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Any two connected degree-two coverings of the torus fit into a commuting square with homeomorphisms of their total and base spaces. They correspond to index-two subgroups of ${\mathbb{Z}}^2$, which are kernels of the three nonzero maps ${\mathbb{Z}}^2\to \mathbb{Z}/2\mathbb{Z}$. Reduction modulo two shows that $G{L}_2(\mathbb{Z})$ acts transitively on these subgroups; the classification of connected covering spaces and the lifting criterion for a covering space then supply the homeomorphism of total spaces.

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 the classical subgroup-covering construction, assume $X$ is path-connected, locally path-connected, and semilocally simply connected. These are the usual hypotheses underlying the classification of connected covering spaces; an arbitrary space, as literally allowed by the wording, need not admit every subgroup covering.

Construct the path-class construction of a subgroup covering directly. Consider paths $\alpha$ starting at $x_0$, and identify $\alpha ,\beta$ when they have the same endpoint and the loop class $[\alpha {\beta }^{-1}]$ belongs to $H$. The subgroup properties make this an equivalence relation. Let $Y$ be its set of classes and define $p([\alpha ])=\alpha (1)$.

Choose a path-connected open neighbourhood $U$ of $\alpha (1)$ whose loops become null-homotopic in $X$. Such neighbourhoods form a local basis under the stated local hypotheses. Define a sheetFor each point of $U$ the resulting class is independent of the chosen path $\eta$, since the difference is a loop in $U$, conjugated by $\alpha$, and is trivial in ${\pi }_1(X,x_0)$. Thus $p$ maps this sheet bijectively onto $U$. Give $Y$ the topology generated by these sheets. Restrictions to smaller such neighbourhoods show that the sheets form a basis and that their endpoint maps are homeomorphisms. Two sheets over $U$ that intersect coincide, by concatenating and reversing paths within $U$; otherwise they are disjoint. Therefore $p^{-1}(U)$ is a disjoint union of copies of $U$, and $p$ is a covering map.

The basepoint is the class of the constant path. The path $\alpha$ lifts to its successive prefix classes, giving a path from this basepoint to $[\alpha ]$; hence $Y$ is path-connected. A loop $\gamma$ lifts closed exactly when its final prefix class agrees with the constant class, namely when $[\gamma ]\in H$. Consequently $p_{\ast }({\pi }_1(Y))=H$. Part (i) supplies injectivity, so

The missing local hypothesis is substantive. In the Hawaiian earring, every neighbourhood of the common basepoint contains a whole sufficiently small circle. Its loop is nontrivial in the whole space, since retracting onto that circle preserves it. An evenly covered neighbourhood would lift that loop to a closed loop in one sheet, so the image subgroup of any covering contains a nontrivial element. Therefore the trivial subgroup cannot be realized by a simply connected covering of this space. The intended result is valid under the local covering-space hypotheses above, not for every topological space.

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.

For $k>0$, the surface presentationhas an epimorphism to $\mathbb{Z}/k$ sending $a_1$ to $1$ and the other generators to zero. Its kernel has index $k$. By the classification of connected covering spaces, it determines a connected $k$-sheeted cover $p:\tilde{\Sigma }\to {\Sigma }_2$. Pull back the orientation, so each local sheet has degree $+1$ and $\mathrm{deg}p=k$. Counting lifts of cells gives $\chi (\tilde{\Sigma })=k\chi ({\Sigma }_2)=-2k$, hence $2-2g=-2k$ and $g=k+1$.

For $k<0$, use the $|k|$-sheeted cover and compose with an orientation-reversing diffeomorphism of ${\Sigma }_2$, for instance reflection in a symmetric embedded genus-two surface. By multiplicativity of mapping degree, the degree becomes $-|k|=k$. For $k=0$ take a constant map from the torus. Thus the arbitrary-degree maps to a surface of genus two can be chosen with

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.

A covering map is a continuous surjection $p:\tilde{X}\to X$ for which each point has an open neighbourhood $U$ with $p^{-1}(U)$ a disjoint union of open sets, each mapped homeomorphically onto $U$. A connected simply connected covering space is a universal cover.

There is a necessary connectedness qualification in the printed conclusion. Let $A_0$ be the path component of $a_0$ in $A$. Then $p(\tilde{A})=A_0$: one inclusion follows by projecting paths, and the other by lifting a path in $A_0$ from $a_0$. Thus the asserted surjection onto all of $A$ requires $A$ to be path connected. Without it, take $p$ to be the identity of $\mathbb{R}$ and $A$ to be two disjoint intervals; a selected component cannot cover both.

To prove the assertion about a component of a pulled-back universal cover, choose an evenly covered neighbourhood $U$ in $X$. Local path connectedness of $A$ supplies a path-connected relatively open neighbourhood $V\subseteq U\cap A_0$. Each inverse sheet over $V$ is path connected, so it lies wholly in $\tilde{A}$ or is disjoint from it. These sheets evenly cover $V$ under $p{|}_{\tilde{A}}$. Therefore $p{|}_{\tilde{A}}:\tilde{A}\to A_0$ is a covering map.

Under the classification of connected covering spaces, its subgroup consists of based loops in $A_0$ whose lifts close at ${\tilde{a}}_0$. A loop viewed in $X$ closes in the universal cover exactly when its class in ${\pi }_1(X,a_0)$ is trivial. Hence the subgroup isHere ${\pi }_1(A,a_0)={\pi }_1(A_0,a_0)$, so the subgroup formula itself needs no change. Injectivity of the covering's induced fundamental-group map is the usual homotopy-lifting property.

A surjective inverse-polynomial germ projection without even covering comes from the inverse of $P(w)=w^3-3w$ over $G=\mathbb{C}$. Consider $R=\mathbb{C}\setminus \{-1,1\}$, with $\pi (w)=P(w)$. Since $P^{\prime }(w)=3(w^2-1)\ne 0$ on $R$, each $w$ defines a holomorphic inverse germ of $P$ at $P(w)$. This identifies $R$ with a component of the space of germs of holomorphic functions: inverse germs continue along paths in $R$, and the holomorphic identity $P(f(z))=z$ persists under continuation. A germ with value $\pm 1$ cannot exist, because differentiating that identity would give $P^{\prime }(\pm 1)f^{\prime }(z)=1$. Since $R$ is connected, all its inverse germs belong to one complete analytic function.

The projection is surjective. A generic value has three distinct regular preimages; the critical values $2,-2$ each retain a simple regular preimage, respectively $2,-2$, even after their double critical preimages $-1,1$ are removed. Thus fibre size drops from $3$ to $1$ at these values, making an evenly covered neighbourhood impossible. This is surjective but not a regular topological covering.

If “regular” is specifically used for deck transitivity among genuine coverings, there is also a genuine nonnormal example. Remove the critical values from the base and all their preimages from the source:This is a connected three-sheeted analytic covering. A small loop about each critical value swaps the two roots meeting there. The two swaps are distinct and generate $S_3$ (connectedness also forces transitive monodromy). The stabilizer of one sheet is an order-$2$ subgroup, which is not normal in $S_3$. The classification of connected covering spaces therefore shows this covering is not regular in the deck-transitive sense. Either convention has a fully explicit example.

Choose ${\hat{x}}_i\in p_i^{-1}(x_0)$. By the classification of connected covering spaces and the degree of a connected covering, each cover corresponds to an index-two subgroupAn index-two subgroup is the kernel of a nonzero homomorphism ${\mathbb{Z}}^2\to \mathbb{Z}/2\mathbb{Z}$. There are three:The group $G{L}_2({\mathbb{F}}_2)$ acts transitively on the three nonzero linear functionals on ${\mathbb{F}}_2^2$, and its elementary matrices lift to the integral general linear group. Hence there is $A\in G{L}_2(\mathbb{Z})$ such that

Take $\psi =f_A$. Since ${\psi }_{\ast }(H_1)=H_2$, the lifting criterion for a covering space gives a based liftsuch thatApplying the same argument to $A^{-1}$ gives a lift in the reverse direction. The composites are based lifts of the appropriate identity maps; uniqueness of based lifts makes them identities. Thus $\varphi$ and $\psi$ are homeomorphisms, and the required square commutes. This proves the equivalence of connected double covers of the torus.

Label the four vertices cyclically by $\mathbb{Z}/4$. Reading the arrows in the diagram, an $a$-edge moves one step in one direction and a $b$-edge one step in the other. Thus the monodromy homomorphism is(up to simultaneously reversing the chosen numbering). By the Galois correspondence for covering spaces, the subgroup associated with the chosen vertex above $x_0$ is

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.

Choose a finite generating set $h_1,\dots ,h_r$ for $H$. Under the standard classification of connected covering spaces, each $h_i$ is represented by a based combinatorial loop ${\gamma }_i$ in $Y$. Let $Z$ be the union of the images of these finitely many finite edge paths. Then $Z$ is a finite connected subgraph containing $y_0$.

Part (b) makes the inclusion-induced mapinjective. Its image contains every $h_i$, because every ${\gamma }_i$ lies in $Z$, and therefore contains the subgroup they generate, namely all of $H$. The map is consequently an isomorphism.

A deck transformation of a covering $p:\tilde{X}\to X$ is a homeomorphism $d:\tilde{X}\to \tilde{X}$ with $p\circ d=p$. A connected covering is a regular covering map if its deck transformation group acts transitively on each fiber; for a path-connected base, transitivity on the fiber over the chosen basepoint is equivalent.

Write $H=p_{\ast }{\pi }_1(\tilde{X},\tilde{b})$ and $G={\pi }_1(X,b)$. A based loop represents an element of $H$ exactly when its lift starting at $\tilde{b}$ is closed. Indeed a closed lift itself projects to a representative of an image element; conversely the image of a closed path lifts to that path, and the closed-endpoint property is unchanged by based homotopy.

Suppose the covering is regular, let $h\in H$, and let $g\in G$. Lift a representative of $g$ from $\tilde{b}$ to an endpoint $v$. By regularity some deck transformation sends $\tilde{b}$ to $v$. Applying it to the closed lift of $h$ gives a closed lift of $h$ at $v$, by uniqueness in the path lifting theorem. The concatenation of the lift of $g$, that closed lift of $h$, and the reverse lift of $g$ is therefore a closed lift of $gh{g}^{-1}$. Thus $gH{g}^{-1}\subseteq H$; applying the same argument to $g^{-1}$ gives equality. Hence a regular covering has a normal image subgroup.

For the explicit cover, label the bottom vertex $B=\tilde{b}$, the top $T$, the left $L$, the right $C=c$, and the rear/front equatorial vertices $U,V$, respectively. Reading the arrows gives the complete monodromy table:

Thus the $x$ edges form the cycle $B,V,L,T,C,U,B$, and the $y$ edges form the cycle $B,C,V,T,U,L,B$. In particular every vertex has one incoming and one outgoing edge of each label, as a covering of the two-circle wedge requires. All word paths below are traversed from left to right.

Choose the spanning tree consisting of the first five $x$ edges from $B$ to $U$. The tree paths to $B,V,L,T,C,U$ have labels $1,x,x^2,x^3,x^4,x^5$. For every directed non-tree edge $v\stackrel{z}{\to }w$, the path from $B$ along the tree to $v$, then along the edge, then back along the tree from $w$, has label $t_{v}z{t}_w^{-1}$. Collapsing a spanning tree turns a connected graph into a wedge of circles, so these paths form a free basis of its fundamental group. There are seven non-tree edges, since $12-6+1=7$. Their projected labels give the explicit generating setThe covering map induces an injection on fundamental groups, so these seven generators are also a free basis for the image subgroup. The first comes from the remaining $x$ edge $U\to B$; the other six come from the outgoing $y$ edge at each vertex in the table.

Now the word $h=y{x}^2$ lifts at $B$ around the triangleso $h\in H$. Starting at $C$, however, its lift iswhich is not closed. A deck transformation sending $B$ to $C$ would take the first closed path to a closed lift of the same word at $C$, contradicting uniqueness of lifts and the displayed nonclosed path. No such deck transformation exists. This is also the closed-lift obstruction expressed by the deck transformation group as a monodromy centralizer.

For a connected covering of a connected, locally path-connected, semilocally simply connected space, the classification of connected covering spaces says that the image subgroup is normal if and only if the covering is regular. The two-circle wedge and this connected finite graph satisfy those hypotheses. Since regularity fails, $H$ is not normal in the base group ${\pi }_1(X,b)\cong F(x,y)$. All normality comparisons here are in that base group.

Finally takeThe lift of $gh{g}^{-1}=y^2{x}^2{y}^{-1}$ from $B$ followsIt ends at $V\ne B$, so $gh{g}^{-1}\notin H$. This is the required explicit witness, and independently proves the failure of normality.