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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 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.