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True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

A group homomorphism satisfies $\phi (ab)=\phi (a)\phi (b)$ for every $a,b\in G$. Its kernel of a group homomorphism is $K=\{g:\phi (g)=e_H\}$. The homomorphism equation gives $\phi (e_G)=e_H$ by cancellation, and $\phi (g^{-1})=\phi (g{)}^{-1}$. Therefore $K$ contains the identity and is closed under multiplication and inverses, so it is a subgroup.

For $k\in K$ and $x\in G$,Thus conjugation carries $K$ into itself, and applying the inverse conjugation gives equality: $K$ is a normal subgroup. If $K=\{e,\xi \}$, conjugation cannot send $\xi$ to the identity because it is injective. ConsequentlyThis proves directly the normal subgroup of order two is central principle for kernels, including the kernel facts needed below.

Write the Heisenberg group element as $M(x,y,z)$. Matrix multiplication and inversion giveAll matrices have determinant one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from matrix multiplication; hence they form a subgroup of $G{L}_3(\mathbb{R})$.

Two such matrices commute precisely when $x{z}^{\prime }=x^{\prime }z$. Requiring this for every $x^{\prime },z^{\prime }$ forces $x=z=0$. ConsequentlyThe map $\pi (M(x,y,z))=(x,z)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$, with kernel of a group homomorphism $Z(H)$. Equivalently, it gives the explicit quotient group isomorphismIt is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the first isomorphism theorem directly.

A torsion-free divisible Abelian group is naturally a vector space over the rational numbers: for $m\in \mathbb{Z}$ and $n>0$, define $(m/n)a$ as the unique $b$ with $nb=ma$. Divisibility supplies existence and torsion-freeness supplies uniqueness.

In the usual group first-order language $\{+,-,0\}$, a model $A$ of the universal part of DAG is a torsion-free group which is Abelian. If the language instead uses only $\{+,0\}$, a substructure can be merely a torsion-free cancellative commutative monoid. Handle this convention by first taking its Grothendieck group $G(A)$: its elements are formal differences $a-b$, withThe cancellative commutative monoid condition makes $A\to G(A)$ injective. If $n(a-b)=0$, then $na=nb$, and torsion-freeness gives $a=b$; hence $G(A)$ is a torsion-free abelian group. In the full group language simply take $G(A)=A$.

For $A\ne \{0\}$ form its rational divisible hullConcretely its elements are fractions $g/n$ with $n>0$, where $g/n=h/m$ if $mg=nh$. The canonical embedding of $A$ into $P$ is injective, and $P$ is nontrivial, divisible, Abelian and torsion-free, so it satisfies DAG.

Let $j:A\to N$ be any structure embedding into a model of DAG. Extend it first to formal differences if necessary. Its unique extension to the rational divisible hull sends $g/n$ to the unique element $b\in N$ with $nb=j(g)$. This is a group homomorphism fixing the given copy of $A$. It is injective: an element mapped to zero has $j(g)=0$, hence $g=0$. Thus every embedding into a DAG model factors through $P$.

The zero case must be treated separately: its rational divisible hull is zero and does not satisfy DAG. Instead choose $P=(\mathbb{Q},+)$. Given any nontrivial DAG model $N$, choose $b\ne 0$; the map $q\mapsto qb$ embeds $P$ into $N$ over zero. Therefore DAG has algebraically prime models, including over the trivial base.

Here a divisor is a modulus of a number field, rather than an arbitrary real-weighted divisor. Writewhere ${\mathfrak{c}}_0$ is a nonzero integral ideal, the finite multiplicities $m_{\mathfrak{p}}$ are nonnegative integers, and ${\mathfrak{c}}_{\infty }$ is a set of real Archimedean places. The multiplicity of a place in a modulus is $m_v(\mathfrak{c})=m_{\mathfrak{p}}$ at the finite place corresponding to $\mathfrak{p}$, is $1$ at a real place in ${\mathfrak{c}}_{\infty }$, and is $0$ at every other infinite place. In particular complex places have multiplicity $0$.

Let $I_{\mathfrak{c}}$ be the group of nonzero fractional ideals prime to ${\mathfrak{c}}_0$. PutLet $P_{\mathfrak{c}}$ consist of the principal ideals $(a)$ with $a\in K_{\mathfrak{c},1}^{\times }$. The generalized ideal class group is the ray class groupLet $U={\mathcal{O}}_K^{\times }$ be the unit group, let $U_{\mathfrak{c}}=U\cap K_{\mathfrak{c},1}^{\times }$ be the group of ray units, and define the residue and signature group of a modulusThe first factor is omitted when ${\mathfrak{c}}_0={\mathcal{O}}_K$. There is a group homomorphism $\rho :U\to G_{\mathfrak{c}}$ recording the unit's finite residues and its signs. Its kernel is $U_{\mathfrak{c}}$.

Let $P^{\mathfrak{c}}$ consist of all principal fractional ideals prime to ${\mathfrak{c}}_0$, and put $R_{\mathfrak{c}}=P^{\mathfrak{c}}/P_{\mathfrak{c}}$. The two short exact sequences areandFor the first short exact sequence, weak approximation for number fields realizes every choice of finite unit residues and real signs by some $a\in K^{\times }$ prime to ${\mathfrak{c}}_0$. Send that data to the class of $(a)$ in $R_{\mathfrak{c}}$. Changing $a$ without changing its residues and signs multiplies it by an element of $K_{\mathfrak{c},1}^{\times }$, so the map is well-defined. Its kernel consists exactly of data arising from units: if $(a)=(b)$ with $b\in K_{\mathfrak{c},1}^{\times }$, then $a/b\in U$. This gives $R_{\mathfrak{c}}\cong G_{\mathfrak{c}}/\rho (U)$. For the second short exact sequence, forget the ray conditions. Its kernel is $R_{\mathfrak{c}}$, and weak approximation for number fields gives a representative prime to ${\mathfrak{c}}_0$ for every ideal class. These are the two parts of the ray class exact sequence.

For $K=\mathbb{Q}(\sqrt{15})$, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{15}]$. The finite modulus is trivial and both real places occur, so $H_{\mathfrak{c}}$ is the narrow ideal class group and $G_{\mathfrak{c}}=\{\pm 1{\}}^2$. The two field embeddings send $\sqrt{15}$ to $\sqrt{15}$ and $-\sqrt{15}$. The given unit $ϵ=4+\sqrt{15}$ is positive at both places, since $4-\sqrt{15}>0$; the unit $-1$ is negative at both. Thus the unit signature map has imageThe given ideal class group has order $2$. The second short exact sequence therefore givesTo determine the group structure, retain the ideal $\mathfrak{p}=(2,1+\sqrt{15})$ that generates the ordinary ideal class group. We have ${\mathfrak{p}}^2=(2)$: all generators $4$, $2(1+\sqrt{15})$ and $(1+\sqrt{15}{)}^2$ lie in $(2)$, whileputs $2$ in ${\mathfrak{p}}^2$. Since $2$ is totally positive, $[\mathfrak{p}{]}^2=1$ in the narrow ideal class group. Its image in the ordinary ideal class group is nontrivial, so $[\mathfrak{p}]$ has order exactly $2$.

The class of the principal ideal $(\sqrt{15})$ is a nontrivial element of $R_{\mathfrak{c}}$. Its two signs are $(+,-)$, and multiplying by a unit can only reverse both signs or neither, so no generator of this ideal is totally positive. Its square is $(15)$, which does have a totally positive generator. Thus $[(\sqrt{15})]$ is another element of order $2$, distinct from $[\mathfrak{p}]$ because its ordinary ideal class is trivial. These two elements are independent and generate all four classes. ConsequentlyThe nontrivial ordinary ideal class already has a lift of order $2$, so the extension in the second short exact sequence splits; it cannot be cyclic of order $4$.

Label the distinct roots ${\alpha }_1,\dots ,{\alpha }_n$. Define $\tau (\sigma )$ by $\sigma ({\alpha }_i)={\alpha }_{\tau (\sigma )(i)}$. Composition respects this rule, so it is a group homomorphism. If its image permutation is the identity, $\sigma$ fixes all the roots; they generate $L$, so $\sigma$ is the identity. Thus $\tau$ is injective.

For $f=t(t^2-2)$, label the roots $0,\sqrt{2},-\sqrt{2}$. The splitting field is $\mathbb{Q}(\sqrt{2})$ and its nontrivial automorphism swaps precisely roots $2,3$. Hence $\boxed{\mathrm{im}\tau =⟨(23)⟩}$.

For $f=t^3-2$, Eisenstein criterion gives irreducibility. The real root $\alpha =\sqrt[3]{2}$ generates a degree-three real field. Adjoining a primitive cube root of unity ${\zeta }_3$ gives a further extension of degree two, since it is nonreal and satisfies $t^2+t+1=0$. Thus its splitting field has degree $6$. In characteristic zero this is a Galois extension, so its Galois group has order $6$. The injection into $S_3$ is therefore an isomorphism.

A one-dimensional continuous group representation is a continuous group homomorphism $\rho :S^1\to {\mathbb{C}}^{\times }$. Its image is compact. If some $|\rho (z)|$ differed from one, its positive or negative powers would be unbounded, so the image lies in $S^1$. Lift $t\mapsto \rho (e^{it})$ to a continuous real argument $a(t)$ with $a(0)=0$. Then $a(s+t)-a(s)-a(t)$ is a continuous integer multiple of $2\pi$ and vanishes at $(0,0)$, hence is always zero. The continuous Cauchy functional equation gives $a(t)=ct$. Periodicity forces $e^{2\pi ic}=1$, so $c=n\in \mathbb{Z}$. Consequently $\boxed{\rho (z)=z^n}$; conversely every such map is a continuous one-dimensional group representation.

No such finite group exists. Cayley theorem embeds every finite group $G$ of order $n$ into $S_n$: the left-regular group action sends $g$ to the permutation $x\mapsto gx$, and this is faithful because the image of the identity element is $g$.

To make every image permutation even, let $t=(n+1n+2)$, with support disjoint from the first $n$ letters. For $\sigma \in S_n$, let $e(\sigma )=0$ for an even permutation and $e(\sigma )=1$ for an odd permutation, and setThe sign homomorphism gives $e(\sigma \tau )\equiv e(\sigma )+e(\tau )(\mathrm{mod}2)$, and $t$ commutes with all permutations of the first $n$ letters. Thus $\iota$ is a group homomorphism. It is injective by restriction to those letters, and its sign is $(-1{)}^{e(\sigma )}(-1{)}^{e(\sigma )}=1$. ConsequentlyThis is the alternating-group embedding of a finite group.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

A permutation representation is a group homomorphism $\rho :G\to \mathrm{Sym}(\Omega )$, equivalently a group action on a set $\Omega$. The left regular action takes $\Omega =G$ and $\rho (g)(h)=gh$. Each such map is a bijection, and $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$. If $\rho (g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\mathrm{Sym}(G)$ with the symmetric group $S_n$, proving Cayley theorem:For a nonabelian simple group, compose this faithful representation with the sign of a permutation homomorphism $S_n\to \{1,-1\}$. Its kernel is a normal subgroup of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and the same representation embeds $G$ in the alternating group $A_n$.

Finally let $S_3$ act on the two cosets of its normal subgroup $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with kernel $A_3$; explicitly, send even permutations to the identity and odd permutations to $(12)$.

The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA nontrivial free product has both factors different from the trivial group.

The normal form theorem for a free product says that every element has a unique expressionwhere the empty expression represents the identity. The $g_j$ are called syllables in a free product. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the free product are injective.

For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal{W}$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a permutation $L_h$ of $\mathcal{W}$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.

The multiplication law inside $G_i$ gives $L_h{L}_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a group action of $G_1\ast G_2$ on $\mathcal{W}$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same permutation, and therefore have identical sequences. This proves the normal form theorem for a free product.

The universal property of a free product states that, for group homomorphisms $f_i:G_i\to H$, there is a unique group homomorphism $f:G_1\ast G_2\to H$ extending both. Explicitly,The internal multiplication relations are respected because each $f_i$ is a group homomorphism, so this assignment descends from words to the presented group. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the free product.

Substitution sends the defining relation toThus the assignment respects the relator and extends to a group homomorphism $\boxed{\theta :S\to S}$.

The base free product for the group encoding of a modular machine isThe letters $x,y$ generate the first factor, and $t$ generates the second. Let $T_K$ be the kernel of the group homomorphism $K\to ⟨x,y⟩\cong {\mathbb{Z}}^2$ that kills $t$. Thenis a free group with this displayed free basis of a group. Indeed every word in $K$ can be rewritten as a product of such conjugates followed by an element of $⟨x,y⟩$, so they generate the kernel. A freely reduced product of these conjugates is nontrivial by the normal form theorem for a free product: after combining adjacent occurrences of the same conjugate, different successive indices give a nonzero intervening $x,y$-syllable. Thus there is no relation among the proposed basis elements.

For clarity, a modular machine of modulus $m>1$ has at most one instruction for each residue pair $(a,b)$, with $0\le a,b<m$ and $0\le c<m^2$. Its right and left transitions are respectivelyIts halting set at a designated terminal configuration $H_0(\mathcal{M})$ consists of the nonnegative pairs whose forward computation reaches $(0,0)$, including $(0,0)$ itself. These formulas explain the exponents in the associated-subgroup maps below.

The auxiliary Higman group $Q$ has no nontrivial finite quotients of a group. One can use this permitted property directly, but there is a short verification. In any finite quotient, let $n_i$ be the order of the image of $a_i$. Since conjugation sends $a_{i+1}$ to its square, $n_{i+1}$ must be odd. If some $n_i>1$, choose the least prime number $p$ dividing any $n_i$. It is odd. Conjugation by the preceding generator givesThe multiplicative order of $2$ modulo $p$ is therefore a divisor of $n_{i-1}$. It is greater than one and divides $p-1$ by the Fermat little theorem, so it has a prime divisor smaller than $p$. That prime also divides $n_{i-1}$, contradicting the choice of $p$. Hence every $n_i=1$ and the quotient is trivial.

Given any finitely presented group $P$, form $\hat{P}=P\ast ⟨v⟩$, and apply part (a) to the word $w=v$. This word is nontrivial by the normal form theorem for a free product, so the output $J=\hat{P}(v)$ contains $P$ by the embedding already proved.

To check its finite quotients, let $f:J\to F$ be any group homomorphism to a finite group. The image of the copy of $Q$ is trivial by the preceding property. The identifications $u_j=a_3^{j+1}{a}_1{a}_3^{-(j+1)}$ make all $f(u_j)$ trivial. Thus the equations $u_j^{-1}c{u}_j=g_j$ give, in the image,Since $f(c)=f(z)$, this forces $f(x_i)=1$ for all generators of $\hat{P}$, including $v$. The defining word $c=[v,z]$ then has trivial image, so $f(z)=f(d)=1$ as well. Every generator of $J$ has trivial image. ThereforeThis uses the concrete construction of part (a), not an assumption that arbitrary quotients preserve an embedding of $P$.

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.

Define the length of a formal tensor expression recursively byThe canonical strict monoidal functor $Q:\mathcal{F}br\to \mathbf{B}$ sends $u$ to $|u|$, sends every associator and unitor to an identity braid, and sends $c_{u,v}$ to the block braiding $c_{|u|,|v|}$. The pentagon and triangle become identity equations; the braiding axioms become the corresponding block-braid equations. Thus the assignment respects the defining relations, andon the nose. It is a braided monoidal functor with identity comparison maps.

To prove that $Q$ is an equivalence of categories, choose a standard parenthesization $x^{[n]}$ of $n$ copies of $x$, with $x^{[0]}=I$. On $x^{[n]}$, define the image of ${\sigma }_i$ by canonically exposing the $i$th and $(i+1)$st factors, applying $c_{x,x}$ there, and restoring the chosen parentheses. The monoidal coherence theorem makes this independent of the structural rebracketing. Crossings on disjoint pairs commute by the tensor interchange law. The adjacent braid group relations follows from the hexagon laws and naturality of the braiding, as in the Yang–Baxter calculation below. Hence these assignments give group homomorphisms $B_n\to \mathrm{Aut}(x^{[n]})$ and a functor $S:\mathbf{B}\to \mathcal{F}br$.

Equip $S$ with the canonical rebracketing maps $x^{[m]}\otimes x^{[n]}\to x^{[m+n]}$. Their monoidal functor axioms follow from the monoidal coherence theorem; the block-braiding compatibility follows by iterating the two hexagon laws. Therefore $S$ is a strong monoidal functor compatible with the braiding.

We have $QS=1_{\mathbf{B}}$, including its comparison maps. For every formal expression $u$, there is a canonical structural isomorphismobtained by rebracketing and inserting the units appearing in $u$. These maps form a natural isomorphism $SQ\Rightarrow 1_{\mathcal{F}br}$. To check naturality, it suffices to check the generating morphisms: for associators and unitors it is exactly monoidal coherence; for $c_{u,v}$ it follows from the hexagon expansion into the elementary crossings defining $S$. Composition and tensor product then preserve the equation. The same structural coherence shows that $\theta$ is monoidal.

Thus $Q$ has a specified quasi-inverse and is the required equivalence:Only ordinary monoidal coherence, together with the defining braiding axioms, has been used; a separate braided coherence theorem is unnecessary.