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Use integral coefficients throughout, allowing a nontrivial orientation character. A dimension-$d$ Poincare duality group has a finite resolution by finitely generated projective group-ring modules andas abelian groups. The action on the top copy of $\mathbb{Z}$ may be by signs. This group-ring criterion is particularly suited to the Lyndon–Hochschild–Serre spectral sequence.

Apply that sequence to the extension with coefficient module $\mathbb{Z}G$:As an $N$-module, $\mathbb{Z}G$ is a direct sum of copies of $\mathbb{Z}N$ indexed by the cosets in $Q$. Cohomology of $N$ commutes with this direct sum because its projective resolution is finitely generated. Hence the inner cohomology vanishes except when $b=n$, where it is a direct sum of rank-one groups indexed by $Q$. The $Q$-action shifts these summands freely and transitively, with possible signs from the orientation module. Choosing a generator in one summand and transporting it by $Q$ identifies the resulting left $Q$-module with its regular module $\mathbb{Z}Q$. The residual right group action records the orientation twist; it need not be trivial.

Now the duality of $Q$ says that the only nonzero $E_2$ term is $(a,b)=(q,n)$ and that this term is $\mathbb{Z}$ as an abelian group. No differential can enter or leave it. Therefore $H^{\ast }(G;\mathbb{Z}G)$ has exactly the duality pattern in degree $n+q$. The usual extension-resolution construction also supplies a resolution by finitely generated projective modules: combine the finite kernel and quotient resolutions, lift the quotient differentials and add the necessary homotopy corrections. Its total length is at most $n+q$. Alternatively, the same spectral sequence for arbitrary coefficient modules gives the dimension bound, while the extension finiteness lemma supplies finite generation of the resolution. The surviving top term makes the bound sharp. This proves the extension rule for Poincare duality groups:Requiring a trivial top orientation action would need an additional orientation condition; the untwisted conclusion is not implicit in the hypotheses.

We next need two precise auxiliary dimension results. The Strebel infinite-index subgroup theorem says that an infinite-index subgroup of an integral $P{D}^d$ group has integral cohomological dimension at most $d-1$. The Stallings-Swan theorem says that a group of integral cohomological dimension at most one is free, including the trivial group. Since $G/N$ contains an element of infinite order, $N$ has infinite index in $G$, so $\mathrm{cd}N\le 2$. If its dimension is at most one, the required free-group conclusion follows, with finite rank because $N$ is finitely presented.

Suppose instead that $\mathrm{cd}N=2$. Finite presentation gives a partial free resolution with finitely generated modules through degree two. The kernel in degree one is finitely generated and, because the integral cohomological dimension is two, projective. Truncating at this kernel gives a finite resolution by finitely generated projective modules. Thus $N$ has the finiteness needed for group-ring cohomology to commute with direct sums.

Choose an infinite-order element $\bar{t}\in G/N$, lift it to $t\in G$, and let $H$ be the preimage of $⟨\bar{t}⟩$. Then $H=N⋊⟨t⟩$. Restriction of $\mathbb{Z}H$ to $N$ yields copies of $\mathbb{Z}N$ indexed by all powers of $t$. For every $j$, the quotient generator acts on $H^j(N;\mathbb{Z}H)$ by shifting this direct sum, possibly with an automorphism on each copy. Its invariants vanish, since an invariant finite-support sequence must be zero. Its coinvariants are one copy of $H^j(N;\mathbb{Z}N)$: the shift relations identify successive copies, and any twists can be removed by transporting the identifications along the integer index.

The infinite cyclic group has integral cohomological dimension one, so only columns zero and one occur in the spectral sequence. Column zero vanishes and column one consists of these coinvariants. We obtain the cyclic-extension shift of group-ring cohomologyas abelian groups. Moreover $H^2(N;\mathbb{Z}N)\ne 0$. Here is a useful proof of that last point: if top cohomology of a finite projective resolution were zero, its last dual differential would be onto. Its target is projective, so that differential splits. Dualizing back splits the last injection of the original resolution, shortening it and contradicting integral cohomological dimension two.

It follows that $H^3(H;\mathbb{Z}H)\ne 0$, and hence $\mathrm{cd}H=3$. Strebel's theorem forces $H$ to have finite index in $G$. It is therefore itself a $P{D}^3$ group, by finite-index invariance of Poincare duality for torsion-free groups. The displayed shift now makes $H^j(N;\mathbb{Z}N)$ zero except for $j=2$, where it is $\mathbb{Z}$. Together with its finite projective resolution, this provesThe classification of two-dimensional Poincare duality groups identifies the latter with the fundamental group of a closed aspherical surface. This classification includes orientation-twisted surface groups; it does not include the two-sphere or real projective plane, whose fundamental groups do not have integral cohomological dimension two.

In this latter case we have also proved that $G/N$ is virtually infinite cyclic, since $H/N=⟨\bar{t}⟩$ has finite index. If necessary replace the surface group by its characteristic index-two orientable subgroup and take the subgroup generated by it and $t$. This still has finite index in $G$ and is a semidirect product of an orientable closed aspherical surface group by $\mathbb{Z}$. The Dehn-Nielsen-Baer theorem for closed orientable surfaces realizes the outer automorphism defined by conjugation by $t$ as a surface homeomorphism $f$. For the torus this is just realization of $G{L}_2(\mathbb{Z})$ by linear homeomorphisms. Changing the chosen lift accounts for an inner automorphism. The mapping torusis a surface bundle over $S^1$ with fundamental group ${\pi }_1(S){⋊}_{f_{\ast }}\mathbb{Z}$. Thus a finite-index subgroup of $G$ is the fundamental group of a surface fibration over the circle. Taking a further double cover of the base if required makes the monodromy orientation-preserving.

A nilpotent group has lower central series ${\gamma }_{1}G=G$, ${\gamma }_{j+1}G=[G,{\gamma }_{j}G]$ terminating at the identity. A polycyclic group has a finite subnormal series with cyclic factors, which may be finite or infinite.

If a nilpotent group is generated by finitely many elements, then ${\gamma }_{j}G/{\gamma }_{j+1}G$ is generated by the finitely many weight-$j$ iterated commutators in these generators. Modulo the next term, commutator identities make these quotients abelian and permit this collection of words. Each quotient is thus a finitely generated abelian group. Refine each by a series with cyclic factors and splice the refinements through the finite lower central series. This proves that a finitely generated nilpotent group is polycyclic.

For the duality-group assertion, first prove the useful finite-index reduction. Induct on the length of a polycyclic series, taking its last proper term $H◃G$. By induction $H$ has a finite-index poly-infinite cyclic group. Because $H$ is finitely generated, it has only finitely many subgroups of any fixed finite index. Intersect all automorphic images of that subgroup to obtain a characteristic finite-index subgroup $H_0\subset H$ which is still poly-infinite cyclic. It is normal in $G$. The quotient $G/H_0$ is finite-by-cyclic. If it is finite, $H_0$ already suffices; if its cyclic quotient is infinite, the subgroup generated by a lift of its generator is infinite cyclic and has finite index. Its preimage in $G$ is an extension of $H_0$ by $\mathbb{Z}$, hence poly-infinite cyclic. This proves polycyclic groups are virtually poly-infinite cyclic.

Take such a finite-index subgroup $P$ of the given dimension-three Poincare duality group. It is again a dimension-three duality group. The extension rule proved in Question 3, applied successively to its infinite cyclic series, makes $P$ a duality group of dimension equal to its Hirsch length. Hence that length is three. Its last quotient giveswhere $H$ has a poly-infinite cyclic series of length two. Thus $H\cong \mathbb{Z}⋊\mathbb{Z}$, with the last generator acting on the first by $+1$ or $-1$. In the first case $H={\mathbb{Z}}^2$. In the second case $H$ is the Klein-bottle group $⟨a,b\mid ba{b}^{-1}=a^{-1}⟩$. Its characteristic subgroup $⟨a,b^2⟩\cong {\mathbb{Z}}^2$ is the centralizer of its commutator subgroup $⟨a^2⟩$. Replace $H$ by this characteristic index-two subgroup if necessary. It remains normal in $P$, and the quotient is finite-by-infinite-cyclic. Taking the preimage of a finite-index infinite cyclic subgroup givesThis extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0={\mathbb{Z}}^2{⋊}_B\mathbb{Z}$ for some $B\in G{L}_2(\mathbb{Z})$.

In additive notation on ${\mathbb{Z}}^2$, $[t,v]=(B-I)v$. All its other internal commutators vanish, so induction givesConsequentlyThis is the nilpotence criterion for an abelian-by-cyclic group. If $B=I$ the group is abelian; otherwise it is two-step nilpotent. The criterion concerns this actual semidirect product. A finite extension of it is only known to be virtually nilpotent: for example ${\mathbb{Z}}^2{⋊}_{-I}\mathbb{Z}$ is virtually abelian but its lower central series never terminates.

Write elements of the real Heisenberg group as $(x,y,z)$. Direct matrix multiplication and inversion giveThese formulas show that $N_k=(k\mathbb{Z}{)}^3$ is a subgroup. It is discrete. For the right quotient, multiplying by $(ka,kb,kc)$ changes the coordinates toThus $(x\mathrm{mod}k,y\mathrm{mod}k)$ is well defined on the quotient. Over a small base rectangle choose real lifts $x,y$ of its coordinates. Every coset above that rectangle has a representative $(x,y,z)$, and two such representatives differ precisely by $z\mapsto z+kc$. Henceis a smooth local trivialization. The fibres are circles, proving the locally trivial fibration over ${\mathbb{R}}^2/k{\mathbb{Z}}^2\cong T^2$. Reducing first $x,y$ and then $z$ also gives a compact fundamental region, so $N_k$ is cocompact.

The ambient Lie group is diffeomorphic to ${\mathbb{R}}^3$, hence simply connected and contractible. Its quotient has fundamental group $N_k$, and the first integral homology is its abelianization. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^m{b}^n{c}^{\ell }$, since its third coordinate is $k^{2}mn+k\ell$. The commutator formula isso $c$ is central and $[a,b]=c^k$. Every commutator lies in $(0,0,k^2\mathbb{Z})$, and this subgroup is generated by $[a,b]$. Thus $N_k^{\prime }=⟨c^k⟩$. The homomorphism $(x,y,z)\mapsto (x/k,y/k,z/k\mathrm{mod}k)$ is surjective onto ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$ and has exactly this kernel; the cross term in multiplication is divisible by $k$ after dividing by $k$. ThereforeFor $k=1$ the last factor is trivial. This is the Scaled Heisenberg lattice, whose fibre is central but whose total group need not split as a direct product of groups.

The possible ranks and the corresponding homeomorphism types areThere is no example for $n=2$ or $n\ge 4$. The uniqueness here is topological, not uniqueness of a Riemannian metric.

To see the restrictions, use the prime decomposition of a closed orientable three-manifold. Its fundamental group is the free product of the prime three-manifold groups. A free product of two nontrivial groups is nonabelian, so at most one summand can have nontrivial fundamental group. The remaining simply connected summands are three-spheres by the Poincaré conjecture. If the nontrivial summand is not irreducible, it is $S^2\times S^1$, giving rank one. If it is irreducible and has infinite fundamental group, the sphere theorem for three-manifolds makes it aspherical. Its fundamental group is then a dimension-three Poincare duality group. But ${\mathbb{Z}}^n$ is a duality group of dimension $n$, as seen from the torus classifying space or its Koszul resolution, so $n=3$.

For rank three, the flat case of the geometrization theorem says that a closed irreducible orientable three-manifold with fundamental group ${\mathbb{Z}}^3$ admits a flat metric. Write it as ${\mathbb{R}}^3/\Gamma$. By the Bieberbach theorem, $\Gamma$ contains a full-rank translation lattice $\Lambda$. If $g=(R,v)\in \Gamma$ commutes with translation by each $\lambda \in \Lambda$, then $R\lambda =\lambda$ for every lattice vector, forcing $R=I$. Since $\Gamma$ itself is abelian, it consists entirely of translations. Its quotient is therefore the three-torus. Rank zero follows directly from the Poincaré conjecture, and rank one was already determined by the prime decomposition of a closed orientable three-manifold. Historically, before the Poincare theorem, the uniqueness conclusions were phrased allowing connected sums with homotopy three-spheres; the modern theorem removes that ambiguity.

The six flat types below mean the six closed orientable flat three-manifold types. There are infinitely many individual flat metrics and there are additional noncompact flat Riemannian manifolds. In each closed case the universal cover is Euclidean space and the deck transformation group is a torsion-free crystallographic group. The holonomy groups of closed orientable flat three-manifolds are $1,C_2,C_3,C_4,C_6,C_2\times C_2$.

The first is a flat torus ${\mathbb{R}}^3/\Lambda$, with trivial holonomy. For each cyclic case choose a planar lattice $L$ admitting the relevant rotation $R_m$, and generate the deck transformation group by translations in $L$ and the screwIts $m$th power is unit translation along the axis. A nontrivial holonomy element has nonzero axial displacement modulo integers, so has no fixed point. All transformations preserve orientation. A compact prism fundamental domain gives compactness. Equivalently, the manifold is a torus mapping torus with finite-order monodromy. In planar lattice bases one may takeThe half-turn uses any planar lattice, the quarter-turn a square lattice, and the third- and sixth-turns a hexagonal lattice. These give respectively the half-turn flat three-manifold, third-turn flat three-manifold, quarter-turn flat three-manifold, and sixth-turn flat three-manifold. Their holonomies have orders $2,3,4,6$, and their first Betti numbers are all one, since the only fixed direction of holonomy is the screw axis.

The sixth type is the Hantzsche-Wendt manifold. A concrete group is generated byTheir linear parts are commuting half-turns about perpendicular axes. Moreover ${\alpha }^2$, ${\beta }^2$ and $(\alpha \beta {)}^2$ are translations by $(1,0,0)$, $(0,1,0)$ and $(0,0,-1)$. Their translation subgroup is the unit cubic lattice and their linear quotient is $C_2\times C_2$. In each nontranslation coset, the displacement along the fixed axis is a half-integer, so the action is free. The quotient is compact and orientable. No vector is fixed by both linear parts, so its first Betti number is zero. Bieberbach classification gives precisely these six affine types; the explicit screw descriptions distinguish them without identifying arbitrary finite quotients with holonomy.

An index-$\lambda$ handle in dimension $d$ is $D^{\lambda }\times D^{d-\lambda }$. A handle attachment glues its attaching region $S^{\lambda -1}\times D^{d-\lambda }$ to a corresponding neighborhood in the boundary of the preceding manifold with boundary. Besides the embedding of the attaching sphere, one specifies a framing of an embedded sphere, namely the normal-disk directions used in this gluing. Corners are then smoothed.

In dimension four, a two-handle is $D^2\times D^2$ and attaches along $S^1\times D^2$. Its attaching circle is a knot in the boundary of the zero-handle $D^4$. Identify that boundary with $S^3$ and remove a point away from the circles to draw them in ${\mathbb{R}}^3$. Disjoint attaching circles give a framed link. Each integer records the twist of its normal framing relative to the zero Seifert framing; the integer is also the linking number of the circle and its framed push-off. The resulting boundary change is Dehn surgery on a framed link. This is the two-handle part of a Kirby diagram.

For $S^2\times S^2$, use the following explicit handle decomposition. Write each factor as two disks, $S^2=D_{-}^2\cup D_{+}^2$, glued along their boundary circles. The four product pieces giveThe first is a four-dimensional zero-handle after rounding its corners. Its boundary is the union of the solid tori $S^1\times D^2$ and $D^2\times S^1$. The two middle product pieces attach along these tori. Their attaching circles $S^1\times \{0\}$ and $\{0\}\times S^1$ form the standard Hopf link, with linking number one after choosing orientations.

Both framings are zero. For example, the normal disk of $S^1\times \{0\}$ is the second disk factor. Its constant normal push-off has linking number zero with the original circle: the circle bounds $D^2\times \{0\}$ in the four-ball, and a parallel push-off bounds a disjoint translated disk. Thus the framed self-intersection is zero; the same argument applies with the factors interchanged. These product framings are the zero Seifert framings.

After these two two-handles, the remaining product piece $D_{+}^2\times D_{+}^2$ is a four-ball. Its boundary is exactly the boundary of the assembled first three pieces, so it attaches as a four-handle. Equivalently, zero surgery on the Hopf link gives $S^3$, which this handle caps. We have therefore constructed the actual product, provingThere are no one- or three-handles. As a check, the two factor spheres have self-intersection zero and meet once, giving the intersection form $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The explicit product construction, rather than the intersection form alone, identifies the smooth four-manifold. This is the zero-framed Hopf-link handle decomposition of the sphere product.

Here projection means Hermitian orthogonal projection: $P=P^{\ast }=P^2$ with $\mathrm{tr}P=k$, as indicated by the unitary conjugation in the question. These matrices represent the complex Grassmannian as orthogonal projections, a compact manifold of complex dimension $k(n-k)$. The constants $c_i$ are real. Let $A^{\ast }=-A$ and use the given tangent directions $[A,P]$. Cyclicity of the matrix trace givesThe commutator $B=[P,C]$ is skew-Hermitian. If this derivative vanishes for every skew-Hermitian $A$, choose $A=B$ to obtainso $B=0$. Conversely $[P,C]=0$ makes every tangent derivative zero. Because the diagonal entries of $C$ are distinct, a commuting matrix $P$ is diagonal. Its idempotent entries are zero or one. Thus the critical projections are exactlywith $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ critical points and values $f(P_I)={\sum }_{i\in I}{c}_i$.

To compute the Hessian matrix, write $E=\mathrm{span}\{e_i:i\in I\}$ and $E^{\perp }=\mathrm{span}\{e_j:j\notin I\}$. A neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. With $W=\left(\genfrac{}{}{0ex}{}{I}{Z}\right)$, its orthogonal projection is $P(Z)=W(I+Z^{\ast }Z{)}^{-1}{W}^{\ast }$. Writing $C_E,C_{E^{\perp }}$ for the diagonal blocks, we getExpanding $(I+Z^{\ast }Z{)}^{-1}=I-Z^{\ast }Z+O(\Vert Z{\Vert }^4)$ yieldsEach complex coordinate has two real components, each with Hessian coefficient $2(c_j-c_i)$. Every coefficient is nonzero, proving that $f$ is a Morse function. Its negative directions correspond to $j<i$. For $I=\{i_1<\cdots <i_k\}$ there are $i_a-a$ complementary indices below $i_a$, givingThis is the diagonal trace Morse function on a complex Grassmannian. Its critical values need not be distinct; ties between sums do not make its critical points degenerate.

All Morse indices are even. Choose a Morse-Smale gradient flow and form its integral Morse chain complex. The odd-degree chain groups vanish, so every differential vanishes. Equivalently the Morse handle-attachment theorem supplies only even-dimensional cells. The integral homology of complex Grassmannians is thereforewhere $N_r$ counts the subsets with ${\sum }_a(i_a-a)=r$, and $0\le r\le k(n-k)$. In particular there is no torsion. The nondecreasing sequence $i_a-a$ lies between zero and $n-k$, so reversing it identifies these subsets with partitions of an integer $r$ fitting in a $k$ by $(n-k)$ rectangle. The Poincaré polynomial is explicitlyFor the permitted example $\mathrm{Gr}(2,4)$, the subsets $12,13,14,23,24,34$ have indices $0,2,4,4,6,8$. ThusThe general calculation above applies throughout $2\le k\le n-2$; it does not reduce to any of the excluded rank-one cases.

Work over $\mathbb{Q}$. A closed connected orientable surface of genus $g$ has Betti numbers $(1,2g,1)$; the circle has Betti numbers $(1,1)$. The Künneth theorem therefore givesWe use the following precise consequence of the Morse handle-attachment theorem: a closed manifold with a Morse function has the homotopy type of a finite CW complex with one index-$j$ cell for each index-$j$ critical point. Its degree-$j$ rational cellular chain group has dimension $c_j$, so its homology has dimension at most $c_j$. Consequently $c_j\ge b_j$, the weak Morse inequalities. Summing yieldsThis counts every critical point, regardless of whether some critical values coincide.

To attain the bound, construct a surface Morse function $h:{\Sigma }_g\to \mathbb{R}$ from the standard orientable handle decomposition: start with a disk, attach $2g$ orientable bands in the usual paired pattern forming the genus-$g$ surface with one boundary component, and cap that boundary with a disk. Give the first disk a minimum, each band one saddle with local form $-x^2+y^2$, and the final cap a maximum. Regular collar coordinates join successive levels without additional critical points. This constructs $h$ with counts $(1,2g,1)$.

On the product takeIts critical points are exactly the pairs of a critical point of $h$ with $\theta =0,\pi$. The Hessian matrix is block diagonal, with surface block $\mathrm{Hess}h$ and circle entry $-\delta \cos \theta$. It is nonsingular and its Morse index is the sum of the two factor indices. Thus this product of Morse functions has countsand exactly $4g+4$ critical points. If distinct critical values are desired, choose $\delta$ outside the finite set producing collisions after first choosing distinct critical values for $h$. The minimal Morse critical count on a circle times an orientable surface is therefore attained for every $g\ge 0$.

A Morse function is a smooth real-valued function all of whose critical points have nonsingular Hessian matrices. We use the descending convention for a gradient-like vector field $V$: away from critical points, $df(V)<0$, and near a critical point of Morse index $\lambda$, in Morse lemma coordinates $f=f(p)-|u{|}^2+|v{|}^2$, the field can be chosen as $V=(2u,-2v)$.

For its flow ${\varphi }_t$, define the unstable manifold $W^u(p)=\{x:{\lim }_{t\to -\infty }{\varphi }_t(x)=p\}$ and the stable manifold $W^s(q)=\{x:{\lim }_{t\to +\infty }{\varphi }_t(x)=q\}$. Their dimensions are $\mathrm{ind}p$ and $m-\mathrm{ind}q$. The Morse-Smale transversality condition isFor a closed manifold, this lets us form the Morse-Smale complex over $\mathbb{Z}$. Orient each unstable manifold and takeThe transverse intersection for an index difference of one is one-dimensional; quotienting by flow time gives a finite set of trajectories, and $n(p,q)$ counts them with orientation signs. Compactifying the one-dimensional spaces of trajectories for an index difference of two adds broken trajectories. Their oriented boundary count is zero, giving ${\partial }^2=0$. The descending cells are the unstable manifolds, and these counts are the incidence numbers of the corresponding cellular chain complex, so the homology of this complex is $H_{\ast }(M;\mathbb{Z})$. Ambient orientability is unnecessary: orientations of the unstable cells provide the requisite coorientations and trajectory signs.

For the Klein bottle, use the quotient of ${\mathbb{R}}^2$ by $(s,t)\sim (s+2\pi ,-t)$ and $(s,t)\sim (s,t+2\pi )$. A model of the standard four-critical-point height function isIt is well defined under both identifications. Its critical points, their Morse indices, and critical values areThe Hessian matrix is diagonal with entries $-\cos s$ and $-\epsilon \cos t$, so each point is nondegenerate. The flat quotient metric gives descending equations $\dot{s}=\sin s$, $\dot{t}=\epsilon \sin t$. Their saddle separatrices are the coordinate lines; the two saddle-to-saddle intersections are empty, and all remaining stable and unstable intersections are transverse. This is a Morse-Smale gradient flow. If one insists on the exact local-model version of a gradient-like vector field above, change the two coordinate speeds near the critical points to the standard quadratic-model speeds. This preserves the separatrices, signs and transversality.

Both branches of either saddle flow to the minimum with opposite endpoint signs, so $\partial a=\partial b=0$. From the maximum to $a$, the two trajectories lie on $s=0$ and approach $t=\pi$ from opposite directions. The identification in $t$ is a translation, so their incidence signs are opposite and cancel. The two trajectories to $b$ lie on $t=0$ and approach $s=\pi$ from opposite directions. The identification in $s$ reverses the transverse $t$-coordinate, reversing one additional orientation sign. These two contributions consequently have the same sign. Choosing the orientation of $b$ gives $\partial q=2b$.

One can check the coefficient directly from the cell attachment of the top cell: its boundary word is $ab{a}^{-1}b$, with exponent sums zero on the base loop $a$ and two on the fiber loop $b$. Thus the integral Morse complex of the Klein bottle isandCounting trajectories without signs, or working only modulo two, would miss the order-two class.

The Morse lemma states that if $p$ is a nondegenerate critical point of a smooth real function $f$ on an $m$-dimensional smooth manifold, there are smooth coordinates $y_1,\dots ,y_m$, centered at $p$, such thatHere $\lambda$ is the Morse index, the number of negative directions of the Hessian matrix at $p$. The equality is exact on a neighborhood, not merely a second-order expansion.

To prove it, start with any coordinates putting $p=0$. Since $df(0)=0$, integral Taylor theorem givesThe matrix $A(x)=(a_{ij}(x))$ is smooth and symmetric, with $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. By Sylvester's law of inertia, a constant invertible linear coordinate change makes $A(0)=\mathrm{diag}(-I_{\lambda },I_{m-\lambda })$.

We perform completion of squares with coefficients depending smoothly on the original $x$. The first pivot $a_{11}(x)$ stays nonzero and keeps its sign on a sufficiently small neighborhood. Algebraically,The remaining coefficient matrix is the Schur complement. At zero it is the remaining signed diagonal matrix, so its first pivot is also nonzero nearby. Continue recursively, shrinking the neighborhood finitely many times. Every pivot $d_i(x)$ is smooth and nonzero, with sign ${ϵ}_i=-1$ for $i\le \lambda$ and $+1$ thereafter. Absorb $|d_i(x)|$ into the square by its smooth positive square root. This produces a smooth triangular matrix $B(x)$ with nonzero diagonal such thatAll coefficients here are functions of the original $x$; no circular coordinate substitution is involved. Define $\Phi (x)=B(x)x$. Its derivative at zero is $D\Phi (0)=B(0)$, which is invertible. The inverse function theorem makes $y=\Phi (x)$ a smooth coordinate system and gives the asserted exact quadratic form. The number of negative squares agrees with the Hessian matrix signature and hence with the Morse index. This is the smooth completing-square proof of the Morse lemma.

The cellular approximation theorem says that a continuous map between CW complexes is homotopic to a cellular map, meaning that it sends each $q$-skeleton into the target's $q$-skeleton. Give $S^n$, for $n>0$, its CW complex structure with one zero-cell and one $n$-cell. Its $n$-skeleton is the whole sphere, so the theorem applies and gives the requested result. For $n=0$ use the two zero-cells of $S^0$. In particular the conclusion isThere is no local-finiteness assumption on $X$.

For completeness, here is a direct proof of the needed case of cellular approximation, including the point-avoidance step. We use three standard facts, stated explicitly. First, a compact subset of a CW complex lies in a finite subcomplex. Second, a continuous map from a smooth manifold into Euclidean space has arbitrarily close smooth approximations, and smooth cutoffs exist around compact subsets of open sets. Third, the Sard theorem says the critical values of a smooth map have measure zero; when the domain dimension is smaller than the target dimension, its entire image has measure zero.

The first fact has a useful short proof. If a compact set met infinitely many open cells, choose one point in each of countably many distinct cells. Closure-finiteness makes every closed cell meet the chosen set, and every subset of it, in finitely many points. The weak topology of a CW complex then makes all those subsets closed. The chosen set is an infinite closed discrete subset of a compact space, a contradiction. Taking the closures of the finitely many cells met proves the claim.

Now $f(S^n)$ lies in a finite subcomplex $Y$. Take a top-dimensional open cell $e^d$ of $Y$ with $d>n$, and identify its interior with the open unit ball in ${\mathbb{R}}^d$ using its characteristic map. On the open set $U=f^{-1}(e^d)$, write $F:U\to B^d$ for the coordinate expression. Choose $0<r^{\prime }<r^{\prime \prime }<1$ and a smooth cutoff $\chi$ supported in $F^{-1}(B_{r^{\prime \prime }})$ and equal to one near the compact set $F^{-1}({\overline{B}}_{r^{\prime }})$. By smooth approximation of maps into Euclidean space, choose a smooth $h:U\to {\mathbb{R}}^d$ with $|h-F|<\delta$, where $\delta <\min (r^{\prime }/2,1-r^{\prime \prime })$. Replace $F$ byThe straight-line interpolation is a homotopy supported away from the boundary of the cell and stays inside that cell. Where $\chi \ne 1$, we have $|F|>r^{\prime }$ and hence $|G|>r^{\prime }/2$. By the Sard theorem, choose $z\in B_{r^{\prime }/4}$ outside $h(U)$. Where $\chi =1$ we have $G=h$, and elsewhere $|G|>r^{\prime }/2$, so the modified map avoids $z$ everywhere.

Radial retraction of $D^d\setminus \{z\}$ onto its boundary fixes that boundary. It therefore descends through the characteristic map, including its possibly noninjective boundary identifications, and fixes all other cells of $Y$. Composing gives a homotopy whose endpoint misses all of $e^d$. Remove every cell of dimension greater than $n$ in descending order. There are finitely many such cells, so their homotopies concatenate to yield $g:S^n\to Y^n\subseteq X_n$. This proves dimension reduction of sphere maps into CW skeleta. The approximation step is essential: an arbitrary continuous image of $S^n$ can fill a higher-dimensional region, so dimension alone would not supply an omitted point.

An ellipse here is a nondegenerate ellipse, hence a smooth embedded circle. Write the first ellipse aswhere $A$ is invertible. Write the second as $Q(x)=0$, wherefor a positive-definite symmetric matrix $B$. Its gradient is nonzero on the ellipse. Let $g(t)=Q(\gamma (t))$. At an intersection, transversality says the tangent vector ${\gamma }^{\prime }(t)$ is not tangent to $Q=0$, soEvery intersection is thus a simple zero of $g$, and $g$ changes sign there. The intersection set is finite: it is closed and compact, and transversality makes each of its points isolated. Start at a parameter with $g\ne 0$ and traverse one full period. Returning to the same sign requires an even number of sign changes. Therefore the number of intersection points is even, including zero when the ellipses are disjoint.

For the upper bound, substituting the parametrization into the quadratic gives the trigonometric polynomialIt is not identically zero: otherwise the first ellipse would lie on the second, and its tangent would be tangent to the second everywhere, violating transversality. Put $z=e^{it}$. Multiplying by $z^2$ turns this function into the polynomialIt is nonzero and has degree at most four, so it has at most four distinct roots. Distinct points on the first ellipse correspond to distinct parameters modulo $2\pi$ and hence distinct points $z$ on the unit circle. ThusBoth the parity and the upper bound follow without using an algebraic intersection theorem. The four-point bound is attained by a unit circle and a concentric ellipse whose semiaxes lie on opposite sides of one.

Here closed oriented surfaces are connected, as in the usual single-genus convention. Let $f:X\to Y$ have nonzero mapping degree $d$. Work over $\mathbb{Q}$, so multiplication by $d$ is invertible and cannot vanish.

The Poincare duality pairing on $H^1(Y;\mathbb{Q})$ is nondegenerate. Thus for every $0\ne a\in H^1(Y;\mathbb{Q})$ there is $b\in H^1(Y;\mathbb{Q})$ with $⟨a⌣b,[Y]⟩\ne 0$. Naturality of the cup product and evaluation on the fundamental class giveIn particular $f^{\ast }a\ne 0$, so $f^{\ast }:H^1(Y;\mathbb{Q})\to H^1(X;\mathbb{Q})$ is injective. The cohomology ring of a closed oriented surface has $\dim H^1({\Sigma }_g;\mathbb{Q})=2g$. Comparing dimensions now givesThis proves genus monotonicity under nonzero-degree surface maps by rational cohomology injectivity of a nonzero-degree map, without requiring $f$ to be a covering map or a smooth map.

Use integral singular homology. The singular chain group $C_n(B)$ is the free abelian group on continuous maps ${\Delta }^n\to B$, and its boundary operator isFor $i<j$, deleting vertex $j$ and then vertex $i$ has sign $(-1{)}^{i+j}$, while deleting $i$ and then the shifted vertex $j-1$ has sign $(-1{)}^{i+j-1}$. Both yield the same codimension-two face. Pairing these terms proves ${\partial }^2=0$. The inclusion $A\subset B$ identifies $C_n(A)$ with the subgroup spanned by those simplices having image in $A$, so it is injective and preserved by $\partial$.

The relative chain complex is $C_n(B,A)=C_n(B)/C_n(A)$ with induced boundary $\partial \bar{b}=\overline{\partial b}$. Thus there is a degreewise short exact sequence of chain complexesInclusion and quotient take homological cycles to homological cycles and boundaries to boundaries, hence define the group homomorphisms $i_{\ast }$ and $q_{\ast }$ on homology. We construct and prove exactness of the resulting sequence directly, rather than invoke an unproved algebraic lemma:ending in $H_0(A)\to H_0(B)\to H_0(B,A)\to 0$.

For the relative homology connecting homomorphism, lift a relative cycle $\bar{b}\in C_n(B,A)$ to $b\in C_n(B)$. Since $\partial \bar{b}=0$, the chain $\partial b$ belongs to $C_{n-1}(A)$; it is a homological cycle because ${\partial }^2=0$. DefineChanging the lift by $a\in C_n(A)$ changes the boundary by $\partial a$, hence does not change its homology class. More generally, if a different relative cycle represents the same relative homology class, lifts satisfy $b^{\prime }-b=\partial c+a$ for some $c\in C_{n+1}(B)$ and $a\in C_n(A)$. Then $\partial b^{\prime }-\partial b=\partial a$. This proves independence both of the lift and of the homology representative. Lifting a sum by the sum of lifts proves additivity, so $\delta$ is a well-defined group homomorphism.

At $H_n(A)$, an element in the image of $\delta$ is represented by $\partial b$ for some $b\in C_{n+1}(B)$, so its image in $H_n(B)$ is zero. Conversely, if a homological cycle in $A$ $a$ has $i_{\ast }[a]=0$, then $a=\partial b$ for a $B$-chain $b\in C_{n+1}(B)$. Its quotient $\bar{b}$ is a relative cycle and $\delta [\bar{b}]=[a]$. ThusFor completeness, the other terms are exact by the same elementary chain argument. At $H_n(B)$, clearly $q_{\ast }{i}_{\ast }=0$. If a homological cycle in $B$ $b$ has zero relative class, then $\bar{b}=\partial \bar{c}$ for a quotient chain $\bar{c}$. Lifting $c$ gives $a=b-\partial c\in C_n(A)$, and $\partial a=0$. Hence $[b]=i_{\ast }[a]$, proving $\ker q_{\ast }=\mathrm{im}{i}_{\ast }$. At $H_n(B,A)$, absolute homological cycles have zero connecting image. Conversely, if $\delta [\bar{b}]=0$, then $\partial b=\partial a$ for some $a\in C_n(A)$; the chain $b-a$ is an absolute homological cycle and maps to $[\bar{b}]$. Therefore $\ker \delta =\mathrm{im}{q}_{\ast }$. In degree zero every quotient chain lifts to an absolute homological cycle, because the differential to degree $-1$ is zero, so $q_{\ast }:H_0(B)\to H_0(B,A)$ is surjective. These proofs establish the whole long exact sequence in relative homology and all the algebraic facts used in its construction.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.