore $1\pm 1+1$, which is either $3$ or $1$. By the Lefschetz fixed-point theorem, every such homeomorphism has a fixed point.

Now ${\mathbb{CP}}^2$ is simply connected: its CW structure has no one-cells and only two- and four-cells. If it covered a connected space with degree $d\ge 2$, it would be the universal cover. The usual covering space hypotheses hold for the base because it is locally homeomorphic to this manifold. Its deck transformation group would act transitively on every fibre and have $d$ elements. A nonidentity deck transformation has no fixed point: uniqueness of lifts makes one fixing a point the identity everywhere on the connected cover. This contradicts the preceding calculation. ThusThus the complex projective plane has no nontrivial covering quotient.

Sending a matrix to its last column gives the special orthogonal sphere fibrationThe action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by Gram-Schmidt process near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the long exact sequence of homotopy groups of a fibration to be applied.

At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify ${\mathbb{R}}^3$ with imaginary quaternions. A unit quaternion acts by $v\mapsto qv{q}^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos (\theta /2)+u\sin (\theta /2)$. The kernel is precisely $\{1,-1\}$, since a quaternion commuting with all imaginary quaternions is real. Thus this gives the twofold covering space $S^3\to SO(3)$. Its simply connected total space is the universal cover, yielding ${\pi }_{1}SO(3)=\mathbb{Z}/2$. Coverings induce isomorphisms on higher homotopy groups, so ${\pi }_{2}SO(3)={\pi }_2{S}^3=0$ and ${\pi }_{3}SO(3)={\pi }_3{S}^3=\mathbb{Z}$.

For $n\ge 4$, both ${\pi }_1{S}^{n-1}$ and ${\pi }_2{S}^{n-1}$ vanish. The Serre fibration sequence therefore identifies ${\pi }_{1}SO(n-1)$ with ${\pi }_{1}SO(n)$ and propagates vanishing of ${\pi }_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer isHere ${\pi }_0$ records the single connected component.

For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth homotopy group, isThe first homomorphism is zero since a torsion group cannot map nontrivially to $\mathbb{Z}$. Thus it reduces to a short exact sequence $0\to \mathbb{Z}\to {\pi }_{3}SO(4)\to \mathbb{Z}\to 0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. ThereforeThese are the low homotopy groups of special orthogonal groups.

Let $p:\tilde{M}\to M$ be the universal cover. Its number of sheets is $|{\pi }_1(M)|$, which is finite, so compactness of $M$ makes $\tilde{M}$ compact. As a covering space of a manifold without boundary, it is also a three-manifold without boundary. It is connected and simply connected by the definition of universal cover.

A simply connected manifold is orientable: transport of a local orientation around loops gives the orientation character ${\pi }_1(\tilde{M})\to \{\pm 1\}$, and the trivial fundamental group leaves no orientation obstruction. The degree-one Hurewicz theorem, equivalently the identification of $H_1$ with the abelianization of the fundamental group, gives $H_1(\tilde{M};\mathbb{Z})=0$. The universal coefficient theorem for cohomology then givesbecause the additional term $\mathrm{Ext}(H_0(\tilde{M};\mathbb{Z}),\mathbb{Z})$ vanishes for $H_0\cong \mathbb{Z}$. Integral Poincare duality gives $H_2(\tilde{M};\mathbb{Z})\cong H^1(\tilde{M};\mathbb{Z})=0$. Connectedness gives $H_0\cong \mathbb{Z}$, and the orientation fundamental class gives $H_3\cong \mathbb{Z}$; higher groups vanish by dimension. ConsequentlyThus the universal cover of a closed three-manifold with finite fundamental group is an integral homology sphere. No homeomorphism classification of three-manifolds is needed for this homological conclusion.

For the closed orientable surface ${\Sigma }_g$, use the fundamental group presentationSince $g\ge 1$, define a homomorphism to $\mathbb{Z}/n$ by sending $a_1$ to $1$ and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index $n$. The classification of connected covering spaces supplies a connected $n$-sheeted covering space, including the identity cover when $n=1$.

Every connected finite covering space of ${\Sigma }_g$ is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has $n$ lifts, so the Euler characteristic under a finite covering gives $\chi (\tilde{\Sigma })=n\chi ({\Sigma }_g)$. If the covering surface has genus $h$, this reads $2-2h=n(2-2g)$. ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.

Work first over ${\mathbb{F}}_2$. The mod-two cohomology ring of real projective space is $H^{\ast }({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero mod-two degree of a map between closed manifolds, so $f^{\ast }(a^n)=0$. The group in degree one has only two elements, so $f^{\ast }a$ is either zero or $a$. The latter would imply $f^{\ast }(a^n)=(f^{\ast }a{)}^n=a^n\ne 0$, a contradiction. Therefore $f^{\ast }a=0$, and multiplicativity makes $f^{\ast }$ zero in every positive cohomological degree. Duality of homology and cohomology over a field consequently makes $f_{\ast }$ zero in every positive degree with mod-two coefficients.

For integral coefficients, cellular homology of real projective space gives $H_k\cong \mathbb{Z}/2$ in odd degrees $0<k<n$, zero in even degrees $0<k<n$, and a top group $\mathbb{Z}$ if $n$ is odd and zero if $n$ is even. For each torsion group, the natural reduction map to mod-two homology is injective by the universal coefficient theorem for homology; naturality and the preceding vanishing imply that its integral induced map is zero. When $n$ is odd, the integral top-degree map is also zero because the nonsurjective map has integer degree of a map between oriented manifolds zero. All groups above dimension $n$ vanish. Hence $f_{\ast }$ is zero on all positive integral homology groups as well.

For $n\ge 2$, ${\pi }_1({\mathbb{RP}}^n)=\mathbb{Z}/2$, and its abelianization is already $H_1({\mathbb{RP}}^n;\mathbb{Z})$. The vanishing just proved therefore implies $f_{\ast }{\pi }_1=0$. By the lifting criterion for a covering space, $f$ lifts through the double covering space $q:S^n\to {\mathbb{RP}}^n$ to a map $\tilde{f}:{\mathbb{RP}}^n\to S^n$ with $q\tilde{f}=f$.

Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\tilde{f}$. In particular the lift has image in $S^n\setminus \{z\}\cong {\mathbb{R}}^n$, for either such point $z$. This punctured sphere is contractible, so contraction there gives a homotopy from $\tilde{f}$ to a constant map into $S^n$. Composing with $q$ contracts $f$. ThusThis proves the nonsurjective self-map of real projective space is null-homotopic conclusion, and also shows that all positive-degree induced maps vanish with any coefficients.

For a singular simplex $\sigma :{\Delta }^i\to Y$, there are exactly $n$ lifts to $X$: choose any of the $n$ points above one vertex, then use the lifting property, since the simplex is simply connected. Defineand extend $R$-linearly to singular chains. The restrictions of these lifts to any face are exactly the $n$ lifts of that face, each appearing once. Indeed a lift is determined uniquely by its value at any point of that face, and every possible value in the corresponding fiber extends to a lift of the whole simplex. Therefore $\partial T_i=T_{i-1}\partial$, so $T$ is a chain map and induces the homological transfer for a finite covering, denoted $f^{\ast }$ in the question.

On chains, every lifted simplex projects back to the original simplex. Thus $f_{\#}{T}_i=n\mathrm{id}$, and on homology this givesOver $\mathbb{Q}$, the map $n^{-1}{f}^{\ast }$ is a right inverse of $f_{\ast }$, proving surjectivity in every degree. Consequently each rational Betti number of the covering space is at least that of its base. This transfer goes in the opposite direction to the usual homological pushforward and is not the ordinary cohomological pullback.

For equality, use the connected double covering space $S^1\to S^1$, $z\mapsto z^2$. Both spaces have $b_0=b_1=1$ and all other Betti numbers zero. For strict increase, take a connected double covering space of the genus-two closed orientable surface, whose existence was proved in 1(b). The genus of a finite cover of a closed orientable surface gives covering genus $1+2(2-1)=3$. Its first Betti number is six, whereas that of the genus-two base is four; their zeroth and second Betti numbers are both one. Thus a connected double cover can preserve every Betti number or strictly increase some of them.

The fundamental group of the figure-eight is the free group on its two circle loops $a,b$. A two-sheeted covering space is specified by their permutations of the two-point fiber. It is connected exactly when this action is transitive, equivalently when at least one permutation exchanges the sheets. Since $S_2$ is abelian, relabeling the two sheets does not identify any further cases. There are therefore three connected double covers, given by $(a,b)=(\tau ,1),(1,\tau ),(\tau ,\tau )$, where $\tau$ exchanges the sheets.

In either of the first two covers there are two vertices joined by two parallel edges, with a loop at each vertex. The two covers differ as covers of the fixed base, but their total spaces are homeomorphic by exchanging the base-circle labels. The third cover has two vertices joined by four parallel edges and no loops.

These last two graph types are not homeomorphic. Their two degree-four branch points are intrinsically identifiable topologically. Removing either branch point in the graph with loops leaves two components: the punctured loop at that point, and the remaining connected graph. Removing either branch point in the four-edge graph leaves a connected four-armed graph. Thus the three covers give two homeomorphism types of total space. This is the distinction between equivalence of covers and homeomorphism of covering spaces.