Let $K$ be a number field and $E/K$ an elliptic curve. Heights measure the arithmetic size of points, and turn the finite quotient supplied by the Weak Mordell-Weil theorem into finite generation. For a projective point $[X:Z]\in {\mathbb{P}}^1(K)$ define the Absolute logarithmic Weil heightusing the usual normalized absolute values and local degrees $n_v=[K_v:{\mathbb{Q}}_v]$. The product formula makes this independent of scaling and of enlarging $K$. For a Short Weierstrass form put $h_x(P)=h([x(P):1])$ and $h_x(O)=h([1:0])=0$. Over $\mathbb{Q}$, if $x=A/B$ in lowest terms, $h(x)=\log \max (|A|,|B|)$.

The finiteness property is Northcott theorem: for bounded height and bounded degree over $\mathbb{Q}$ there are only finitely many algebraic numbers. One sees the mechanism from the height-Mahler measure formula. If $\alpha$ has degree $d\le D$ and height at most $H$, its primitive integral minimal polynomial has Mahler measure $e^{dh(\alpha )}\le e^{DH}$. Expanding its product of linear factors bounds every coefficient by $2^D{e}^{DH}$. Only finitely many integral polynomials satisfy this bound, and each has finitely many roots. Thus a height bound gives only finitely many $x$-coordinates in the fixed number field $K$, and each gives at most two points of $E$, besides $O$.

To understand multiplication, the doubling formula givesIts homogeneous numerator and denominator have degree four. They have no common projective zero: at infinity the numerator is nonzero, and at a root of $f(t)=t^3+at+b$ the numerator equals $f^{\prime }(t{)}^2$, since it is $f^{\prime }(t{)}^2-8tf(t)$. Nonsingularity excludes a common root of $f,f^{\prime }$. The coordinate doubling map is therefore a degree-four morphism of the projective line.

The general height growth under a morphism of the projective line is $h(\Phi (t))=dh(t)+O(1)$ for a fixed degree-$d$ morphism. Here is why the error is bounded independently of $t$. At a place $v$, normalize its homogeneous coordinates to have maximum norm $1$. The fixed homogeneous forms have uniformly bounded norm. They also have a positive lower bound: the normalized coordinate space is compact over $K_v$, and their common zero set is empty. Outside finitely many places their coefficients are integral and their reduction has no common zero, making both bounds $1$. Homogeneity and summation over places then give the asserted height estimate. Consequently there is a fixed $C$ with

Define the canonical height of an elliptic curve by the conventional normalizationThe successive approximants differ by at most $C/(2\cdot 4^{n+1})$, so the limit exists andThus canonical height retains Northcott finiteness, while replacing the bounded-error doubling rule by an exact one.

We also need the height parallelogram identity. It follows from the symmetric-square proof of the naive height parallelogram estimate, which we give explicitly using the supplied addition formulae. For $x_i=X_i/Z_i$, the unordered pair $x(P+Q),x(P-Q)$ is represented by the binary quadratic $C{T}^2-BTS+A{S}^2$, whereThese forms define the biquadratic morphism for elliptic sums and differences. They have no common zero. If $C=0$, the two projective coordinates agree. On the finite diagonal $x_1=x_2=t$, $B=4f(t)$ and $A=f^{\prime }(t{)}^2-8tf(t)$, so simultaneous vanishing would violate nonsingularity. At $(\infty ,\infty )$ the homogeneous $A$ is nonzero. This also handles $P=\pm Q$ and sums or differences equal to $O$, for which the affine fractions alone would be inappropriate.

The same local norm argument as above, now using bidegree $(2,2)$, givesOn the other hand the binary quadratic is the product of two linear factors for $x(P+Q)$ and $x(P-Q)$. At nonarchimedean places coefficient maximum norms multiply exactly: scale each factor to norm $1$ and reduce; the product of the two nonzero residue polynomials remains nonzero. This is multiplicativity of the nonarchimedean polynomial norm. At archimedean places the norms are comparable up to fixed constants, by the coefficient triangle inequality and compactness of normalized linear factors. ThereforeCombining the estimates proves the naive-height parallelogram relation with a uniform error. Apply it to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and pass to the limit to obtainAs in the polarization argument of (a), this makes $\hat{h}$ quadratic, with the canonical height pairingA torsion point has canonical height zero because its multiples have bounded naive height. Conversely, if $\hat{h}(P)=0$, then all $2^{n}P$ have bounded naive height. Northcott finiteness makes two equal, giving $(2^j-2^i)P=O$ for some $j>i$. Thus canonical height vanishes exactly on torsion. In particular the torsion subgroup is finite, since the comparison with naive height bounds the heights of all its points.

We can now prove the Mordell-Weil theorem. The Weak Mordell-Weil theorem, as explained in Question 2(ii), makes $E(K)/2E(K)$ finite. Choose representatives $R_1,\dots ,R_s$ and put $M={\max }_i\hat{h}(R_i)$. Every $P$ can be written $P=2Q+R_i$. The parallelogram identity and nonnegativity giveIterating this height descent makes the height at step $n$ at most $M+2^{-n}\max (\hat{h}(P)-M,0)$, so after finitely many steps it is at most $M+1$. There are only finitely many points of that height by Northcott and the naive-height comparison. Reversing $P=2Q+R_i$ shows that every point is an integer combination of the finitely many representatives and this finite bounded-height set. ThereforeOn the resulting free part the canonical height pairing is positive definite. Its nonnegativity extends from rational vectors to real vectors by density. If a nonzero real vector $v$ were in its kernel, in a basis of the free part round $nv$ to integer vectors $z_n$. Their rounding errors are uniformly bounded, and $\hat{h}(z_n)=\hat{h}(z_n-nv)$ would be uniformly bounded. The vectors $z_n$ are unbounded and therefore include infinitely many distinct vectors, contradicting Northcott finiteness. Thus there is no real null vector. Heights consequently support quantitative point searches, tests of independence, and the regulator of an elliptic curve, in addition to providing the descent that proves finite generation.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

Work with the usual commutative unital ring convention. Write $f(T)=a_{1}T+a_2{T}^2+\cdots$, where $a_1=f^{\prime }(0)$ is a unit, and seek $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. Since $g$ has zero constant term, every coefficient of the composition of these formal power series involves only finitely many terms.

The linear coefficient forces $b_1=a_1^{-1}$. Suppose $b_1,\dots ,b_{n-1}$ have been chosen. The coefficient of $T^n$ in $f(g(T))$ isIndeed the term involving $b_n$ can only come from $a_{1}g$: in $g^j$ with $j\ge 2$, using a degree-$n$ term forces total degree at least $n+j-1>n$. Choose $b_n=-a_1^{-1}{C}_n$. This recursive construction of a compositional inverse uniquely gives $f\circ g=T$.

To check the other composition rather than assume it, construct $h(T)={\sum }_{n\ge 1}{c}_n{T}^n$ satisfying $h\circ f=T$. At degree $n$, its new coefficient is $c_n{a}_1^n$ plus terms already determined; $a_1^n$ is a unit, so the same recursion works. Composition is associative, as can be checked coefficientwise, or after truncation modulo $T^{N+1}$ for every $N$. ThereforeConsequently the compositional inverse of a formal power series is unique and satisfies

Set $d=-7$. The affine rational pair $(u,v)=(2,-1)$ gives $P=(84,756)$ on the elliptic curve $y^2=x^3-21168$. It is a point of infinite order: in the rational torsion of a diagonal cubic classification, neither $-7$ nor $-7/2$ is a rational cube, so this curve has trivial rational torsion subgroup.

There is also a short direct torsion check specific to this curve. The primes $5$ and $11$ are good and have point counts $6$ and $12$, respectively, because cubing is bijective in their finite fields. Reduction at these two primes bounds the entire rational torsion by order dividing $6$, including its possible $5$- and $11$-primary parts by reduction at the other prime. The flex calculation excludes order $3$, and the inversion calculation excludes order $2$. Hence $P$ is indeed nontorsion.

Every nonzero multiple $nP$ is distinct, and its inverse projective image has $Z\ne 0$: the only rational point at infinity on the diagonal cubic is its identity. Thus each yields a distinct rational pair. More explicitly, if $nP=(x_n,y_n)$ on the Weierstrass model, thenHere $x_n\ne 0$, since $x_n=0$ would require a rational square equal to $-21168$. For example the elliptic-curve addition formula gives $2P=(28,28)$, which produces $(u_2,v_2)=(5/3,4/3)$. This proves there are infinitely many rational pairs of cubes summing to seven.

Use the invertible linear projective transformationThe identitygivesThe transformation is invertible because $d\ne 0$, and it sends $O$ to $(0:1:0)$. Hence it is an isomorphism of the pointed smooth projective curves, giving the required Weierstrass equation of an elliptic curve:For affine points $u=X/Z,v=Y/Z$, the transformation and its inverse areThese formulae are interpreted in their stated affine charts; the projective transformation handles the exceptional points without any ambiguity.

The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

We take $d\in {\mathbb{Q}}^{\times }$, as required for this rational elliptic curve. The tangent at $O=(1:-1:0)$ is $X+Y=0$. Its intersection with the curve has $d{Z}^3=0$, so it meets the curve three times at $O$: the identity is an inflection point of a plane cubic.

For $P,Q$, let their chord meet the diagonal cubic elliptic curve at the third point $R$, using the tangent if $P=Q$ and counting intersection multiplicities. Then join $R$ to $O$ and take the third intersection $S$. Define $P+Q=S$. Equivalently, three collinear intersection points satisfy $P+Q+R=O$.

Here inversion has the particularly simple formIndeed $P$, its displayed inverse, and $O$ lie on the same line, since swapping $X,Y$ preserves both the curve and $X+Y$. Thus addition is the chord or tangent operation followed by swapping the first two coordinates. The rule includes limiting cases involving $O$ and repeated intersections. Its associativity follows from the elliptic curve group law from Riemann-Roch: the identification $P\mapsto [P-O]$ with the degree-zero Picard group turns the line-intersection Weil divisor relation into addition of divisor classes. Since a line cuts a Weil divisor linearly equivalent to $3O$, that relation is exactly the chord rule just described.

Suppose a nonempty set $S$ had no minimal member for the intersection-power-set predecessor relation. Put $b=\bigcap S$, which is a set by separation. For every $y\in S$ there is an $x\in S$ with $\mathcal{P}(x\cap y)\subseteq y$. Since $b\subseteq x\cap y$, this gives $\mathcal{P}(b)\subseteq y$. Intersecting over every $y\in S$ yields $\mathcal{P}(b)\subseteq b$.

Now form the diagonal subset $d=\{u\in b:u\notin u\}$. It belongs to $\mathcal{P}(b)$, hence to $b$, so its defining condition implies $d\in d⟺d\notin d$, a contradiction. Thus $\boxed{R\text{ is well-founded}.}$ This is a Cantor diagonal argument; it uses neither the Axiom of foundation nor the axiom of choice. In particular, no selection of an infinite descending sequence is needed.

Order the collection $\mathrm{Ch}(P)$ of all chains in a partial order, including the empty chain, by inclusion. Suppose that $f:\mathrm{Ch}(P)\to P$ were an order-preserving injection. The chain-poset nonembedding lemma follows by a transfinite recursion of length the Hartogs ordinal $h(P)$:Inductively, the previously constructed points form a strictly increasing chain. If $\beta <\alpha$, then $C_{\beta }⊊C_{\alpha }$, since $p_{\beta }$ belongs to the latter but not the former. Monotonicity gives $p_{\beta }\le p_{\alpha }$, and injectivity makes this inequality strict. Thus the recursion continues at every $\alpha <h(P)$ and gives an injection $h(P)\to P$, contradicting the defining property of the Hartogs ordinal. Therefore $\boxed{\mathrm{Ch}(P)\text{ has no order-preserving embedding into }P.}$ The empty chain is part of the convention: excluding it would make a one-point poset a counterexample.

Suppose $F=pJ=p^{\prime }{J}^{\prime }$ are two such factorizations, with $J,J^{\prime }$ final and $p,p^{\prime }$ discrete fibrations. Apply the unique diagonal construction to the square with left side $J$, top $J^{\prime }$, bottom $p$ and right side $p^{\prime }$. It gives a unique functor $L:\mathcal{E}\to {\mathcal{E}}^{\prime }$ such that $LJ=J^{\prime }$ and $p^{\prime }L=p$. Reversing the two factorizations gives $M:{\mathcal{E}}^{\prime }\to \mathcal{E}$ with $M{J}^{\prime }=J$ and $pM=p^{\prime }$.

Both $ML$ and $1_{\mathcal{E}}$ fill the square with left side $J$, top $J$, bottom $p$ and right side $p$. Uniqueness from part (i) gives $ML=1_{\mathcal{E}}$. Similarly $LM=1_{{\mathcal{E}}^{\prime }}$. ThusIn particular the conclusion is a canonical isomorphism of intermediate categories, not just an unspecified equivalence.

For each $d\in \mathcal{D}$, let $P(d)={\pi }_0(d\downarrow F)$, the collection of connected components, where connectedness means connection by a zigzag of arrows. Precomposition with $u:d\to d^{\prime }$ gives a function $P(u):P(d^{\prime })\to P(d)$. Identity and composition are inherited from precomposition.

Construct a category $\mathcal{E}$ whose objects are pairs $(d,c)$ with $c\in P(d)$. An arrow $(d,c)\to (d^{\prime },c^{\prime })$ is an arrow $u:d\to d^{\prime }$ satisfying $c=P(u)c^{\prime }$. Composition is that of $\mathcal{D}$, and the component equation is preserved by composition. The projection $p:\mathcal{E}\to \mathcal{D}$ is a discrete fibration: an arrow $u:d\to d^{\prime }$ has precisely one lift into $(d^{\prime },c^{\prime })$, namely the arrow from $(d,P(u)c^{\prime })$ represented by $u$.

Define $J:\mathcal{C}\to \mathcal{E}$ byThis is well defined because $(a,1_{Fa})$ and $(a^{\prime },Fj)$ lie in the same component of $(Fa\downarrow F)$ whenever $j:a\to a^{\prime }$. Clearly $pJ=F$.

For an object $(d,c)$ of $\mathcal{E}$, an object of $((d,c)\downarrow J)$ is exactly a pair $(a,u:d\to Fa)$ representing the component $c$. Its morphisms are exactly the arrows in $(d\downarrow F)$ between such objects. Thus this comma category is the component $c$ itself, which is nonempty and connected. This proves that $J$ is a final functor, and henceThis is the final-discrete-fibration factorization. For small categories the collections $P(d)$ are sets and this construction is the category of elements of the presheaf $P$. For categories of a larger size, the construction is interpreted in the same chosen ambient size framework.

Write the square as $GH=KF$, with $F:\mathcal{A}\to \mathcal{B}$ final and $G:\mathcal{C}\to \mathcal{D}$ a discrete fibration. For $b\in \mathcal{B}$, choose an object $(a,h:b\to Fa)$ of the nonempty comma category $(b\downarrow F)$. Lift $K(h):Kb\to GHa$ uniquely to an arrowover $K(h)$, and tentatively put $L(b)=c_h$.

A morphism $(a,h)\to (a^{\prime },h^{\prime })$ in this comma category is an arrow $j:a\to a^{\prime }$ with $F(j)h=h^{\prime }$. The composite $H(j){\ell }_h$ is a lift of $K(h^{\prime })$ with codomain $H{a}^{\prime }$. Uniqueness of lifts gives $H(j){\ell }_h={\ell }_{h^{\prime }}$, including equality of their source objects. Since the comma category is connected by zigzags, $c_h$ is independent of the choice. Thus $L(b)$ is well defined and $GL(b)=Kb$.

For $f:b\to b^{\prime }$, take $(a^{\prime },h^{\prime }:b^{\prime }\to F{a}^{\prime })$. Lift $K(f)$ into $L(b^{\prime })$, obtaining $t:c\to L(b^{\prime })$. Composing with ${\ell }_{h^{\prime }}$ gives a lift of $K(h^{\prime }f)$ into $H{a}^{\prime }$, whose source must be $L(b)$. Hence $t:L(b)\to L(b^{\prime })$. Set $L(f)=t$. Uniqueness of this lift makes the definition independent of the chosen comma object and proves $L(1_b)=1_{L(b)}$ and $L(gf)=L(g)L(f)$.

At $b=Fa$, choose $(a,1_{Fa})$. Its lift is $1_{Ha}$, so $L(Fa)=Ha$. For $j:a\to a^{\prime }$, $Hj$ is the unique lift of $K(Fj)$ into $H{a}^{\prime }$, so $L(Fj)=Hj$. We have constructed the required functor with $LF=H$ and $GL=K$.

If $L^{\prime }$ is another filler, $L^{\prime }(h):L^{\prime }b\to Ha$ must equal ${\ell }_h$ by unique lifting. Thus $L^{\prime }b=L(b)$ for every object. Its arrows are then forced by unique lifting into those objects. ThereforeThis proves the orthogonality of final functors and discrete fibrations.

Take $\Gamma ={\mathrm{PSL}}_2(\mathbb{Z})$ as a subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$. It is discrete because representatives have integer entries, so a bounded set of matrices contains only finitely many such representatives, modulo the two choices of sign. It is finitely generated byThe Euclidean algorithm on the first column of an integer determinant-one matrix reduces it using these two transformations, proving the generation statement.

It is non-elementary. The powers of $T$ move an interior upper-half-space point towards $\infty$, so $\infty \in \Lambda (\Gamma )$. Conjugating $T$ by $S$ gives a parabolic with fixed point $0$, and conjugating that parabolic by $T$ gives one with fixed point $1$. Their interior orbits approach those respective points. Thus $0,1,\infty$ are three distinct points in the Kleinian limit set.

However $S^2=1$ in the projective group, and $S$ is a nonidentity elliptic Möbius transformation fixing the hyperbolic geodesic with boundary endpoints $i,-i$. It is a half-turn around that axis. Near an ordinary point of the axis, quotienting identifies the transverse angle modulo $\pi$, giving cone angle $\pi$ instead of $2\pi$. The elliptic stabilizer obstruction to a hyperbolic manifold quotient therefore givesThe obstruction is the nonfree action, not finite generation or elementary behavior. An underlying topological quotient can still be a manifold; the conical quotient metric is what prevents this example from being a hyperbolic manifold quotient.

Write $N={\Gamma }^{\prime }$. As a subgroup of a discrete group, it is a Kleinian group, even if it is not finitely generated. Since it is infinite, part (b) gives a nonempty closed Kleinian limit set $\Lambda (N)$. For $\gamma \in \Gamma$, conjugation and base-point independence giveOne can see this directly by applying $\gamma$ to $No$: the image is the orbit of $\gamma o$ under the conjugate subgroup, and the choice of interior point does not affect its limit set. Normality makes the conjugate subgroup equal to $N$, so $\Lambda (N)$ is $\Gamma$-invariant. Applying part (c) yields $\Lambda (\Gamma )\subseteq \Lambda (N)$.

Conversely $No\subseteq \Gamma o$, so every boundary accumulation point of the smaller orbit is one of the larger orbit: $\Lambda (N)\subseteq \Lambda (\Gamma )$. Therefore the limit set of an infinite normal Kleinian subgroup satisfiesInfinitude supplies nonemptiness, normality supplies invariance, and non-elementarity supplies minimality. Properness of the subgroup is not needed, and no finite-generation assumption on it has been inserted.

Let $F\subseteq S_{\infty }^2$ be nonempty, closed and $\Gamma$-invariant. We first show it cannot be a singleton when $\Gamma$ is a non-elementary Kleinian group.

Suppose every element fixes one boundary point, conjugated to $\infty$. Each boundary map then has the affine form $z\mapsto az+b$. If all elements have $|a|=1$, their action in upper half-space keeps the height of $o$ fixed. An orbit sequence escaping compact interior sets must have its horizontal coordinate tending to infinity along any convergent boundary subsequence, so its only possible boundary limit is $\infty$. If some element has $|a|\ne 1$, move its second fixed point to $0$, making it $f(z)=az$. For any $g(z)=cz+d$, its conjugates are $f^{n}g{f}^{-n}(z)=cz+a^{n}d$. Choose $n\to +\infty$ or $-\infty$ so $a^n\to 0$. If $d\ne 0$, these are distinct group elements converging to $z\mapsto cz$, contradicting discreteness. Thus $d=0$ for every group element: all fix both $0$ and $\infty$, and their only possible interior-orbit boundary limits are these two points. In both cases the Kleinian limit set has at most two points. This contradicts non-elementarity. Therefore $F$ has at least two distinct points.

Use the Klein ball model, in which hyperbolic geodesics are straight chords, and take the closed Euclidean convex hull $C$ of $F$ in the closed ball. It contains an interior point $o$, since the open chord between two distinct points of $F$ lies in the open ball. The interior part of this hull is hyperbolically convex and invariant under $\Gamma$: isometries preserve geodesic segments, so they preserve the hyperbolic convex hull of the invariant set. Thus $\Gamma o\subseteq C$.

The only points of $C$ on the boundary sphere are the points of $F$. Indeed, if $\xi \in S_{\infty }^2\setminus F$, compactness of $F$ gives $\xi \cdot z\le c<1$ for every $z\in F$, because $\xi \cdot z=1$ on the unit sphere only when $z=\xi$. The same linear inequality holds throughout $C$, excluding $\xi$ from it. Conversely $F\subset C$, so $C\cap S_{\infty }^2=F$.

Every boundary accumulation point of $\Gamma o$ consequently lies in $F$. Base-point independence from part (b) identifies that accumulation set with $\Lambda (\Gamma )$, proving minimality of the Kleinian limit set:This proves the stated containment for every nonempty closed invariant set, not merely for sets containing an arbitrarily chosen orbit on the boundary.

Fix an interior point $o\in {\mathbb{H}}^3$. The limit set of a Kleinian group is the set of boundary accumulation points of its orbit, orThe boundary compactification may be realized as the closed unit ball. A discrete isometry group has only finitely many orbit points in each compact interior region: isometries moving $o$ into a compact set form a compact family after accounting for the compact rotation stabilizer, and infinitely many distinct group elements there would contradict discreteness. The stabilizer of $o$ is finite for the same reason. Thus an infinite group has an infinite orbit escaping every compact interior set and a nonempty boundary accumulation set. A finite group has empty limit set. This also shows the displayed intersection does not introduce ordinary interior accumulation points.

For another point $o^{\prime }$, the corresponding points satisfy $d(\gamma o,\gamma o^{\prime })=d(o,o^{\prime })=:C$. In the Poincaré ball model the distance identity isIf ${\gamma }_{j}o\to \xi \in S_{\infty }^2$, bounded distance impliesSo ${\gamma }_j{o}^{\prime }\to \xi$ as well. Exchanging $o,o^{\prime }$ proves equality of the two accumulation sets. Hence the definition is independent of the interior base point.

It is closed as an intersection of the compact orbit closure with the closed boundary, and it is $\Gamma$-invariant because multiplying an orbit by an element of $\Gamma$ permutes that orbit. Changing ball coordinates or conjugating the geometric realization transports the set by the induced boundary homeomorphism. Different nonconjugate representations of an abstract group are not asserted to have the same subset of one fixed sphere. None of these arguments requires finite generation.

A Kleinian group is a discrete subgroup of the orientation-preserving isometry group of ${\mathbb{H}}^3$, identified with ${\mathrm{PSL}}_2(\mathbb{C})$. For an abstract finitely generated group this means a faithful representation into ${\mathrm{PSL}}_2(\mathbb{C})$ whose image is discrete; the chosen image is the geometric realization. Its boundary action is by Möbius transformations on $S_{\infty }^2\cong \mathbb{C}\cup \{\infty \}$.

Discreteness means the identity is isolated in the group topology, and for a subgroup it also excludes an infinite sequence of distinct elements converging anywhere in the isometry group. Torsion-freeness is not part of this definition: elliptic Möbius transformations of finite order are allowed. Therefore a Kleinian quotient need not be a smooth hyperbolic manifold.

For the intended standard Hopf link, identify $S^3$ with the unit sphere in ${\mathbb{C}}^2$ and its two components with the coordinate circles $z_1=0$ and $z_2=0$. Outside small neighborhoods of those circles, both coordinates are nonzero, and their arguments together with $|z_1|$ giveOn this product the longitude of either component is the meridian of the other. Therefore the two attaching meridians become the two coordinate curves of $T^2$, intersecting exactly once. Adding the product collar does not change a gluing of solid tori. The meridian-intersection-one argument in Question 2(c) identifies the result with the standard genus-one splitting of the sphere. Thus zero surgery on the Hopf link gives

There is a genuine link-type qualification in the wording: merely specifying two linked but individually unknotted components does not determine the Hopf link. For example the two components of the torus link $T(2,4)$ are unknotted but have linking number $2$. Filling their preferred longitudes is zero surgery with linking matrixso the resulting first homology is its cokernel, $\mathbb{Z}/2\oplus \mathbb{Z}/2$, rather than the zero group of $S^3$. The cokernel formula follows because zero longitudes impose twice the other meridian as the homology relations. Thus the sphere answer applies to the standard once-linked Hopf interpretation; the literal unspecified link description alone is insufficient to force a single manifold.

Each component is an unknot, so its closed exterior $S^3\setminus \mathrm{int}{U}_i$ is a solid torus. Choose the usual boundary basis $({\mu }_i,{\lambda }_i)$, where ${\mu }_i$ is a meridian of the component and ${\lambda }_i$ its Seifert longitude. In the exterior's fundamental group, ${\mu }_i$ generates $\mathbb{Z}$ and ${\lambda }_i$ is trivial, since it bounds a meridian disk of that exterior.

Every essential embedded circle on a torus represents a primitive homology class $a{\mu }_i+b{\lambda }_i$ with $\gcd (a,b)=1$. If it bounds a disk in the exterior, it is null-homotopic there, so $a=0$. Primitivity then forces $b=\pm 1$. Essential simple curves of the same primitive unoriented slope on a torus are isotopic: lift them to the plane and straighten their parallel primitive lines, or cut the torus along one curve to obtain an annulus. ConsequentlyThis is the preferred longitude of an unknot. The disk is only required to lie in the complement of its own solid torus; it may meet the other link component, so no additional disjoint-disk assertion is being used.