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Use shifted degreesIts graded Lie bracket is the sign-adjusted Whitehead bracket from Question 2, equivalently the Samelson product on loops. This regrading is essential: the geometric bracket has degree minus one in the unshifted sphere grading. A free graded Lie algebra on a graded vector space $V$ means $\mathbb{L}(V)$ with the universal extension property for linear maps from $V$ to any graded Lie algebra.

Here are the general results used. The Bott–Samelson theorem says that, for a connected based CW complex $X$, the inclusion $X\to \Omega \Sigma X$ induces an isomorphism of graded algebraswith the Pontryagin product. The Milnor–Moore theorem says that for simply connected $Y$, rational loop Hurewicz homomorphism identifies $L_Y$ with the primitives of that loop-homology Hopf algebra and induces $U(L_Y)\cong H_{\ast }(\Omega Y;\mathbb{Q})$. The graded Poincaré-Birkhoff-Witt theorem identifies the associated graded algebra of $U(L)$, for its word-length filtration, with the graded symmetric algebra on $L$; hence its primitive elements of a Hopf algebra are exactly $L$ over $\mathbb{Q}$. Finally the rational Whitehead theorem identifies rational homology equivalences of simply connected CW complexes with rational homotopy equivalences. Rationalization of a topological space induces the corresponding isomorphisms of loop-space homotopy and rational homology.

Suppose first $Y$ has the rational homotopy type of $\Sigma X$. A reduced suspension is simply connected when $X$ is connected; a disconnected $X$ has $H_1(\Sigma X;\mathbb{Q})={\tilde{H}}_0(X;\mathbb{Q})\ne 0$, so it cannot have the rational homology of the simply connected $Y$. Put $V={\tilde{H}}_{\ast }(X;\mathbb{Q})$, in strictly positive degrees. Bott–Samelson theorem gives $H_{\ast }(\Omega Y;\mathbb{Q})\cong T(V)$ as an algebra. To deduce freeness of the primitive Lie algebra, its coproduct must also be considered: the initial Bott–Samelson theorem generators can have reduced-diagonal terms when $X$ has nontrivial cup products.

By Milnor–Moore theorem write this Hopf algebra as $U(L_Y)$. Its algebra indecomposables satisfyThe first identity follows from the enveloping relation: after all positive products are killed, every bracket is killed and no additional linear relation is introduced. Hence each homogeneous basis vector of $V$ has a primitive representative in $H_{\ast }(\Omega Y;\mathbb{Q})$. Choose such representatives $p_v$. Sending $v$ to $p_v$ defines an algebra map from $T(V)$ to itself whose effect on indecomposables is the identity. It is an isomorphism: $p_v=v+$ decomposable terms, and every factor in those terms has smaller positive degree than $v$. Degree induction solves recursively for each original generator and constructs the inverse. This works even with arbitrarily many generators in a degree.

Equip the source tensor algebra with primitive generators. The map is now a Hopf-algebra isomorphism, because the chosen images are primitive. The primitives in this tensor Hopf algebra are exactly $\mathbb{L}(V)$: $U(\mathbb{L}(V))=T(V)$ by the two universal properties, and the graded Graded Poincaré–Birkhoff–Witt theorem gives its primitive subspace. Consequently $L_Y\cong \mathbb{L}(V)$, proving the forward direction without assuming a primitive coproduct for the original generators.

Conversely suppose $L_Y\cong \mathbb{L}(W)$ for a positively graded $W$. Choose a homogeneous basis $w_{\lambda }$ of $W$. Represent its corresponding homotopy class by a based sphere map $S^{|w_{\lambda }|+1}\to Y_{\mathbb{Q}}$, and combine these maps intoThe wedge sum is a reduced suspension, namely $Z=\Sigma ({\bigvee }_{\lambda }{S}^{|w_{\lambda }|})$. The preceding direction, applied to this wedge sum, gives its free rational homotopy Lie algebra on the sphere inclusions. Indeed these particular tensor generators are primitive because the reduced diagonal of each sphere is zero, and loop Hurewicz homomorphism sends the adjoint sphere inclusion to that generator. Naturality of the bracket shows that $F_{\ast }$ sends the free generating basis to the prescribed free generating basis of $L_Y$. It is therefore an isomorphism of free graded Lie algebras, hence an isomorphism in every rational homotopy degree. The rational Whitehead theorem makes $F$ a rational homotopy equivalence. This proves the reverse implication and also shows that a wedge sum of spheres can be chosen as the reduced suspension.

For every chosen connected $X$ with $Y{\simeq }_{\mathbb{Q}}\Sigma X$, the required algebra statement is precisely the Bott–Samelson theorem isomorphism, transported along rational homotopy equivalence:Multiplication is concatenation of tensors and the degree-zero tensor is the unit. All freeness assertions here are graded, with the Koszul signs described above.

Use tangent vectors, rather than the affine tangent plane, to differentiate. Every skew-symmetric matrix $A$ gives a curve $X\exp (tA)$ in the special orthogonal group. Its first derivative under $f$ is $\mathrm{tr}(CXA)$. The trace pairing annihilates every skew-symmetric matrix precisely when $CX$ is symmetric: choosing $A$ with only entries $a_{ij}=-a_{ji}$ tests each off-diagonal antisymmetric entry. Thus criticality is equivalent to $Y=CX$ being symmetric.

Since $X$ is orthogonal, $Y{Y}^T=C^2$, so at a critical point $Y^2=C^2$. Consequently $Y$ commutes with $C^2$. The diagonal entries $c_i^2$ are distinct, and the equation $Y{C}^2=C^{2}Y$ gives $(c_j^2-c_i^2)y_{ij}=0$ for $i\ne j$. Hence $Y$ and $X$ are diagonal. Orthogonality and the determinant constraint give exactlyConversely each such $D$ makes $CD$ symmetric, so this list is complete.

To calculate the Hessian matrix, use local coordinates $X=\exp (A)D$ with $A^T=-A$; the matrix exponential is a local coordinate map near zero. Its quadratic term givesbecause $(A^2{)}_{ii}=-{\sum }_{j\ne i}{a}_{ij}^2$. There are no mixed quadratic terms. For $i<j$, $c_j>c_i>0$, so $c_i{\epsilon }_i+c_j{\epsilon }_j$ has the sign of ${\epsilon }_j$ and is never zero. Every listed critical point is therefore nondegenerate, with Morse indexThis is the trace height on the special orthogonal group calculation; the rotations in the hint select the individual coordinates $a_{ij}$.

For $n=3$, the signs $(+--),(-+-),(--+),(+++)$ have indices $0,1,2,3$ respectively. A Morse function on a compact closed manifold has Euler characteristic equal to the alternating number of critical points, by its Morse handle-attachment theorem and resulting finite CW complex. Hence

The usual compact closed-manifold hypothesis is needed for the first assertion, or a proper bounded-below Morse exhaustion in an appropriate noncompact version. As literally printed without any such condition, $f(x)=x$ on $\mathbb{R}$ is a Morse function with no critical points, while $b_0=1$. We prove the intended closed-manifold statement, which applies to the surface product later in the question.

By the Morse handle-attachment theorem, crossing an index-$i$ critical point attaches an $i$-handle, and no topological change occurs between critical levels. Retraction to the core disks of handles gives a CW complex model with $c_i$ cells in dimension $i$. Its cellular chain complex has $C_i\cong {\mathbb{Z}}^{c_i}$. Tensor with $\mathbb{Q}$, which computes the ranks $b_i$ of integral homology. Since homology is a quotient of a vector subspace of $C_i\otimes \mathbb{Q}$, $b_i\le c_i$. This argument has not used the Morse inequalities.

The strong Morse inequalities and top-dimensional equality areTheir weak form is $c_i\ge b_i$. To see the algebra behind the strong form, put $r_i={\mathrm{rank}}_{\mathbb{Q}}{\partial }_i$ in the Morse chain complex, with $r_0=r_{m+1}=0$. The kernel-image dimensions give $c_i=b_i+r_i+r_{i+1}$, so the alternating difference at $j$ is exactly $r_{j+1}\ge 0$. Conversely, given the strong inequalities, denote those alternating differences by $S_j$, with $S_{-1}=0$. Then $c_j-b_j=S_j+S_{j-1}\ge 0$, proving that they imply every weak inequality.

They are not equivalent as numerical restrictions. For the Betti numbers $b=(1,0,0,1)$ of $S^3$, the formal counts $c=(2,0,0,2)$ satisfy all weak inequalities and the Euler equality, but at $j=1$ give $c_1-c_0=-2<-1=b_1-b_0$. They therefore cannot be the counts of a Morse function on $S^3$; the strong inequalities detect that additional obstruction.

If all the strong inequalities are equalities, all $r_i$ vanish and $c_i=b_i$. An integer boundary matrix with rational rank zero is the zero matrix. Thus the integral complex itself has zero differential, not merely its rationalization. Consequentlyand higher homology vanishes. In particular there is no homology torsion. This is an integral consequence of a Perfect Morse function, rather than just an assertion about ranks.

For ${\Sigma }_g$, the integral homology is free with Betti numbers $(1,2g,1)$. The Künneth theorem therefore has no torsion terms, and the product's Poincaré polynomial isSumming the weak inequalities yields the lower boundChoose a surface Morse function $h$ with one minimum, $2g$ saddles and one maximum. It can be built from one closed disk, the standard $2g$ one-handles, and a closing closed disk, using the local forms $u^2+v^2$, $-u^2+v^2$, and $-u^2-v^2$ at the handle centres. Regular collars interpolate between the handles without new critical points. Now set $F(x,y)=h(x)+h(y)$. Its critical points are the ordered pairs of those of $h$, and its Hessian matrix is a block direct sum; the indices add. This product of Morse functions has exactly $(2g+2{)}^2$ critical points and realizes every displayed Betti count. Small bumps constant near the critical pairs can separate their values without altering the Hessians or creating new critical points.

The same minimum does not apply to arbitrary smooth functions. Here is an explicit finite-critical-point counterexample, avoiding any assumption that degeneracy leaves the Morse lower bound intact. On $T^2=(\mathbb{R}/\pi \mathbb{Z}{)}^2$, takeIt is well defined because shifting either coordinate by $\pi$ reverses two of its factors. Its derivatives are $\sin v\sin (2u-v)$ and $\sin u\sin (u-2v)$. If either $\sin u$ or $\sin v$ vanishes, both derivatives vanish only at $(0,0)$. Otherwise $2u-v$ and $u-2v$ must be multiples of $\pi$, giving exactly $(\pi /3,2\pi /3)$ and $(2\pi /3,\pi /3)$ in addition to the origin. At the first point the Hessian is $\sqrt{3}\left(\begin{array}{cc}1& -1/2\\ -1/2& 1\end{array}\right)$, positive definite; at the second it is its negative. They are a minimum and maximum. The origin is an isolated degenerate monkey saddle, with leading term $uv(u-v)$. Thus this three-critical-point function on a torus has exactly three critical points.

For $g>1$, perform connected-sum gluing of functions between this torus and a perfect Morse surface of genus $g-1$: delete a small neighbourhood of the torus maximum and of the other surface's minimum, and join their regular boundary circles by a monotone cylinder. Collar coordinates ensure nonzero differential throughout the neck, so precisely those two critical points are lost and none are added. The result on ${\Sigma }_g$ has $3+[2(g-1)+2]-2=2g+1$ critical points, retaining the degenerate point. For $g=1$, simply use $h_0$. Calling this smooth function $\tilde{h}$ and the earlier perfect Morse function $h$, the smooth product $\tilde{h}(x)+h(y)$ hascritical points. This explicitly proves the negative answer to the unrestricted smooth-minimum question; it does not assert that this smaller illustrative count is sharp.

Use the standard CW complex structure on the real projective plane, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the cellular chain complex has $d{e}_2=2e_1$ and $d{e}_1=0$.

The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le 2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The cellular chains of a product of finite CW complexes have boundary $d(e_i\times e_j)=d{e}_i\times e_j+(-1{)}^i{e}_i\times d{e}_j$. In the ordered basesthe nonzero matrices areIn particular $d_2{d}_3=d_3{d}_4=0$. The degree-one homology is ${\mathbb{Z}}^2/2{\mathbb{Z}}^2$. The degree-two cellular cycles are $\mathbb{Z}{e}_{11}$, and the cellular boundaries there are $2\mathbb{Z}{e}_{11}$. The degree-three cycles are $\mathbb{Z}(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the integral homology of two real projective planes:The degree-three torsion is the Tor functor contribution in the Künneth theorem; it would be missed by retaining only tensor products of the two factors' homology groups.

A simply connected space is path-connected and has trivial fundamental group, equivalently every based loop is null-homotopic. The Whitehead theorem says that a map of CW complexes inducing isomorphisms on all homotopy groups is a homotopy equivalence. Its homological Whitehead theorem form says that a map between simply connected CW complexes, in particular finite ones, is a homotopy equivalence if it induces isomorphisms on integral homology. The latter implication follows from the relative Hurewicz theorem and the ordinary Whitehead theorem; homotopy equivalences necessarily induce homology isomorphisms.

For the attached-cell space $X$, the cellular homology chain complex in positive degrees isDualizing gives the cellular cohomology complex $\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{t\mapsto (2t,3t)}{\to }{\mathbb{Z}}^2$. The last map is injective and its image is primitive because $\gcd (2,3)=1$. Its cokernel is $\mathbb{Z}$, explicitly via $(a,b)\mapsto 3a-2b$. HenceThe van Kampen theorem gives ${\pi }_1(X)=⟨a\mid a^2=1,a^3=1⟩=1$. Moreover $H_2(X)=\ker (23)$ is generated by $(3,-2)$. Since $X$ is simply connected, the Hurewicz theorem gives ${\pi }_2(X)\cong H_2(X)$. Choose a map $S^2\to X$ representing that generator. It induces an isomorphism on $H_2$, and on $H_0$, while all other reduced homology groups vanish on both sides. The homological Whitehead theorem proves $\boxed{X\simeq S^2}$. This is a special case of coprime two-cell attachments to a circle.

It is not homeomorphic to the sphere. Indeed, if it were, its embedded one-skeleton would be a Jordan circle in $S^2$, with the two open cell interiors as its complementary regions. The Jordan curve theorem and its Schoenflies form identify the closures of those regions with discs. A characteristic map, homeomorphic on the interior of such a disc, has local degree $+1$ or $-1$ at an interior point; the boundary homomorphism in relative homology then makes its attaching map on the circle have degree $+1$ or $-1$. This contradicts degrees $2$ and $3$. This argument does not require choosing special covering representatives of the attaching maps.

Finally, two compact examples with the same additive cohomology areBoth have $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^2$, $H^2=\mathbb{Z}$ and zero higher cohomology, by their cellular chain complexes. But their fundamental groups are respectively ${\mathbb{Z}}^2$ and the nonabelian free group on two generators, so they cannot be homotopy equivalent. This illustrates that cohomology groups do not determine homotopy type.

For singular cochains $a\in C^p(X;R)$ and $b\in C^q(X;R)$, define the cup product on a singular $(p+q)$-simplex byThe identity $\delta (a⌣b)=\delta a⌣b+(-1{)}^{p}a⌣\delta b$ makes this descend to cohomology. The induced product is associative, unital and graded-commutative: $ab=(-1{)}^{pq}ba$ in degrees $p,q$.

For finite CW complexes, the integral cohomological Künneth theorem for finite cell complexes is the split short exact sequenceThe splitting need not be natural. If the factor cohomology is torsion-free, the cohomological cross product is an isomorphism, with multiplication $(a\times b)(c\times d)=(-1{)}^{|b||c|}ac\times bd$.

A genus-$g$ closed oriented surface has cellular groups $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^{2g}$, $H^2=\mathbb{Z}$, and no higher cohomology. Choose the positive orientation class $\omega$ and a symplectic basis $a_i,b_i$ of $H^1$. The cohomology ring of a closed oriented surface is specified bywith unit $1\in H^0$. These products follow by representing their Poincare duality classes by oriented handle curves: the $i$th meridian meets the $i$th longitude once positively, curves on distinct handles have zero intersection, and swapping two oriented curves reverses the intersection sign. In general, for closed oriented $n$-manifolds and transverse oriented submanifolds of codimensions $p,q$, the cup product of their Poincare-dual classes is the dual class of their oriented intersection. Complementary-dimensional intersections give the signed point count $⟨\alpha \beta ,[M]⟩$.

For ${\Sigma }_2$, the four curves $A_1,B_1,A_2,B_2$ give $a_1{b}_1=a_2{b}_2=\omega$ and all other pairings zero apart from their negatives in reversed order. Thus the degree-one intersection pairing on an oriented surface is two symplectic blocks, and the point class is dual to $\omega$.

For ${\Sigma }_2\times {\Sigma }_2$, let $a_i,b_i,{\omega }_1$ come from the first factor and $c_j,d_j,{\omega }_2$ from the second. The Künneth theorem gives their graded tensor-product ring, whose additive ranks in degrees $0,1,2,3,4$ areThe top orientation class is $\Omega ={\omega }_1{\omega }_2$. The two fiber surfaces $\{p\}\times {\Sigma }_2$ and ${\Sigma }_2\times \{q\}$ have dual classes ${\omega }_1,{\omega }_2$, intersect once positively and have zero self-intersection. The other sixteen degree-two classes are products of degree-one classes and are dual, up to the chosen orientation convention, to products of handle curves. Their intersections are encoded byThe minus sign in the first mixed formula comes from commuting $c_j$ past $b_k$. These classes together with the fiber classes give the intersection form of a product of two closed oriented surfaces, here nine hyperbolic blocks. Complementary degrees one and three are similarly paired, for example $a_i(b_k{\omega }_2)={\delta }_{ik}\Omega$; geometrically this pairs a three-dimensional product submanifold with a transverse loop. This exhibits the full duality, not just the additive groups.

The Lefschetz fixed-point theorem says that a self-map of a compact polyhedron with nonzero Lefschetz number has a fixed point, where $L(f)={\sum }_q(-1{)}^q\mathrm{tr}(f_{\ast }:H_q(-;\mathbb{Q})\to H_q(-;\mathbb{Q}))$. A map homotopic to the identity has $L(f)=\chi$, and Euler characteristic of a product givesConsequentlyThe product $g_1{g}_2$ itself is not determined by the printed assumptions. For every $h\ge 0$, translation by a nonzero torus point, times the identity on ${\Sigma }_h$, is a fixed-point-free map of ${\Sigma }_1\times {\Sigma }_h$ homotopic to the identity, and its genus product is $h$. Thus the exact conclusion is the criterion for fixed-point-free self-maps homotopic to the identity on products of surfaces, rather than a single value of $g_1{g}_2$.

Use integral cellular homology. All three vertices become a single vertex, and all three oriented edges become one oriented circle, with generator $a$. The face goes round that circle three times in the same direction. Thus the quotient has a CW complex structure with one cell in each of dimensions zero, one and two, and cellular chain complexThe degree-two boundary is injective and its cokernel is $\mathbb{Z}/3$. ThereforeThe attaching word is $a^3$, so the Seifert-van Kampen theorem also gives ${\pi }_1(X)\cong \mathbb{Z}/3$. This is the degree-three circle-attachment model of a Moore space.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

Take ${\mathbb{CP}}^3$. It is a closed smooth manifold and is simply connected: its CW complex structure has no one-cells, and attaching cells of dimension at least two to a point cannot create a fundamental group. Its cohomology ring is $\mathbb{Z}[x]/(x^4)$ with $|x|=2$, so $x^3\ne 0$.

The theorem used is the Lusternik–Schnirelmann critical point theorem: every smooth real-valued function on a closed connected smooth manifold has at least its unnormalized Lusternik–Schnirelmann category many distinct critical points, without a nondegeneracy assumption. Here category is the minimum number of open subsets covering the manifold whose inclusions are null-homotopic. The cup length bound gives category at least four. To justify that bound, a cover by three categorical sets would make each positive-degree class vanish on its assigned set. The long exact sequence of each pair lifts three copies of $x$ to relative classes, and their relative cup product would lie in $H^6(M,M;\mathbb{Z})=0$, contradicting $x^3\ne 0$. Thus every smooth function has at least four critical points, including functions with degenerate critical points. The critical-point theorem in this formulation is described in Conley pairs in geometry—Lusternik–Schnirelmann theory and more.

Under the intended convention that the finite group is nontrivial, take ${\mathbb{CP}}^2$. It is simply connected by its CW complex structure. The Lefschetz fixed-point theorem says that a continuous self-map of a compact triangulable space with nonzero rational Lefschetz number has a fixed point.

For any continuous $f:{\mathbb{CP}}^2\to {\mathbb{CP}}^2$, write $f^{\ast }x=dx$ for the integral degree-two generator, with $d\in \mathbb{Z}$. Preservation of the cup product gives $f^{\ast }(x^2)=d^2{x}^2$. Only degrees zero, two and four contribute to the rational Lefschetz number, soConsequently every such map has a fixed point. In a free group action, each nonidentity group element must act by a fixed-point-free homeomorphism. This is impossible here, so no nontrivial finite group can act freely. This is the fixed-point property of even-dimensional complex projective space for complex dimension two. If the trivial group is included, the literal request has no example, since the trivial group acts freely on every space.

False. Take $M=S^4$ and $N=S^3$. Both are closed oriented simply connected manifolds and $4>3$. A cohomological obstruction rules out even a topological fiber bundle.

Suppose a fiber bundle $F\to S^4\to S^3$ existed. The long exact sequence of homotopy groups of a fibration, including its component terms, makes $F$ path-connected. A local trivialization over a small open ball in $S^3$ identifies its preimage, an open subset of $S^4$, with $F\times B^3$. Projection onto $F\times \{0\}$ gives a retraction. An open smooth four-manifold has the homotopy type of a CW complex of dimension at most four, so its rational cohomology vanishes in degrees above four. A retract's cohomology injects into that of the ambient space. Therefore

Use the multiplicative cohomological Serre spectral sequence with rational coefficients. Since $S^3$ is simply connected, there is no nontrivial coefficient action, and only columns zero and three occur on $E_2$. Thus $E_3=E_2$, and the only possibly nonzero differential is $d_3$. Let $u$ be the degree-three base generator. It cannot survive, because the sequence converges to $H^{\ast }(S^4;\mathbb{Q})$ and that ring has zero degree-three part. It has no outgoing differential and its only possible incoming differential is from $E_3^{0,2}$. Hence there is $a\in H^2(F;\mathbb{Q})$ withThe class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\ge 2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,But $E_3^{3,2r-2}=H^3(S^3;\mathbb{Q})\otimes H^{2r-2}(F;\mathbb{Q})$, in which $u{a}^{r-1}\ne 0$ by minimality of $r$; also $r$ is nonzero over $\mathbb{Q}$. This is a contradiction. Thus no such fiber bundle exists. This is the special case of a sphere cannot fibre over a lower-dimensional odd sphere required here.

Take a CW complex with one vertex, loops labelled $a,b$, and one two-cell attached along $a^2{b}^3{a}^3{b}^2$. The Seifert-van Kampen theorem gives the required fundamental group. Connected double covering spaces correspond to index-two subgroups, equivalently kernels of surjective homomorphisms $G\to C_2$. The relation imposes $5a+5b=0$ modulo two, so the unique nonzero possibility sends both $a,b$ to one. Thus there is a unique connected double cover up to covering isomorphism.

Construct it with two vertices; both labelled edges switch sheets, and attach the two lifts of the two-cell. Contract one of the $a$ edges to a spanning tree. With coset representatives $1,a$, the remaining Schreier generators can be taken as $A=a^2$, $B=b{a}^{-1}$ and $C=ab$. Reading the two lifts of the attaching word, starting on the two respective sheets, gives $ABCB{A}^{2}BC$ and $ACBCACB$. ThusFor example, a letter $a$ contributes nothing on the first sheet and $A$ on the second; a letter $b$ contributes $B$ on the first and $C$ on the second. Both letters then switch sheets, which directly verifies the two relators.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

Use integral homology throughout. The singular chain group $C_k(X)$ is free on the continuous maps from the standard $k$-simplex to $X$. Since the boundary operator preserves chains lying in $A$, these form a chain subcomplex. The relative chain complex and its relative homology areThe short exact sequence of chain complexes gives the long exact sequence in relative homology.

A sufficient hypothesis for the quotient comparison is a nonempty good pair: $A$ is closed and has an open neighbourhood $U$ which admits a deformation retraction onto $A$ while fixing $A$ throughout. A CW pair is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular collar neighbourhood. The induced quotient map $q:X\to Y=X/A$ collapses $A$ to a point $a$.

Here is the proof of the collapsing a pair theorem. Put $V=U/A\subset Y$. The deformation retraction makes $H_{\ast }(U,A)=0$, so the long exact sequence of the triple gives $H_{\ast }(X,A)\cong H_{\ast }(X,U)$. It also contracts $V$ onto $a$, giving $H_{\ast }(Y,\{a\})\cong H_{\ast }(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, excision givesSimilarly, excision of the closed point $a$ in the open set $V$ givesThe quotient map restricts to a homeomorphism between the two punctured pairs, so these are the same groups and the identifications commute with $q_{\ast }$. Finally $H_{\ast }(Y,\{a\})\cong {\tilde{H}}_{\ast }(Y)$, including degree zero. ThusThe neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.

For the genus-two closed orientable surface with its chosen orientation, $H_0(X)=\mathbb{Z}$, $H_1(X)={\mathbb{Z}}^4$, $H_2(X)=\mathbb{Z}$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb{Z}$. Its relevant long exact sequence in relative homology isThe final map is an isomorphism because both spaces are connected. Consequently $H_1(X,A)=\mathrm{coker}{i}_{\ast }$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The quotient topological space is connected, so its unreduced $H_0$ is $\mathbb{Z}$.

In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its homology class in $X$ is therefore zero, so $i_{\ast }=0$. The exact sequence becomesThe first sequence splits since its quotient is a free abelian group. Hence the homology after collapsing a separating surface curve isGeometrically, collapsing the boundary of each once-punctured torus fills its puncture with a cone on the circle, which is a disk. The quotient is a wedge sum of two tori at the collapsed point. Their two independent fundamental classes explain the extra second-homology generator.

In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed intersection pairing on an oriented surface therefore provides an integer homomorphism $H_1(X)\to \mathbb{Z}$ taking $[A]$ to $\pm 1$. Thus $[A]$ is a nonzero primitive homology class, and $i_{\ast }:\mathbb{Z}\to {\mathbb{Z}}^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is ${\mathbb{Z}}^3$. The same exact sequence gives $H_2(X,A)\cong H_2(X)\cong \mathbb{Z}$. Therefore the homology after collapsing a nonseparating surface curve isAnother description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed torus with two marked points, then identify the two points. Identifying two distinct points in a connected CW complex adds a loop up to homotopy, so this quotient has the homotopy type of $T^2\vee S^1$. The preceding relative homology calculation proves the groups without requiring that homotopy description.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.