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Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

Represent a class in $H_k(K)$ by a chain cycle $z\in Z_k(K)$. Since every simplex of $K=M\cup N$ belongs to at least one of the two simplicial subcomplexes, split the chain in the simplicial chain complex asBecause $\partial z=0$, we haveThe left side lies in $C_{k-1}(M)$ and the right side in $C_{k-1}(N)$, so this common chain is supported in $L=M\cap N$. It is a cycle because ${\partial }^2=0$. The connecting homomorphism is thereforeChanging the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to homology classes.

The exact sequenceinduces a long exact sequence in sheaf cohomology. Under $H^0({\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast })=\mathrm{Div}(X)$ and $H^1({\mathcal{O}}_X^{\ast })=\mathrm{Pic}(X)$, its connecting homomorphism is the divisor-to-Picard map $D\mapsto \mathcal{O}(D)$. Exactness identifies its kernel with divisors of global nonzero meromorphic functions, namely principal divisors:

Inclusion and passage to the quotient give a short exact sequence of chain complexesDefine the connecting homomorphism as follows. If $[c]\in H_k(K,L)$, then $\bar{\partial }[c]=0$, so $\partial c\in C_{k-1}(L)$. Since ${\partial }^{2}c=0$, it is a chain cycle in $L$, and we setChanging $c$ by a relative boundary or a chain in $L$ changes $\partial c$ by a boundary in $L$, so $\delta$ is well defined.

The resulting sequence isWe verify exactness directly. If a cycle of $K$ maps to zero in the relative homology, it differs from a boundary by a cycle in $L$, so it lies in $\mathrm{im}{i}_{\ast }$. If a relative cycle $[c]$ satisfies $\delta [c]=0$, then $\partial c=\partial \ell$ for some $\ell \in C_k(L)$; hence $c-\ell$ is a cycle in $K$ mapping to $[c]$. Finally, if a cycle $z\in C_{k-1}(L)$ becomes a boundary in $K$, say $z=\partial c$, then $[c]$ is a relative cycle with $\delta [c]=[z]$. The reverse containments follow immediately from consecutive composites being zero. This proves the long exact sequence in relative homology.

Suppose $X=A\cup B$ and that the interiors of $A$ and $B$ cover $X$. The Mayer–Vietoris theorem gives the long exact sequence in homologywhere all four displayed maps except the connecting homomorphism $\partial$ are induced by the relevant inclusion maps.

To prove it, let $C_{\ast }^{A,B}(X)$ be the chain complex generated by those singular simplices whose images lie wholly in $A$ or wholly in $B$. The sequenceis a short exact sequence of chain complexes. Repeated barycentric subdivision makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is chain homotopic to the identity. The inclusion $C_{\ast }^{A,B}(X)\hookrightarrow C_{\ast }(X)$ therefore induces an isomorphism on every homology group. The long exact homology sequence of a short exact sequence of chain complexes now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial [z]$.

Apply part i to the short exact sequence of trivial $G$-modulesThe relevant segment isBoth outer groups vanish by parts a(iv) and a(v), so the connecting homomorphism is an isomorphism:

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the coefficient of either incident $n$-simplex from the other, up to the orientation sign. Connectivity through such faces therefore determines every top-simplex coefficient from one integer. If the signs are globally compatible, their oriented sum is a cycle and generates $H_n(K)\cong \mathbb{Z}$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting fundamental class, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

The long exact sequence in homology of the triple $A\subseteq U\subseteq X$ containsBecause $A\hookrightarrow U$ is a homotopy equivalence, $H_{\ast }(U,A)=0$. Hence the middle map is an isomorphism in every degree:

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesA good pair $(X,A)$ has $A$ closed and a neighborhood $U$ that deformation retracts onto $A$. The collapsing a pair theorem states that the quotient map givesTo prove it, choose such a neighborhood $U$. The first result identifies $H_{\ast }(X,A)$ with $H_{\ast }(X,U)$. Excision identifies the latter with $H_{\ast }(X/A,U/A)$, while $U/A$ is contractible because the deformation retraction can be chosen relative to $A$ using the homotopy extension property. The long exact sequence of the pair $(X/A,U/A)$ then identifies this relative group with ${\tilde{H}}_{\ast }(X/A)$.

For $f:(D^n,\partial D^n)\to (D^n,\partial D^n)$, define $\mathrm{deg}f$ byin $H_n(D^n,\partial D^n;\mathbb{Z})\cong \mathbb{Z}$. In the long exact sequence of the pair, the connecting homomorphismis an isomorphism. Naturality shows that $f_{\ast }$ and $(f{|}_{\partial D^n}{)}_{\ast }$ multiply the corresponding generators by the same integer, so

Finally,Under this identification, the map induced by $g(x,t)=(f(x),t)$ is the suspension of the map induced by $f$. By degree under suspension,This argument proves the needed product assertion directly from the natural suspension isomorphism in reduced homology.

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of chain complexesIts long exact sequence in homology isIf $\bar{z}$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the connecting homomorphism is $\hat{\beta }[\bar{z}]=[w]$.

The coefficient sequencesimilarly givesAt chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the Bockstein homomorphism is $\beta [\bar{z}]=[\bar{w}]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes \mathbb{Z}/n$. Naturality of connecting homomorphisms givesExactness of the first long exact sequence says $\hat{\beta }{\rho }_{\ast }=0$, and hence

For the standard cellular chain complex of Real projective space ${\mathbb{RP}}^3$, there is one copy of $\mathbb{Z}$ in degrees $0,1,2,3$, with $d_2=2$ and $d_1=d_3=0$. Modulo two all cellular differentials vanish, while the Bockstein is the identity from degree two to degree one and zero elsewhere. Therefore

Finally, Smith normal form decomposes a bounded chain complex of finitely generated free abelian groups, up to chain isomorphism and contractible summands, into one-term complexes $\mathbb{Z}[i]$ and two-term complexesin degrees $i+1$ and $i$. The former represents a free homology summand, and the latter represents $\mathbb{Z}/m$ in degree $i$. The two requested conclusions now follow from the next two parts.

Collapsing the boundary identifies the top relative homology of a disk pair with the top reduced homology of its quotient sphere. Together with the boundary connecting homomorphism, this shows that the induced quotient map and the boundary map have the same mapping degree, using the corresponding orientations.

For a map of oriented disk pairs, the multiplier on top relative homology is its relative mapping degree. Naturality of the connecting homomorphism identifies it with the mapping degree of the boundary map.

On a Riemann surface, the quotient sheaf $\mathcal{P}={\mathcal{M}}_X/{\mathcal{O}}_X$ records the finite negative Laurent series tails of meromorphic germs. At $x$, a local coordinate $t$ identifies its stalk with ${⨁}_{m\ge 1}\mathbb{C}{t}^{-m}$. As a sheaf it is the direct sum of sheaves ${⨁}_x(i_x{)}_{\ast }{\mathcal{P}}_x$. Global sections prescribe locally finite families of principal parts, possibly infinite on a noncompact surface. The connecting homomorphism to $H^1(X,{\mathcal{O}}_X)$ is exactly the obstruction to a global meromorphic function with those parts and no extra poles.

A degreewise short exact sequence of cochain complexes induces a long exact sequence from a coefficient sequence in the coefficient case, and a long exact cohomology sequence generally. Its degree-one connecting homomorphism lifts a quotient cocycle, takes its coboundary, and identifies that result in the subcomplex.