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Replace the open knot complement by the compact knot exterior $E_K$; they have the same homotopy type. The abelianizationsends an oriented meridian of a knot to $1$. The connected covering space corresponding to its kernel is the infinite cyclic cover of a knot exterior ${\tilde{E}}_K$. Choose a generator $t$ of its deck transformation group. Thenis the Alexander module of a knot over $\Lambda =\mathbb{Z}[t,t^{-1}]$. It is a finitely presented torsion module, and its order, equivalently the greatest common divisor of the appropriate maximal matrix minors of an Alexander matrix, is the Alexander polynomial of a knot, up to a unit $\pm t^m$.

For a fibered knot, the infinite cyclic cover of a knot exterior is $\Sigma \times \mathbb{R}$. Write the mapping torus with the gluing convention in the question. The deck transformationhas exactly that quotient: $(x,1)$ and $(\varphi (x),0)$ are identified. Choose $t=D$. Its action on $H_1(\Sigma ;\mathbb{Z})$ is ${\varphi }_{\ast }$. If one instead chooses the generator projecting in the positive circle direction, its action is ${\varphi }_{\ast }^{-1}$; this replaces $t$ by $t^{-1}$ and produces an associate for a knot.

Choose an integral basis for $H_1(\Sigma )\cong {\mathbb{Z}}^{2g}$ and let $\Phi$ be the matrix of ${\varphi }_{\ast }$. The Alexander module of a knot has the finite free module presentationThere are no further relations: using $tv=\Phi v$ and the invertibility of $\Phi$, every Laurent combination reduces uniquely to an integral combination of the chosen basis elements. Taking the order of this square presentation gives the monodromy formulaFor a disk fiber both sides are $1$, using the empty determinant convention.

Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.

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.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.

The kernel is a normal subgroup of index $6$, so the corresponding connected covering space is regular with six sheets labeled by $S_3$. Write $a=(12)$, $b=(123)$ and use right multiplication for path lifting. Above the real projective plane, right multiplication by $a$ pairs the labels, giving three copies of its two-sheeted covering space $S^2$. Each sphere has two marked antipodal points corresponding to its pair $\{b^i,b^{i}a\}$, $i=0,1,2$. Above the circle, right multiplication by $b$ gives two cycles of length three. Attach their six marked points to the identically labeled points of these three spheres.

The upper circle follows $e\to b\to b^2\to e$; the lower follows $a\to ab\to a{b}^2\to a$. The point paired with $b^i$ is $b^{i}a=a{b}^{-i}$, explaining the reversed lower labeling in the figure. On each summand these are genuine covering maps, and the attachment matches the six lifts of the wedge point. A small neighborhood of that point lifts to six disjoint copies of its neighborhood in the wedge. This verifies the covering space locally, and the action of $a,b$ is transitive, so it is connected. Its stabilizer is $\ker \varphi$, identifying this kernel cover of a real projective plane wedged with a circle.

A covering map induces an injection on fundamental groups. There is no injection ${\mathbb{Z}}^2\hookrightarrow \mathbb{Z}/2$, and there is no injection $\mathbb{Z}/2\hookrightarrow {\mathbb{Z}}^2$ because ${\mathbb{Z}}^2$ is torsion-free. Thus neither proposed covering exists.

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.

Finally, $\det K=|{\Delta }_K(-1)|$ is a nonzero odd integer, so the minus complex is acyclic over $\mathbb{Q}$. Since a knot exterior has the rational homology of a circle,

For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would forceContinuity on the connected interval $[0,1/2)$ makes $\varphi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\varphi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.

Now let $g:[0,1]\to \mathbb{C}\setminus \{0\}$ and put $h=g/|g|$. This is a continuous path in the unit circle. The exponential map $\theta \mapsto e^{i\theta }$ is a covering map, so the path lifting theorem supplies a continuous lift $\theta :[0,1]\to \mathbb{R}$ after one value of $\theta (0)$ is chosen. HenceOne can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.

If $\theta$ and $\tilde{\theta }$ are two such phases, thenThis integer-valued function is continuous and hence constant. Thereforedoes not depend on the chosen lift.

For $u(t)=g(t^2)$, the phase $\theta (t^2)$ givesFor $v(t)=g(t{)}^2$, the phase $2\theta (t)$ gives

Finally, takeThey have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the winding number of the closed path.

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichand every restriction $p:U_{\alpha }\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The Lebesgue number lemma supplies a subdivisionfor which each $\gamma ([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing ${\tilde{x}}_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\varphi$ for the quotient map. The identity $p\circ Q=q\circ {\mathrm{pr}}_Y$ on $Y\times \mathbb{Z}$, where $Q$ is the quotient defining $\widetilde{Y/\varphi }$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small open set lifts to one copy in every level. Near a glued point choose $U⋐A$ and use the matched pairAfter the prescribed identifications, each such pair maps homeomorphically to $q(U\cup \varphi (U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\varphi }$ is path-connected. Translationis a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The covering-space subgroup and deck-group quotient then gives

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichwith every restriction $p:U_{\lambda }\to U$ a homeomorphism.

Let $N=⟨⟨a⟩⟩$ in
${\pi }_1(Y)=⟨a,b\mid a^2{b}^{-3}⟩$. ThenThe subgroup is normal of index three, so the degree of a connected covering shows that every fibre $p^{-1}(y)$ has exactly three points.

The cell complex of a covering from a coset graph gives an explicit model. Take vertices $v_0,v_1,v_2$ with indices modulo three; put an $a$-loop $A_i$ at every $v_i$, and a directed $b$-edge $B_i:v_i\to v_{i+1}$. Attach a 2-cell at each $v_i$ alongMap every $v_i$ to $y_0$, every $A_i$ to the $a$-cell, every $B_i$ to the $b$-cell, and each 2-cell homeomorphically to the unique 2-cell of $Y$. Its attaching word is the lift of $a^2{b}^{-3}$ beginning at $v_i$. This defines the required connected three-sheeted covering.

Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.

The universal cover $\tilde{Y}$ of a connected graph is a tree. The lift $\tilde{\beta }$ is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\tilde{\beta }$ is injective unless $\beta$ is constant.

Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\tilde{Y}$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving thatis an injective group homomorphism.

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.

The two-fold branched cover of a knot is the double cover of $S^3$ branched along the knot. Its first homology has a nonsingular linking form and, when finite, has order $\det K$.