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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.

Let ${\pi }_i:{\mathbb{C}}^{g_i}\to X_i$ be the quotient covering maps. Since the source universal cover is simply connected, the lifting criterion for a covering space gives a lift $F:{\mathbb{C}}^{g_1}\to {\mathbb{C}}^{g_2}$ of $f\circ {\pi }_1$. The lift is a holomorphic map because ${\pi }_2$ is locally a biholomorphism.

For every $\lambda \in {\Lambda }_1$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete Euclidean lattice ${\Lambda }_2$. It is a continuous function of $z$ on a connected space, hence is constant. Differentiating shows that each coefficient of $DF$ is ${\Lambda }_1$-periodic. Such a coefficient is bounded on a compact fundamental parallelepiped, and periodicity bounds it on all of ${\mathbb{C}}^{g_1}$. The several-variable Liouville theorem, obtained by applying the one-variable theorem on coordinate lines, makes every coefficient constant.

Consequently $DF=A$ for a constant complex-linear map $A$ and $F(z)=Az+x$. The period differences become $A\lambda \in {\Lambda }_2$. Thus the affine lift of a holomorphic map between complex tori givesEvery holomorphic map between complex tori is a homomorphism followed by a translation. The translation vector $x=F(0)$ is determined modulo ${\Lambda }_2$, while $A$ is the derivative of any lift and is unaffected by that ambiguity.

For connected oriented closed manifolds of dimension $n$, the degree of a map between oriented manifolds is the integer determined bywhere the brackets are the chosen fundamental classes. Functoriality gives $\mathrm{deg}(g\circ f)=\mathrm{deg}g\mathrm{deg}f$.

Give $S^n$ its boundary orientation as the boundary of the unit ball in ${\mathbb{R}}^{n+1}$. The antipodal map is the restriction of $-I$. Its ambient determinant is $(-1{)}^{n+1}$, and it takes the outward normal at $x$ to the outward normal at $-x$. Thus its effect on the boundary orientation has this same sign, givingThe original PDF has $A:S^n\to S^n$, correcting the extra prime on the target in the TeX transcription.

For even dimension $n=2k\ge 2$, cellular homology of real projective space gives $H_n({\mathbb{RP}}^n;\mathbb{Z})=0$. Thus the composite map on top homology factors through zero, andZero is attained by constant maps.

For odd dimension $n=2k+1\ge 3$, Real projective space is orientable: the antipodal deck transformation has degree $+1$. Orient it so that the double covering map $q:S^n\to {\mathbb{RP}}^n$ has degree $2$. Since $S^n$ is simply connected, the lifting criterion for a covering space gives $f=q\circ \tilde{f}$ for a map $\tilde{f}:S^n\to S^n$. Hencewhich is even. Every even integer occurs: collapse the complement of an oriented embedded disk in ${\mathbb{RP}}^n$ to obtain a degree-one map $g:{\mathbb{RP}}^n\to S^n$, and choose $\tilde{f}$ of any prescribed integer degree $m$. Maps of all integer degrees on spheres are obtained, for example, by suspending the circle maps $z\mapsto z^m$. Setting $f=q\circ \tilde{f}$ gives degree $2m$. ThusThese are the degrees of maps factoring through real projective space. If $k=0$ is permitted in the odd-dimensional clause, there is an exception: ${\mathbb{RP}}^1\cong S^1$, so every integer degree occurs. The lifting argument requires dimension at least two. The initial connected-manifold degree definition excludes $S^0$; with the usual reduced-homology definition for self-maps of $S^0$, a factorization through the one-point ${\mathbb{RP}}^0$ has degree zero.

Finally suppose $\mathrm{deg}f=p$ is prime. Fix any prime $q\ne p$ and work over $F={\mathbb{F}}_q$. We claim that $f^{\ast }:H^r(N;F)\to H^r(S^n;F)$ is injective. If $a\ne 0$, Poincare duality supplies $b\in H^{n-r}(N;F)$ with $⟨a⌣b,[N{]}_F⟩\ne 0$. Naturality of the cup product and evaluation givesbecause $p$ is invertible in $F$. Hence $f^{\ast }a\ne 0$, proving cohomological injectivity of a map of invertible degree.

The intermediate cohomology of the sphere is zero, so $H^r(N;F)=0$ for $0<r<n$. Over a field, the universal coefficient theorem for cohomology identifies this with the dual of $H_r(N;F)$, so these homology groups vanish as well. The universal coefficient theorem for homology then givesEach integral group is a finitely generated abelian group, by the permitted finite CW complex model. Its decomposition can have no free summand, since that would survive modulo $q$, and no torsion summand divisible by any prime $q\ne p$. Thus it is a finite $p$-primary group. There are only finitely many intermediate degrees, so the exponents of their finite cyclic summands have a common bound $p^a$. Taking $a\ge 1$ also covers the case in which all the groups vanish. ThereforeThis proves that a prime-degree sphere map forces primary torsion, including one uniform exponent for all the degrees.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

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.

Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,is a homotopy from $f$ to the constant map with value $F(0)$. Thus $f$ is null-homotopic. Conversely, if $H:S^{n-1}\times [0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times \{1\}$ and hence descends to the quotientwhich is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the extension-null-homotopy criterion for a sphere.

A universal cover of $X$ is a covering map $p:\tilde{X}\to X$ whose total space is path-connected and simply connected. Let $p_1:({\tilde{X}}_1,{\tilde{x}}_1)\to (X,x)$ and $p_2:({\tilde{X}}_2,{\tilde{x}}_2)\to (X,x)$ be two universal covers. The lifting criterion for a covering space applies becauseso $p_1$ has a based lift $h:{\tilde{X}}_1\to {\tilde{X}}_2$ satisfying $p_{2}h=p_1$. Similarly there is a based lift $k:{\tilde{X}}_2\to {\tilde{X}}_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the uniqueness of a universal covering space.

Now let $p:\tilde{X}\to X$ be universal with $\tilde{X}$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the fundamental group givesThus the required factorization holds with $\Phi =F_{\ast }$.

Conversely, suppose $f_{\ast }=\Phi i_{\ast }$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_{\ast }$. The factorization givesHence $f{|}_{\partial \sigma }$ is null-homotopic in $X$ and extends over the disc $|\sigma |$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\ge 3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the contractible space $\tilde{X}$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma |$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the extension criterion into a space with contractible universal cover.

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.