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If the one-point compactification $X^{+}$ is Hausdorff and has a basis of contractible neighbourhoods at the added point, thenThe Excision theorem identifies $H^{\ast }(X,X\setminus K)$ with $H^{\ast }(X^{+},X^{+}\setminus K)$. The contractible neighbourhoods form a cofinal family of the complements of compact $K$, and their pair long exact sequences identify each relative group naturally with reduced cohomology. Passing to the direct limit gives the comparison. Local contractibility matters for singular cohomology; the Hawaiian earring illustrates its failure.

A Gysin sequence of a sphere bundle is the long exact sequence induced by the Gysin homomorphism and the Euler class of a vector bundle.

For the unit sphere bundle $S(E)\to B$ of an oriented rank-$d$ vector bundle, the Gysin sequence contains

Let $S=k[X_0,\dots ,X_n]$ with its usual grading, and put $S(m{)}_d=S_{m+d}$. Define the twisting sheaf on projective space byMore explicitly, on $U_i=D_{+}(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one module over $(S_{X_i}{)}_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule isThe ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an invertible sheaf. In particular $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$.

For $n\ge 1$, a global section is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the fraction field of $S$,To justify this, write a fraction in lowest terms in the unique factorization domain $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. ThereforeThe degree-$m$ monomials are indexed by nonnegative integers $a_0,\dots ,a_n$ with sum $m$. The stars and bars count givesHere and below, an all-integer binomial shorthand must use the counting convention $\left(\genfrac{}{}{0ex}{}{a}{n}\right)=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, ${\mathbb{P}}^0$ is one point and every ${\mathcal{O}}_{{\mathbb{P}}^0}(m)$ is trivial, so its space of global sections is $k$ for every $m$; this case is treated separately.

For a closed inclusion $i:H\hookrightarrow {\mathbb{P}}^n$, direct image sheaf formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves flabby sheaves, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a flasque resolution of $\mathcal{F}$; exactness gives a flasque resolution of $i_{\ast }\mathcal{F}$, and its complex of global sections is the original one becauseTaking cohomology of these identical complexes proves sheaf cohomology under a closed inclusion:This argument applies even without quasi-coherence.

We derive the remaining sheaf cohomology dimensions using the given eventual vanishing, the long exact sequence in sheaf cohomology, and induction on dimension. Multiplication by the linear equation of a hyperplane $H\cong {\mathbb{P}}^{n-1}$ gives the hyperplane exact sequence for twisting sheavesExactness can be checked in each chart by quotienting its polynomial ring by the equation of $H$; multiplication is injective because that equation is a non-zero-divisor. We use the closed-inclusion cohomology identification to replace the cohomology of the final sheaf by that on $H$.

The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher sheaf cohomology is zero. For $n=1$, the relevant long exact sequence isWhen $m\ge 0$, restriction of homogeneous polynomials to the hyperplane point is surjective: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge 0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomesStarting with $H^1(\mathcal{O}(-1))=0$, induction downward gives $h^1(\mathcal{O}(m))=-m-1$ for $m\le -2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for ${\mathbb{P}}^1$.

Now let $n\ge 2$ and assume the formulas for $H\cong {\mathbb{P}}^{n-1}$. For every $m$, the restrictionis surjective: for $m\ge 0$ extend a polynomial in the hyperplane coordinates to the same polynomial on ${\mathbb{P}}^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an injectionFor $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,{\mathcal{O}}_H(m))=0$, and gives the same injection in degree $i$. For each fixed $m$, composing finitely many injections reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. ThereforeFor $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,{\mathcal{O}}_H(m))$ and $H^i(H,{\mathcal{O}}_H(m))$ are zero by induction. Consecutive $H^i$ groups on ${\mathbb{P}}^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.

For top cohomology the exact sequence now reduces toWrite $b_n(m)=h^n({\mathcal{O}}_{{\mathbb{P}}^n}(m))$. The induction hypothesis and eventual vanishing give the recurrenceConsecutive top groups agree for $m\ge -n+1$, so descending from a large twist gives $b_n(m)=0$ for $m\ge -n$. Below this range, iterating the recurrence gives a finite sum:The last equality follows by summing Pascal's identity. These short exact sequences also prove that every group whose dimension is being counted is finite-dimensional. Thus, for $n\ge 1$,With the stated counting convention, the latter is the requested single binomial formula. The base ${\mathbb{P}}^0$ has $h^0=1$ for all twists and no higher cohomology; it is not governed by the negative-twist $H^0$ rule for positive-dimensional projective space.

Finally calculate the canonical line bundle of a smooth variety from its transition functions. On $U_0$ write $t_j=X_j/X_0$ for $1\le j\le n$. On $U_i$, $i>0$, take the coordinates $u_j=X_j/X_i$ in increasing order of $j\ne i$. ThusIn their wedge product, $d{u}_0=-t_i^{-2}d{t}_i$ and
$d{u}_j=t_i^{-1}d{t}_j-t_j{t}_i^{-2}d{t}_i$. Every term involving a second $d{t}_i$ vanishes, so reordering the remaining differentials givesThis computation is valid in every characteristic. Multiply the ordered local generator on $U_i$ by $(-1{)}^i$. The resulting generators ${\eta }_i$ satisfy ${\eta }_i=(X_i/X_0{)}^{-n-1}{\eta }_0$ on $U_i\cap U_0$, and hence the corresponding transition rule on every overlap. They are exactly the transitions of $\mathcal{O}(-n-1)$. Therefore the canonical bundle of projective space is

For a line bundle, Serre duality statesThe intermediate groups on both sides vanish. In the two outer degrees, the top group at twist $m$ has the same dimension as the global-section group at twist $-m-n-1$:The polynomial degree on the right is nonnegative precisely when $m\le -n-1$, and its dimension is $\left(\genfrac{}{}{0ex}{}{-m-1}{n}\right)$. Interchanging $m$ and $-m-n-1$ checks the other outer degree. On ${\mathbb{P}}^0$ both sides are one-dimensional. Thus all the formulas are consistent with Serre duality, which was used only for this final consistency check, not to prove the cohomology formulas.

The exactness axiom of a generalized cohomology theory gives a natural long exact sequence for each CW pair:For $(X,X)$, the restriction $h^n(X)\to h^n(X)$ is the identity. Its preceding relative-to-absolute map has zero image. The previous connecting map is zero too, since it follows the surjective identity in degree $n-1$. Exactness at the relative group therefore makes its map both injective and zero, givingin particular $h^n(x_0,x_0)=0$. No dimension axiom is being used; a generalized theory can have nonzero $h^n(x_0)$ in many degrees.

The wedge-product notation for pairs here means the smash product of CW pairs,Its quotient is the based smash product $(X/A)\wedge (Y/B)$. A multiplicative theory has an external productIf the two spaces are the same, pullback by the diagonal gives the relative cup product in $h^{k+m}(X,A\cup B)$. This is different from the wedge sum of two based spaces.

First read the covering subcomplexes as pointed subcomplexes, so $x_0\in A_j$. Contractibility makes the inclusion $\{x_0\}\hookrightarrow A_j$ a homotopy equivalence, and thus $h^r(A_j,x_0)=0$ in every degree, by exactness and homotopy invariance of cohomology. The relative exact sequence for the triple $\{x_0\}\subseteq A_j\subseteq X$ givesConsequently each homogeneous $w_j$ lifts to $v_j\in h^{r_j}(X,A_j)$. Repeated relative multiplication givesNaturality maps this product to the product of the $w_j$ in $h^{\ast }(X,x_0)$. ThereforeFor nonhomogeneous classes the same conclusion follows by distributivity. This is nilpotence from a contractible subcomplex cover.

There is an alternative sufficient convention: if $X$ is path connected, the $A_j$ need not contain $x_0$. The absolute image $\bar{w}$ of a reduced class restricts to zero at $x_0$, hence at any other point since the point inclusions are homotopic along a path. On a nonempty contractible $A_j$ it therefore restricts to zero, so exactness lifts ${\bar{w}}_j$ to $h^{\ast }(X,A_j)$. Their relative product vanishes as above. The map $h^{\ast }(X,x_0)\to h^{\ast }(X)$ is injective: restriction to the basepoint is split surjective by the map $X\to \mathrm{pt}$, making the previous connecting map zero. Thus vanishing of the absolute product gives vanishing of the reduced product too.

Without either the pointed-subcomplex or connectedness convention, the printed conclusion is false. Take $X=\{x_0,x_1\}$ covered by its two singleton subcomplexes and $h^{\ast }=H^{\ast }(-;\mathbb{Z})$. Both subcomplexes are contractible, but $H^0(X,x_0)$ contains the function $w$ with values $0,1$. Pointwise multiplication gives $w^2=w\ne 0$. This states precisely the needed interpretation rather than assuming a contractible component lies in the basepoint component.

Let $R=M{U}^{\ast }(\mathrm{pt})$ be the coefficient ring of complex cobordism. We use its standard coefficient theorem: $R\cong \mathbb{Z}[b_1,b_2,\dots ]$ with $|b_j|=-2j$. In particular $R$ is concentrated in nonpositive even degrees. Let $H$ be the quaternionic tautological line bundle on ${\mathbb{HP}}^2$, regarded as a complex rank-two bundle by right multiplication by a fixed imaginary quaternion. SetWe will proveThe argument must establish the multiplication as well as the additive groups, since collapse of a spectral sequence alone does not settle ring extensions.

The quaternionic projective space has one cell in each dimension $0,4,8$, giving integral cohomology $\mathbb{Z}$ in exactly these three degrees. The sphere bundle of $H$ is the quaternionic Hopf fibrationWrite $u=e(H_{\mathbb{R}})=c_2(H)$ for its ordinary degree-four class, fixing its sign by this equality. The Gysin sequence of a sphere bundle gives surjections, in fact isomorphisms,Indeed the corresponding neighboring cohomology groups of $S^{11}$ vanish. Both source and target groups are $\mathbb{Z}$, so these maps make $u$ and $u^2$ generators. There are no higher cohomology groups. This proves the cohomology ring of quaternionic projective space in the form $H^{\ast }({\mathbb{HP}}^2;\mathbb{Z})=\mathbb{Z}[u]/(u^3)$.

Apply the multiplicative Atiyah-Hirzebruch spectral sequence for complex cobordism:The skeletal construction obtains this sequence from the long exact sequences of successive skeleta; the quotients are wedges of spheres, and the first differential is the ordinary cellular differential with coefficient groups $R^t$. Its multiplication on the associated graded group is the ordinary cup product with these coefficients. The finite eight-dimensional skeleton makes the filtration finite, so there is no convergence ambiguity.

Only columns $s=0,4,8$ and even rows $t$ are nonzero. If $r$ is odd, the target column $s+r$ is odd and zero. If $r$ is even, the target coefficient degree $t-r+1$ is odd and zero. The same observations exclude incoming differentials. Consequently $E_{\infty }=E_2$, with associated graded $R$-module freely generated by $1,u,u^2$ in filtrations $0,4,8$.

The canonical map from complex cobordism to integral cohomology preserves the chosen complex orientation and hence sends $q$ to $u$. The class $q$ vanishes on the zero-dimensional skeleton since $R^4=0$, and its leading skeletal term is $u$ in filtration four. Multiplicativity makes the leading term of $q^2$ the generator $u^2$ in filtration eight. Thus $1,q,q^2$ lift the associated graded basis.

This resolves all module extensions explicitly. For a class of total degree $d$, subtract its coefficient $a_0\in R^d$ in filtration zero. From the remainder subtract $a_{1}q$, where $a_1\in R^{d-4}$ is its leading coefficient in filtration four, and then subtract $a_2{q}^2$ with $a_2\in R^{d-8}$. The remaining filtration is beyond dimension eight and is zero. Conversely a relation $a_0+a_{1}q+a_2{q}^2=0$ has zero leading coefficient in each of the three successive filtrations, forcing all three $a_j$ to vanish. ThereforeFinally $q^3$ has degree twelve. Every associated graded contribution in that total degree would have coefficient degree $12-s>0$ for $s=0,4,8$, and all such coefficients vanish. Thus $M{U}^{12}({\mathbb{HP}}^2)=0$ and $q^3=0$. The polynomial map $R[t]/(t^3)\to M{U}^{\ast }({\mathbb{HP}}^2)$ taking $t$ to $q$ is surjective by the basis argument and injective by its uniqueness. This proves the complex cobordism ring of the quaternionic projective plane as a graded ring over the stated coefficient ring, including the absence of any additional coefficient-dependent relation.

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.

The geometric argument uses the Mixed Heegaard Floer invariant defined in Question 5(iii). First suppose $g(\Sigma )>0$; the genus-zero qualification is discussed below. Write $r=\Sigma \cdot \Sigma \ge 0$. Blow up $r$ distinct points of the surface. Its proper transform $\tilde{\Sigma }$ in $\tilde{M}=M\#r{\overline{\mathbb{CP}}}^2$ has classChoose the extension of the Spin-c structure withSince $E_j^2=-1$, its exceptional evaluations are $-1$, so the Blowup formula for the mixed Heegaard Floer invariant has no extra power of $U$. Thus ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}\ne 0$, andThis calculation is why the self-intersection appears with a plus sign.

It remains to bound the Chern pairing for a square-zero surface $S$. Its oriented normal bundle has Euler number zero and is trivial, so $\partial \nu S=S\times S^1$. Choose an embedded positive-square surface $T$ disjoint from $S$, with positive direction left in its complement, and cut along $N=\partial \nu T$. To see why the choice is possible, impose orthogonality to $[S]$ in the intersection form. A nonzero isotropic class uses at most one positive direction; $b_2^{+}\ge 3$ therefore leaves at least two. Choose an integral positive-square class in that complement, represent it by a surface, and remove algebraically cancelling intersections with $S$ by tubing along $S$. This alters genus but not the class. The neighborhood of $T$ supplies one positive direction and the complement supplies another. Since its normal Euler number is nonzero, $H^1(\nu T)\to H^1(N)$ is onto, proving that the cut is admissible.

Put the outgoing puncture inside $\nu S$, so that the plus Heegaard Floer cobordism map on the side containing $S$ passes through the Floer group of $\partial \nu S$. On the chain level this means splitting the cobordism along that boundary and composing its handle maps; with the fixed determinant class, every intermediate generator has restriction $\mathfrak{t}=\tilde{\mathfrak{s}}{|}_{S\times S^1}$. If the Chern pairing exceeded $2g(S)-2$, the given completed minus group would be zero. As in Question 3, this is the completed convention: for a non-torsion structure, ordinary infinity may retain $(1-U^N)$-torsion, whereas $1-U^N$ is invertible after completion. Such a pairing is positive for $g\ge 1$, so $\mathfrak{t}$ is non-torsion and the completed infinity group ${\mathbf{HF}}^{\infty }(S\times S^1,\mathfrak{t})$ is zero as well. The completed fundamental long exact sequence, whose plus group is the usual plus theory, then makes that plus group zero. The Heegaard Floer cobordism map, and consequently the Mixed Heegaard Floer invariant, would vanish, contradicting nonvanishing. ThereforeSubstituting the preceding blowup calculation provesReversing the orientation of the surface gives the usual absolute-value form of the Adjunction inequality for the mixed Heegaard Floer invariant.

The printed assumptions omit the positive-genus qualification. With the actual Floer groups, the asserted three-dimensional vanishing is false for $S=S^2$: the torsion structure on $S^2\times S^1$ has two nonzero minus towers, although its pairing is $0>-2$. Correspondingly, a small sphere embedded in a ball of a K3 surface has $[S]=0$, $S^2=0$ and Chern pairing zero, while the canonical Mixed Heegaard Floer invariant is nonzero; the desired inequality would read $0\le -2$. Thus the displayed proof establishes the intended positive-genus statement. If the supplied vanishing is instead taken literally as a formal additional axiom for every genus, the same factorization would force the inequality even for a sphere, but that axiom contradicts the actual $S^2\times S^1$ computation.

Write $R^{i}F$ for the cohomological indexing of the right derived functor. For the contravariant construction, assume $F$ is additive and left exact, so an exact sequence $M^{\prime }\to M\to M^{\prime \prime }\to 0$ gives $0\to F(M^{\prime \prime })\to F(M)\to F(M^{\prime })$. Module categories have enough projective modules: a free module on a generating set surjects onto any module, and repeating the construction on its kernel produces a projective resolutionContravariance turns it into the cochain complexThe composite of consecutive differentials is zero because $F$ is additive and reverses composition. Applying left exactness to $P_1\to P_0\to M\to 0$ gives $F(M)=\ker (F(d_1))$, so $R^{0}F\simeq F$.

Here are the comparison and exactness arguments behind this construction. A map $M\to N$ lifts to a chain map between projective resolutions. In degree zero, lift through the surjection $Q_0\twoheadrightarrow N$ by projectivity of $P_0$. Inductively, the required composite from $P_i$ lands in the cycle module of the target resolution, which is the image of $Q_i$ by exactness, so projectivity supplies the next lift. Two such lifts are chain homotopic. Their difference in degree zero lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1$, so it lifts to a homotopy map $P_0\to Q_1$. Subtracting the corresponding homotopy term in degree one gives another map into the next cycle module; continue inductively. Applying additive contravariant $F$ turns this chain homotopy into a cochain homotopy, so the maps on cohomology agree. Applying this to lifts of identity maps between two resolutions gives mutually inverse maps on cohomology. Thus the contravariant right derived functor is well-defined, contravariant and natural in $M$.

For $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, construct compatible projective resolutions by the horseshoe lemma. Its construction is elementary: lift the projective cover of $M^{\prime \prime }$ into $M$ and add the projective cover of $M^{\prime }$; their sum surjects onto $M$. The kernels fit into a new short exact sequence of the corresponding kernels, so repeat at each degree. This gives a degreewise split sequenceAn additive contravariant functor takes split sequences to split sequences with their arrows reversed. The resulting short exact sequence of cochain complexes givesThe connecting map is obtained by lifting a cocycle from the quotient complex, taking its differential, and viewing the result in the subcomplex; changing a lift changes the result by a coboundary. This construction proves both exactness and naturality of the long exact sequence.

Positive derived functors vanish on projective modules, since such a module has a resolution concentrated in degree zero. For $0\to K\to P\to M\to 0$ with $P$ projective, the preceding sequence gives $R^{i+1}F(M)\simeq R^{i}F(K)$ for $i\ge 1$. This is dimension shifting. If $F$ is exact, all positive derived functors vanish because applying $F$ preserves exactness of the resolution. A natural transformation of additive left-exact contravariant functors induces natural transformations of their derived functors by applying it degreewise. These derived functors form the universal contravariant cohomological delta functor: successive projective surjections, on which the positive functors vanish, and the connecting maps determine any extension of a degree-zero natural transformation uniquely.

The displayed example in the PDF has a genuine variance mismatch. The functor $G(M)={\mathrm{Hom}}_A(A/(a),M)$ is covariant in $M$, not contravariant. For a map $u:M\to N$, its induced map sends $h$ to $u\circ h$, in the same direction. We compute that displayed example with its correct variance, and then explain the contravariant variant.

For an additive covariant left-exact functor, use an injective resolution $0\to M\to I^0\to I^1\to \cdots$ instead of a projective resolution, and set $R^{i}G(M)=H^i(G(I^{\bullet }))$. Such resolutions exist for arbitrary modules. In fact $J={\mathrm{Hom}}_{\mathbb{Z}}(A,D)$, with $(a\varphi )(b)=\varphi (ab)$, is injective: the adjunctionsends a map to its value at $1$, and has inverse $\lambda \mapsto [n\mapsto (b\mapsto \lambda (bn))]$. Extension on the right was proved in Question 5, so it gives extension on the left. Products of injective modules are injective by extending every component separately. The mapis $A$-linear and injective, because a nonzero $m$ is detected by some character at $b=1$. Applying the same embedding to successive cokernels produces the stated injective resolution. Its comparison and homotopy arguments are dual to the projective ones, and covariant derived functors have the long exact sequence in the original order of $M^{\prime },M,M^{\prime \prime }$.

Let $B=A/(a)$. Evaluation at $1+(a)$ identifiesa naturally defined $B$-module. It is left exact: a compatible element in the kernel of $G(M)\to G(M^{\prime \prime })$ is exactly an element of $M^{\prime }$ killed by $a$.

Because $a$ is a non-zero-divisor on $A$, multiplication by $a$ is surjective on every injective module $I$. Indeed, for $m\in I$, the map $aA\to I$ given by $ax\mapsto xm$ is well-defined and $A$-linear. It extends to $A\to I$, and if the extension takes $1$ to $v$, then $av=m$. Thus for an injective resolution we have a short exact sequence of cochain complexesThe cohomology of $I^{\bullet }$ is $M$ in degree zero and zero in higher degrees. Its long exact sequence therefore givesConsequently the Ext from a principal non-zero-divisor quotient computation isAll these groups are naturally $B$-modules. Equivalently they are ${\mathrm{Ext}}_A^i(B,M)$; the two-term free resolution $0\to A\stackrel{a}{\to }A\to B\to 0$ gives the same kernel and cokernel after applying ${\mathrm{Hom}}_A(-,M)$.

If one instead wants a genuinely contravariant example, take $H(N)={\mathrm{Hom}}_A(N,B)$. Then $R^{i}H(N)={\mathrm{Ext}}_A^i(N,B)$, obtained by resolving the variable $N$. There is no claim of vanishing above degree one for arbitrary $N$. At $N=B$, the same two-term resolution has zero differential after applying ${\mathrm{Hom}}_A(-,B)$, and gives ${\mathrm{Ext}}_A^0(B,B)=B$, ${\mathrm{Ext}}_A^1(B,B)=B$, and zero higher groups. This keeps the printed covariant calculation distinct from the general contravariant 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.

Let $q:{\mathbb{CP}}^8\to Q$ collapse the indicated Complex projective space. Write $x\in H^2({\mathbb{CP}}^8;\mathbb{Z})$ for the positive generator, so the cohomology ring of complex projective space is $\mathbb{Z}[x]/(x^9)$. The inclusion of ${\mathbb{CP}}^3$ sends $x$ to its corresponding generator and is an isomorphism in degrees $0,2,4,6$.

The CW complex structure makes the inclusion a cofibration, so the positive-degree cohomology of $Q$ identifies with the relative cohomology of the pair. Its long exact sequence shows that $q^{\ast }$ is injective in positive degrees, with image the additive subgroup generated by $x^4,x^5,x^6,x^7,x^8$. Equivalently, the surviving cellular cochain complex has one copy of $\mathbb{Z}$ in each of degrees $8,10,12,14,16$ and zero differential. Let $u_i$ denote the class with $q^{\ast }{u}_i=x^i$, for $4\le i\le 8$.

Naturality of the cup product, together with the injectivity of $q^{\ast }$, now determines all multiplication:Every other product of positive-degree basis elements is zero. Indeed, their degrees exceed $16$. In particular, the surviving classes do not form a polynomial algebra on a degree-eight generator: the independent classes in degrees $10,12,14$ must also be retained. This is an instance of the cohomology ring of a collapsed projective subspace.

Work first with coefficients ${\mathbb{F}}_2$. For a rank-$r$ real vector bundle $E\to B$ over a CW complex, choose a fibre metric and its disk bundle $D(E)$ and sphere bundle $S(E)$. The mod-two Thom class is the unique class $U\in H^r(D(E),S(E);{\mathbb{F}}_2)$ restricting to the nonzero generator on every fibre pair. The Thom isomorphism theorem asserts thatNo orientation of the vector bundle is needed with these coefficients.