Cup Cap Cross Products and Cohomology Rings — Examples

By [F1], $H_0(S^1;\mathbb{Z})=\mathbb{Z}$, $H_1(S^1;\mathbb{Z})=\mathbb{Z}$ and all higher groups are zero. Radial projection identifies the oriented triangle boundary with the circle. Its three successively oriented edges have primitive all-ones cycle: the simplicial one-cycle condition forces their coefficients to agree and there are no two-simplices in the boundary complex. The simplicial-to-singular comparison used in [F1] carries this positive generator to $[z]$. The Ext groups in [F2] vanish in every degree: for first variable $0$ use the zero resolution; for first variable $\mathbb{Z}$ use the resolution with $\mathbb{Z}$ in degree zero augmented by identity and no higher terms, whose Hom has no degree-one cohomology. Thus evaluation identifies $H^0(S^1;\mathbb{Z})=\mathbb{Z}\mathbf{1}$ and $H^1(S^1;\mathbb{Z})=\mathbb{Z}u$, with the unique $u$ satisfying $⟨u,[z]⟩=1$, and higher cohomology is zero. In particular $u^2=0$ because its target is $H^2(S^1;\mathbb{Z})=0$.

All the homology groups in step 1.1 are finite free, so [F3] applies. Put $x={\mathrm{pr}}_1^{\ast }u=u\times 1$ and $y={\mathrm{pr}}_2^{\ast }u=1\times u$. The graded tensor source has basis $1\otimes 1$ in degree zero, $u\otimes 1,1\otimes u$ in degree one, and $u\otimes u$ in degree two, with no other degrees. Its ring multiplication sends the squares of the degree-one basis elements to zero, their ordered product to $u\otimes u$, and their reversed product to $-u\otimes u$. Hence the target has basis $1,x,y,xy$ and the displayed multiplication relations. In particular $xy$ is a generator, rather than merely a nonzero class. [F3, step 1.1] 2.2 The shuffle $Z=S(z\otimes z)$ is a cycle by [F5]. By [F6], it is a generator of $H_2(T^2;\mathbb{Z})$: the only nonzero tensor term in total degree two is $\mathbb{Z}[z]\otimes \mathbb{Z}[z]$, and every Tor term vanishes. To see the latter directly, each first variable is $0$ or $\mathbb{Z}$ by step 1.1, and tensoring its zero or length-zero identity resolution has zero degree-one homology. The generator $[Z]$ has the product orientation. Write the triangle-boundary chain as $z={\sum }_i{ϵ}_i{\sigma }_i$, where ${ϵ}_i=\pm 1$ is the orientation sign of its edge parameterization relative to the counterclockwise direction. On the square parameterized by ${\sigma }_i\times {\sigma }_j$, the coefficient ${ϵ}_i{ϵ}_j$ converts its parameter orientation to the positive product orientation. In the parameters $(s,t)$, its shuffle triangles have vertex lists $((0,0),(1,0),(1,1))$ with coefficient $+1$ and $((0,0),(0,1),(1,1))$ with coefficient $-1$. Their ordered edge determinants are respectively $+1$ and $-1$, so both signed triangles carry the positive $ds\wedge dt$ orientation. Their diagonal faces cancel; along arc boundaries the circle-cycle endpoint cancellations cancel the outer square faces. Thus $Z$ is precisely the sum of the positively oriented triangles in the product decomposition of the torus. Adjacent triangles induce opposite orientations on their shared edge, giving the same product orientation across their seams.

For the formal degree-one module $\mathbb{Z}e\oplus \mathbb{Z}f$, the map $e\mapsto x$, $f\mapsto y$ kills every square: $(rx+sy{)}^2=r^2{x}^2+rs(xy+yx)+s^2{y}^2=0.$ It therefore induces a map from [F7]'s exterior quotient to the cohomology ring. In that quotient $e^2=f^2=0$ and $ef+fe=(e+f{)}^2-e^2-f^2=0$, so move every $f$ past every $e$ and delete repetitions to express every word in the span of $1,e,f,ef$. Their four images are independent by step 2.1. Thus the induced map is both surjective and injective, proving the claimed exterior-algebra presentation without assuming an abstract basis theorem.

Let $\phi$ be a singular cocycle representing $u$, and let $J=J(\phi ,\phi )$ be tensor evaluation. Then $\phi (z)=1$, and the signed tensor differential gives $Jd=0$. The AW external cochain representing $xy$ therefore satisfies $⟨xy,[Z]⟩=JAS(z\otimes z)=J(z\otimes z)+J(dL+Ld)(z\otimes z)=1.$ Here $d(z\otimes z)=0$ and $Jd=0$ kill the two homotopy terms separately. This proves the asserted positive normalization on the actual oriented cycle of step 2.2. No cellular cochain has been mistaken for a singular representative.

For arbitrary degree-one classes $a=rx+sy$ and $b=r^{\mathrm{\prime }}x+s^{\mathrm{\prime }}y$, the multiplication table gives $ab=(r{s}^{\mathrm{\prime }}-s{r}^{\mathrm{\prime }})xy$, including zero coefficients and repeated inputs. Products of $xy$ with a positive-degree class are zero because degrees above two vanish. The unit multiplies every class unchanged. Reversing one circle orientation replaces its generator and the oriented product cycle by their negatives, so the normalization changes consistently; interchanging the factors gives the sign $-1$ in degree two. The space and coefficients here are fixed and nonempty, so no empty-torus or zero-ring assertion is made. The shuffle and AW calculation retains degenerate simplices and checks the vertex and top degrees. AC is used exactly for the UCT cycle projections, the additive Künneth PID sections and the homological Künneth cycle/boundary constructions of [F8]. All ring arithmetic and orientation signs are the explicit finite calculations above.

Parameterize each paired boundary edge by its positive generator $A_i$ or $B_i$, giving a singular loop in the quotient based at $v$. Put ${ϵ}_k=+1$ on the positive traversal of an edge in the polygon word and ${ϵ}_k=-1$ on its negative traversal. Let $r_k$ be the radial singular edge from $c$ to $v_k$, followed by the quotient map. For ${ϵ}_k=+1$, let $T_k$ have ordered vertices $(c,v_k,v_{k+1})$; for ${ϵ}_k=-1$, use $(c,v_{k+1},v_k)$. The affine maps of these triangles into $P$ followed by the quotient are genuine singular simplices. The last edge of either ordered triangle is the same positively parameterized loop $E_k=A_i$ or $B_i$, so paired occurrences have literally identical last singular edges, rather than only homotopic ones. The two boundary formulas are $\mathrm{\partial }{T}_k=E_k-r_{k+1}+r_k({ϵ}_k=1),\mathrm{\partial }{T}_k=E_k-r_k+r_{k+1}({ϵ}_k=-1)\mathrm{.}$ Thus $Z={\sum }_k{ϵ}_k{T}_k$ has boundary ${\sum }_k{ϵ}_k{E}_k$: radial terms telescope, and every $A_i,B_i$ occurs once with each sign. Consequently $\mathrm{\partial }Z=0$.

The positive polygon-boundary cycle is a generator of $H_1(\mathrm{\partial }P;\mathbb{Z})$. Indeed, the polygon boundary has $4g$ vertices and successively oriented edges; its cellular boundary sends the edge from $v_k$ to $v_{k+1}$ to $v_{k+1}-v_k$ by [F2], so a one-chain is a cycle exactly when all its edge coefficients are equal, and there are no two-cells. The all-ones chain is therefore a primitive positive generator, with its sign fixed by the given boundary order.

Before taking the quotient, the signed fan of step 1.1 is a relative cycle $\tilde{Z}$ of $(P,\mathrm{\partial }P)$, with boundary the positive polygon-boundary cycle. The polygon contracts linearly to $c$, so [F4] gives $H_1(P)=H_2(P)=0$; explicitly the point singular complex has one generator in every degree with differential identity in positive even degrees and zero in odd degrees, hence no positive homology. Therefore [F3] identifies $H_2(P,\mathrm{\partial }P)$ with $H_1(\mathrm{\partial }P)$ by boundary. Since its connecting image is the primitive positive generator computed in step 1.2, $[\tilde{Z}]$ is precisely the positive relative generator, with no orientation sign left unspecified.

In the surface CW structure there is one vertex, $2g$ oriented edges and one oriented two-cell. Each edge has coincident endpoints, so $d_1=0$ by [F2]. Under the quotient map of pairs $(P,\mathrm{\partial }P)\to ({\mathrm{\Sigma }}_g,{\mathrm{\Sigma }}_g^1)$, the relative chain $\tilde{Z}$ becomes $Z$ modulo ${\mathrm{\Sigma }}_g^1$; hence step 2.1 and [F2] identify its relative class with the positive face generator. Its connecting boundary is zero by step 1.1, so $d_2=0$. There are no higher cells. Applying [F1] gives $H_0({\mathrm{\Sigma }}_g)=\mathbb{Z},H_1({\mathrm{\Sigma }}_g)={⨁}_i(\mathbb{Z}[A_i]\oplus \mathbb{Z}[B_i]),H_2({\mathrm{\Sigma }}_g)=\mathbb{Z},$ and zero higher homology. More precisely, [F1] applied to the one-skeleton gives $H_2({\mathrm{\Sigma }}_g^1)=0$; exactness of the pair sequence [F3] then injects $H_2({\mathrm{\Sigma }}_g)$ into $H_2({\mathrm{\Sigma }}_g,{\mathrm{\Sigma }}_g^1)$, while the vanishing connecting image just proved makes the positive face generator lie in its image. Since $Z$ maps to that generator, $H_2({\mathrm{\Sigma }}_g)=\mathbb{Z}[Z]$, with no unlicensed comparison normalization. The stated edge loops have their specified cellular edge coordinates by their relative characteristic-interval classes. The signed fan carries the given surface orientation: positive triangles agree with the polygon orientation, negatively ordered triangles have coefficient minus one, and the prescribed edge identifications glue opposite boundary orientations.

Every homology group in step 3.1 is finite free. Its Ext term in [F5] is zero, by using the identity augmentation as a length-zero free resolution; for the zero group use the zero resolution. Hence evaluation identifies $H^n({\mathrm{\Sigma }}_g;\mathbb{Z})$ with $\mathrm{Hom}(H_n({\mathrm{\Sigma }}_g;\mathbb{Z}),\mathbb{Z})$ in every degree. Define $a_i,b_i$ by the coordinate duals of $[A_i],[B_i]$, and define $\omega$ by $\omega ([Z])=1$. These classes exist and are unique. In particular there are actual singular cocycles representing all these classes; no cellular cochain is being evaluated by a singular formula. Evaluation is injective in degree two, so a product there is determined by its value on $Z$.

Take any two singular one-cocycles $u,w$. Put $u_i=u(A_i)$, $u_i^{\mathrm{\prime }}=u(B_i)$, $w_i=w(A_i)$, $w_i^{\mathrm{\prime }}=w(B_i)$ and $t_k=u(r_k)$. Since $u(\mathrm{\partial }{T}_k)=0$, the two boundary formulas of step 1.1 give $t_{k+1}=t_k+{ϵ}_{k}u(E_k)\mathrm{.}$ By [F6], the contribution of ${ϵ}_k{T}_k$ to $(u⌣w)(Z)$ is $t_{k}w(E_k)$ when ${ϵ}_k=+1$ and $-t_{k+1}w(E_k)$ when ${ϵ}_k=-1$. This uses the actual first edge of the ordered triangle in each case. Coincident quotient vertices do not make its radial or boundary singular edge constant. [F6, step 1.1, step 4.1] 5.2 For $g=0$, use $S^2=D^2\mathrm{/}\mathrm{\partial }{D}^2$ with one vertex and one oriented two-cell. The cellular complex has $\mathbb{Z}$ in degrees zero and two and zero in all other degrees, so [F1] and the length-zero resolution argument of step 4.1 give $H^0=H^2=\mathbb{Z}$, $H^1=0$ and all higher groups zero. The positive cell orientation of [F2] specifies the generator dual to $\omega$; the empty list of degree-one classes has no asserted pair products, and ${\omega }^2=0$ by degree.

In the $i$th block $A_i,B_i,A_i^{-1},B_i^{-1}$, write $t$ for the initial radial value. The recurrence in step 5.1 gives successive values $t,t+u_i,t+u_i+u_i^{\mathrm{\prime }},t+u_i^{\mathrm{\prime }},t$. The four cup contributions are consequently $t{w}_i+(t+u_i)w_i^{\mathrm{\prime }}-(t+u_i^{\mathrm{\prime }})w_i-t{w}_i^{\mathrm{\prime }}=u_i{w}_i^{\mathrm{\prime }}-u_i^{\mathrm{\prime }}{w}_i\mathrm{.}$ The radial value returns to $t$ at the block's end, and in any event it has canceled from the formula. Summing all blocks gives the full calculation $⟨[u]⌣[w],[Z]⟩={\sum }_{i=1}^g(u_i{w}_i^{\mathrm{\prime }}-u_i^{\mathrm{\prime }}{w}_i)\mathrm{.}$ It holds for any representatives of the two classes, since their values on the edge cycles are their homology evaluations.

Insert the coordinate duals from step 4.1 into step 6.1. For $u=a_i,w=b_j$ the sum is ${\delta }_{ij}$; for $u=b_i,w=a_j$ it is $-{\delta }_{ij}$; for two $a$'s or two $b$'s it is zero. The injectivity of degree-two evaluation in step 4.1 gives all four asserted equalities. These signs also agree with [F7]'s graded commutativity. Every product involving $\omega$ and a positive-degree class lands above degree two and is zero by step 4.1. The vertex-value unit from [F7] gives the remaining products and associativity. Thus the listed bases and multiplication specify the entire ring.

At $g=1$, step 6.1 is the single determinant $u_1{w}_1^{\mathrm{\prime }}-u_1^{\mathrm{\prime }}{w}_1$, with the positive product sign. Zero or repeated degree-one inputs give zero by that formula, and reversing orientation negates $Z$ and its normalized dual $\omega$ consistently. There is no empty surface or zero coefficient ring in this example. Radial edges are allowed to repeat as maps after the quotient, and all singular simplices, including degeneracies, are retained. AC occurs only in [F5]'s integral cycle projections and sections, as supplied by [F8]; the finite fan and its cocycle recurrence need none.

By [F1], the mod-two cellular complex of $P^n$ consists of one copy of ${\mathbb{F}}_2$ in each degree $0,\dots ,n$ and zero differentials. A standard skeletal inclusion $P^m\to P^n$ sends each characteristic cell in dimensions at most $m$ to the same cell; its cellular map is therefore the identity in those dimensions. The natural comparison in [F1] gives $H_k(P^n)={\mathbb{F}}_2$ for $0\le k\le n$, zero otherwise, and inclusion is an isomorphism for $k\le m$. By [F2], $H^k(P^n)$ has exactly the same dimensions, and restriction is an isomorphism for $k\le m$. This uses the field dual of mod-two homology, not the integral Hom term with its possible Ext contribution discarded.

Coordinate projective subspaces are closed: their inverse images in the sphere are zero sets of specified coordinates, and the quotient topology tests closed sets by their inverse images. Coordinate permutations induce homeomorphisms, with inverse the opposite permutation, taking each such subspace to the corresponding standard skeleton. Thus step 1.1 also makes restriction to any coordinate $P^m$ an isomorphism in degrees at most $m$. The affine set $U_k=\{x_k\mathrm{\ne }0\}$ is open and homeomorphic to ${\mathbb{R}}^n$ by ratios $x_l\mathrm{/}{x}_k$, $l\mathrm{\ne }k$. These functions descend continuously from the open inverse image in the sphere; the quotient map is open because saturation of an open set is its union with its antipodal image. The inverse assigns the line of the vector whose $k$th coordinate is one. These formulas establish both continuity directions.

Fix $i,j\ge 1$ with $i+j=n$. Let $E=P^i$ use coordinates $x_0,\dots ,x_i$, and let $F=P^j$ use $x_i,\dots ,x_n$. Then $E\cap F=\{p\}$, where $p=[e_i]$. Put $V=P^n\setminus F$ and $W=P^n\setminus E$. In $V$ the vector $(x_0,\dots ,x_{i-1})$ is nonzero. The formula $[x_0:\cdots :x_n]⟼[x_0:\cdots :x_{i-1}:t{x}_i:\cdots :t{x}_n],1\ge t\ge 0,$ defines a strong deformation retraction of $V$ onto the coordinate $Q=P^{i-1}$: its vector is nonzero, it commutes with scaling, and it fixes $Q$. Continuity follows in the quotient charts of step 2.1, jointly with $t$. The same homotopy restricts to a retraction of $E\setminus p$ onto $Q$. Interchanging first and last coordinates gives the analogous retractions of $W$ and $F\setminus p$ onto a $P^{j-1}$. Scaling just coordinate $x_i$ to zero retracts $P^n\setminus p$ onto the coordinate hyperplane $P^{n-1}$ avoiding $p$.

The map $H^i(P^n,V)\to H^i(P^n)$ is an isomorphism. Indeed [F4] and step 3.1 identify $H^{\ast }(V)$ with $H^{\ast }(Q)$, compatibly with restriction from $P^n$. Step 1.1 and step 2.1 give $H^i(V)=0$ and make $H^{i-1}(P^n)\to H^{i-1}(V)$ onto (in fact an isomorphism). Exactness in [F3] first makes the connector into $H^i(P^n,V)$ zero, then makes the displayed map injective and surjective. The same argument for $(E,E\setminus p)$ shows $H^i(E,E\setminus p)\to H^i(E)$ is an isomorphism. The absolute restriction $H^i(P^n)\to H^i(E)$ is an isomorphism by step 2.1. Its commuting square from [F3] therefore makes $H^i(P^n,V)⟶H^i(E,E\setminus p)$ an isomorphism. At $i=1$, the preceding groups are degree-zero constants on the nonempty $P^0$ retract; their restriction is still onto, so no reduced-degree convention has been omitted.

In $U_i={\mathbb{R}}^i\times {\mathbb{R}}^j$, the intersections with $E,F$ are the two coordinate planes. Thus $V\cap U_i=({\mathbb{R}}^i\setminus 0)\times {\mathbb{R}}^j$ and $W\cap U_i={\mathbb{R}}^i\times ({\mathbb{R}}^j\setminus 0)$. Excision [F5] makes $H^i(E,E\setminus p)\to H^i({\mathbb{R}}^i,{\mathbb{R}}^i\setminus 0)$ an isomorphism: remove the closed coordinate hyperplane $E\setminus U_i$, which avoids $p$ and lies inside the open set $E\setminus p$. Also restriction from the pair $(U_i,V\cap U_i)$ to its first coordinate plane is an isomorphism. To verify the latter assertion directly, contract the unused second coordinate. This is a homotopy equivalence on ambient spaces and on the relative subspaces by [F4]. Both ambient spaces are nonempty contractible. Their pair sequences [F3] identify relative degree one with $H^0$ of the subspace modulo constant functions, higher relative degree $k$ with $H^{k-1}$ of the subspace, and degree zero with zero. Naturality and the subspace isomorphisms therefore prove the assertion in all degrees, including $i=1$. The square formed by these two maps and the restriction of step 4.1 commutes by [F3]. Three of its sides are isomorphisms, so the fourth $H^i(P^n,V)⟶H^i(U_i,V\cap U_i)$ is an isomorphism as well. The same proof with $j,F,W$ gives the other factor isomorphism.

Apply the argument of step 4.1 to $(P^n,P^n\setminus p)$, using its $P^{n-1}$ retract from step 3.1. Since $H^{n-1}(P^n)\to H^{n-1}(P^{n-1})$ is onto and $H^n(P^{n-1})=0$, the map to absolute $H^n(P^n)$ is an isomorphism. Excision [F5], removing the closed hyperplane $P^n\setminus U_i$ inside the open punctured space, also gives an isomorphism $H^n(P^n,P^n\setminus p)⟶H^n(U_i,U_i\setminus p)\mathrm{.}$

Take the nonzero classes $a\in H^i(P^n)$ and $b\in H^j(P^n)$. By step 4.1 they lift uniquely to relative classes for $V$ and $W$. By step 5.1 their local restrictions are generators of the two coordinate relative groups, identified by the coordinate projections. Their product is nonzero in $H^n(U_i,U_i\setminus p)$ by [F6]. The sets $V,W$ are open and $V\cup W=P^n\setminus p$, so [F7] applies both to restriction to $U_i$ and to passage to the absolute pair. Step 5.2 identifies both maps out of the top relative group as isomorphisms. Hence $ab\mathrm{\ne }0$ in $H^n(P^n)$: otherwise the relative product, and then its local restriction, would be zero. This proves the top product for every $i,j\ge 1$ with $i+j=n$.

For $n=0$, $P^0$ is a point and its ring is ${\mathbb{F}}_2$, with $x=0$. For $n=1$, step 1.1 gives one nonzero degree-one class $x$ and no groups in degree two or higher, so the ring is ${\mathbb{F}}_2[x]\mathrm{/}(x^2)$ by the unit in [F7]. Proceed by induction on $n\ge 2$. Restriction carries the unique nonzero $x\in H^1(P^n)$ to its namesake in $P^{n-1}$ by step 1.1. Thus its powers $x^k$, $0\le k\le n-1$, restrict to the nonzero powers from the preceding dimension, by [F7], and so are nonzero. Step 6.1 applied to $i=n-1,j=1$ now gives $x^n\mathrm{\ne }0$. All higher powers vanish by the group calculation of step 1.1. Each $x^k$ is the unique generator in its degree. Consequently the polynomial evaluation homomorphism is onto, and its kernel consists exactly of polynomials with no terms of degrees $0,\dots ,n$, namely the ideal $(x^{n+1})$. This proves the asserted graded ring isomorphism.

For $1\le m\le n$, degree-one restriction is the isomorphism in step 1.1, so it sends $x$ to $x$; for $m=0$ its target group is zero. Naturality and the unit in [F7] give every power and the constant term, including identity restriction at $m=n$. Zero inputs and powers above the truncation vanish by step 7.1. There are no empty projective spaces here, and the zero-dimensional point has been treated without introducing $P^{-1}$. All complement deformations were used only with $i,j\ge 1$ and checked at $t=0,1$ in step 3.1; coincident or degenerate singular simplices are retained by the relative suppliers. The assumed AC is used only through [F2] and [F6], as their statements record; the coordinate formulas and finite induction introduce no further choice.

These two quotient descriptions agree: normalization $z\mapsto z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ is continuous and a nonzero scaling changes the normalized vector by a unit phase; inclusion of the sphere provides the inverse on quotients. The sphere quotient is compact and Hausdorff. For Hausdorffness, the map $z\mapsto z{z}^{\ast }$ from the unit sphere to the finite-dimensional Hausdorff space of complex matrices has exactly the phase orbits as fibres: equality of these rank-one matrices implies equality of their images, hence $w=\lambda z$, and the unit norms give $\mathrm{\mid }\lambda \mathrm{\mid }=1$. The induced map of the quotient onto its matrix image is a continuous bijection from a compact space to a Hausdorff space and is a homeomorphism (images of closed sets are compact and therefore closed). Each affine chart $U_k=\{z_k\mathrm{\ne }0\}$ is open and has coordinates $z_l\mathrm{/}{z}_k$, $l\mathrm{\ne }k$, with inverse the line represented by $z_k=1$. The quotient map is open because saturation is a union of translates by phases, so these ratios descend continuously; the displayed inverse is also continuous. Coordinate subspaces are closed by their coordinate-zero inverse images in the sphere.

Attach a $2r$-disk to $P^{r-1}$ by the map $D^{2r}⟶P^r,w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}]\mathrm{.}$ The boundary lands in $P^{r-1}$. Every line outside $P^{r-1}$ has a unique unit representative whose last coordinate is positive real, so the disk interior maps bijectively onto its complement. The induced attachment-quotient map is a continuous bijection from a compact space to the Hausdorff $P^r$ of step 1.1, hence a homeomorphism. Starting with $P^0=\ast$ constructs a finite CW complex with one cell in each dimension $0,2,\dots ,2r$. On the open cell its affine coordinates are $w\mathrm{/}\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}$. This radial map preserves the ordered real orientation: its derivative has positive tangential eigenvalue $(1-\mathrm{\parallel }w{\mathrm{\parallel }}^2{)}^{-1\mathrm{/}2}$ and positive radial eigenvalue $(1-\mathrm{\parallel }w{\mathrm{\parallel }}^2{)}^{-3\mathrm{/}2}$, including the identity derivative at zero. Orient each characteristic disk accordingly.

For $i,j\ge 1$, $i+j=n$, let $E=P^i$ use coordinates $z_0,\dots ,z_i$ and $F=P^j$ use $z_i,\dots ,z_n$; their intersection is $p=[e_i]$. Set $V=P^n\setminus F$, $W=P^n\setminus E$. Scaling coordinates $z_i,\dots ,z_n$ by $t$, with $t$ decreasing from $1$ to $0$, retracts $V$ onto the coordinate $P^{i-1}$ using $z_0,\dots ,z_{i-1}$. It also retracts $E\setminus p$ onto that subspace. Throughout, the first $i$ coordinates are not all zero, so the formula is defined, commutes with complex scaling, and fixes the retract. The affine charts of step 1.1 verify joint continuity. Interchanging the two coordinate blocks gives the corresponding retractions for $W$ and $F\setminus p$. Scaling only $z_i$ to zero retracts $P^n\setminus p$ onto its coordinate hyperplane $P^{n-1}$. The latter formula is defined because a point other than $p$ has a nonzero coordinate other than $z_i$.

No two occupied cellular dimensions are adjacent, so every differential is zero, its source or target being zero. By [F1], $H_{2k}(P^n)=\mathbb{Z}$ for $0\le k\le n$, every other group is zero, and the generator is the image of the positive top cell of the standard $P^k$. Standard inclusions preserve these generators, because their maps on those relative characteristic disks are identities. All groups are free, so every Ext term in [F2] vanishes: use the identity augmentation as a length-zero free resolution for $\mathbb{Z}$, and the zero resolution for zero. Evaluation therefore gives $H^{2k}(P^n)=\mathbb{Z}{a}_{n,k}$, with $a_{n,k}$ evaluating to $+1$ on that generator, and zero other degrees. Restrictions preserve $a_{n,k}$ whenever the target dimension is at least $k$. These are actual singular cohomology classes, obtained by evaluation, not cellular cochains substituted into a singular product.

A coordinate permutation on $P^r$ acts as the identity in cohomology. To prove this, realize an adjacent interchange in two coordinates by first using the real rotation matrix with columns $(\cos t,\sin t)$ and $(-\sin t,\cos t)$ for $0\le t\le \pi \mathrm{/}2$, then multiplying the one column with the extra minus sign by a phase varying from $1$ to $-1$. These are complex invertible matrices and give a continuous path from the identity to the interchange. Finite compositions handle every permutation; projectivizing the path gives a homotopy, so [F4] applies. If a coordinate $P^k$ is placed in $P^n$ in any chosen coordinate order, an ambient permutation takes that inclusion to the standard one. Consequently its restriction also sends $a_{n,k}$ to the normalized top generator of the ordered $P^k$. A permutation of complex coordinates preserves their real orientation: each interchange switches two blocks of length two and has real determinant $+1$. The rotations and phase multiplications above likewise have positive real determinant, the latter being $\mathrm{\mid }\lambda {\mathrm{\mid }}^2=1$ on its block.

First consider the top class of $P^r$ at the point $[e_r]$ of its standard open top cell. The map $H^{2r}(P^r,P^{r-1})\to H^{2r}(P^r)$ is an isomorphism by [F3] and step 3.1. Its characteristic-disk pullback evaluates to $+1$ on the positive disk by [F1], [F2] and the definition of $a_{r,r}$.

By [F4], the first retract in step 2.2 gives $H^{2i-1}(V)=H^{2i}(V)=0$, using step 3.1 and step 4.1. The pair sequence [F3] therefore makes $H^{2i}(P^n,V)\to H^{2i}(P^n)$ an isomorphism. The same is true for $H^{2i}(E,E\setminus p)\to H^{2i}(E)$. Their natural square and the absolute restriction isomorphism from step 4.1 imply that $H^{2i}(P^n,V)\to H^{2i}(E,E\setminus p)$ is an isomorphism. The symmetric conclusions hold in degree $2j$ for $F,W$. Finally the punctured-space retract in step 2.2 gives $H^{2n-1}(P^n\setminus p)=H^{2n}(P^n\setminus p)=0$, so $H^{2n}(P^n,P^n\setminus p)\to H^{2n}(P^n)$ is an isomorphism as well. All the odd-degree vanishings used here hold for $i=1$ or $j=1$, where the retract is a point.

Shrinking to a centered smaller disk in its interior retains that positive relative generator: the radial annulus retracts to its boundary, and excision [F5] identifies the resulting punctured-disk groups; the positive radial parameter has positive scaling. The affine map in step 2.1 is radial with positive scale and takes the center to zero, so the corresponding local class is exactly the cube-normalized positive generator used in [F6]. This proves positivity at $[e_r]$.

Identify $U_i$ with ${\mathbb{C}}^i\times {\mathbb{C}}^j$ by its ordered ratios. The intersections $E\cap U_i,F\cap U_i$ are its coordinate planes, and $V\cap U_i=({\mathbb{C}}^i\setminus 0)\times {\mathbb{C}}^j$, $W\cap U_i={\mathbb{C}}^i\times ({\mathbb{C}}^j\setminus 0)$. Excision [F5] makes $H^{2i}(E,E\setminus p)\to H^{2i}({\mathbb{C}}^i,{\mathbb{C}}^i\setminus 0)$ an isomorphism: the removed hyperplane is closed and avoids $p$, hence is contained in the open punctured space. Restricting $(U_i,V\cap U_i)$ to the first coordinate plane also induces an isomorphism. Indeed contraction of the unused coordinate gives homotopy equivalences on ambient spaces and subspaces. For a nonempty contractible ambient space and nonempty relative subspace, [F3] identifies relative degree zero with zero, degree one with the subspace's $H^0$ modulo constants, and degree $k\ge 2$ with subspace $H^{k-1}$. By [F4] and naturality these identifications prove the asserted relative isomorphism. In the square with the map proved in step 5.1, these two isomorphisms force $H^{2i}(P^n,V)\to H^{2i}(U_i,V\cap U_i)$ to be an isomorphism too. Repeat with $j$. Finally excision of the closed hyperplane $P^n\setminus U_i$ gives the isomorphism from $H^{2n}(P^n,P^n\setminus p)$ to $H^{2n}(U_i,U_i\setminus p)$.

Move any coordinate point $[e_l]$ to $[e_r]$ by a coordinate permutation. Its global pullback fixes $a_{r,r}$ by step 4.1. On the local ratio coordinates it merely permutes the remaining complex coordinates, which preserves their real orientation by step 4.1. Naturality of the pair maps therefore proves the same positivity at $[e_l]$. Apply this to $E,F$ at $p$ and to $P^n$ at $p$. In all three cases the ordered complex coordinates induce exactly the ordered real orientations used in [F6]; swapping complex blocks introduces sign $(-1{)}^{(2i)(2j)}=+1$.

Lift $a_{n,i}$ and $a_{n,j}$ uniquely through the two relative-to-absolute isomorphisms of step 5.1. By step 6.1 their local restrictions are coordinate relative generators, and step 6.2 makes them positive. The local cup product is the positive top generator by [F6], with real factor dimensions $2i,2j$. The complements $V,W$ are open, their union is $P^n\setminus p$, and their local intersections are open. Thus [F7] makes both restriction of this relative product and its passage to the absolute product commute. The top comparison isomorphisms in step 5.1 and step 6.1, with positivity from step 6.2, give $a_{n,i}{a}_{n,j}=a_{n,n}\mathrm{.}$ In particular this is a primitive generator, not merely a nonzero integer multiple.

For $n=0$ there is only $H^0=\mathbb{Z}$, and $u=0$. For $n=1$, choose $u=a_{1,1}$; its square is zero by step 3.1, so the ring is $\mathbb{Z}[u]\mathrm{/}(u^2)$. Inductively for $n\ge 2$ set $u=a_{n,1}$. Restriction to $P^{n-1}$ is an isomorphism through degree $2n-2$ and preserves the normalized classes by step 3.1. Naturality in [F7] and the induction hypothesis show $u^k=a_{n,k}$ for $k<n$. Step 7.1 with $i=n-1,j=1$ gives $u^n=a_{n,n}$. Higher powers vanish by step 3.1. The polynomial evaluation map is therefore onto, and its kernel is exactly $(u^{n+1})$: each degree up to $2n$ has the independent infinite-order generator $u^k$, so all its coefficients must vanish for an evaluated polynomial to be zero. The constant class is the unit in [F7]. Standard restrictions preserve $u$ for positive-dimensional targets by its normalization and the degree-two restriction isomorphism, and send it to zero for the point target. They preserve every power and its positive evaluation.

The case $k=0$ evaluates the constant unit as $+1$ on the positive point. The cases $n=0,1$ and identity or point restrictions were checked in step 8.1; no $P^{-1}$ is used. There are no empty projective spaces under the stated $n\ge 0$ hypothesis. Zero inputs and all products above dimension vanish, and singular degeneracies remain included by the actual relative cochain suppliers. Step 2.2 verifies the deformation endpoints and nonzero vector domains; step 6.2 fixes the orientation signs rather than suppressing an integer unit ambiguity. AC is precisely [F8]'s inherited UCT projections and relative additive splittings, with no choice needed for the finite coordinate constructions.

The two endpoints of $\sigma$ equal $x=(1,0)$, so $\mathrm{\partial }z=x-x=0$. The normalization says $a(z)=1$. This does not depend on the representative: replacing $a$ by $a+\delta u$ changes the evaluation by $u(\mathrm{\partial }z)=0$.

The front face for degree one is all of $\sigma$, and its back face is its last vertex $x$. Hence $a\cap z=a(\sigma )x=x\mathrm{.}$ By [F2] this identity passes to $\alpha \cap [z]=[x]$.

Every point of $S^1$ can be joined to $x$ by an arc $t\mapsto (\cos (t\theta ),\sin (t\theta ))$, using any angle $\theta$ for that particular point. Thus there is one path component, and [F3] sends $[x]$ to $1$, not $-1$. This is the asserted positive generator. The coincident endpoints cause no cancellation of the retained vertex: only the boundary subtracts them. The input and coefficient ring are fixed and nonzero; neither an empty space nor a constant loop can satisfy this evaluation normalization. No simultaneous choice of angles or representatives is required, and no AC is used.

The wedge is the finite CW complex obtained by attaching one disk in each of dimensions two, four and six to a single vertex by the constant boundary maps. This is exactly the stated wedge quotient: both quotients identify each disk boundary and all resulting basepoints, and leave each disk interior unchanged. The finite quotient topologies agree by this description. Its cellular groups are $\mathbb{Z}$ in degrees $0,2,4,6$, zero elsewhere; every boundary is zero because one of its two adjacent chain groups is zero. By [F2], these are also its singular homology groups. The inclusion of each sphere summand is cellular and sends its sole positive-dimensional characteristic disk to the identically parameterized disk in $Y$. Thus it induces the identity generator map in that dimension and zero into the other positive-dimensional homology groups.

These homology groups, and those of each sphere computed from its same two-cell complex, are free. Therefore every Ext term of [F3] is zero, using the length-zero identity free resolution for $\mathbb{Z}$ and the zero resolution for zero. Evaluation identifies cohomology with the integral dual of homology in every degree. Hence $Y$ has the additive groups asserted. Moreover for each $r>0$ the restriction map $H^r(Y;\mathbb{Z})⟶H^r(S^2;\mathbb{Z})\oplus H^r(S^4;\mathbb{Z})\oplus H^r(S^6;\mathbb{Z})$ is an isomorphism: for $r=2,4,6$ its sole nonzero coordinate is the dual of the identity generator map from step 1.1, and in every other degree both sides are zero. This is natural singular evaluation, so these are the actual restriction maps. In degree zero restriction is the diagonal $\mathbb{Z}\to {\mathbb{Z}}^3$, not an isomorphism; we do not use it as one.

Let $a\in H^p(Y)$ and $b\in H^q(Y)$ with $p,q>0$. On any sphere summand $S^d$, a positive-degree class can be nonzero only in degree $d$. If either $p$ or $q$ differs from $d$, one restriction is zero. If both equal $d$, their product lies in degree $2d>d$ and is zero. In every case [F4] gives $(ab){\mathrm{\mid }}_{S^d}=a{\mathrm{\mid }}_{S^d}b{\mathrm{\mid }}_{S^d}=0$. The jointly injective restrictions of step 2.1 in degree $p+q>0$ imply $ab=0$. Bilinearity handles finite sums of positive-degree homogeneous classes, proving the asserted vanishing for the whole positive-degree ideal.

By [F1], the classes $1,u,u^2,u^3$ are infinite-order generators of $H^{\ast }(X)$ in degrees $0,2,4,6$, respectively, and $u^2\mathrm{\ne }0$. Pairing these bases with the corresponding degree bases from step 2.1 gives the claimed additive isomorphisms. If a graded ring isomorphism $f:H^{\ast }(X)\to H^{\ast }(Y)$ existed, it would send $u$ to a degree-two class. Step 3.1 gives $f(u{)}^2=0$, while multiplicativity gives $f(u^2)=f(u{)}^2$. Injectivity would force $u^2=0$, a contradiction. Thus no graded ring isomorphism exists.

The constant unit survives in both rings, so the vanishing assertion is explicitly restricted to two positive-degree inputs. Zero inputs, repeated positive-degree inputs and degrees above six are all covered by step 3.1 and the computed groups. Neither space is empty or a point; the single common vertex is only its zero-skeleton and contributes one copy of $\mathbb{Z}$. The wedge basepoint identifications were built into its characteristic maps in step 1.1, without treating singular degeneracies as zero. The assumed AC is used only through [F1] and [F3], as their statements record; the finite CW maps and product restrictions require no additional choice.

Both operations are bilinear and commutative. The first is associative coordinatewise and has unit $(1,1)$. For the second, if $z=(e,f)$, both $(x\ast y)\ast z$ and $x\ast (y\ast z)$ equal $(ace,acf+ade+bce)$; its unit is $(1,0)$. Thus these define commutative unital rings on the same additive group. They are respectively $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}[ϵ]\mathrm{/}({ϵ}^2)$, with $(a,b)$ corresponding to $a+bϵ$ in the latter.

More strongly, let $\mu :G\times G\to G$ be bilinear and invariant under every additive automorphism, meaning $f(\mu (x,y))=\mu (f(x),f(y))$. With $f=-\mathrm{i}\mathrm{d}$, bilinearity gives $-\mu (x,y)=\mu (-x,-y)=\mu (x,y)$. Thus $2\mu (x,y)=0$, and torsion-freeness of ${\mathbb{Z}}^2$ forces $\mu =0$. A zero multiplication cannot have a unit on a nonzero group, since $\mu (u,(1,0))=0\mathrm{\ne }(1,0)$. Consequently no unital multiplication can be recovered in a manner invariant under all additive automorphisms. The zero bilinear pairing is indeed canonical; the refuted claim concerns a unital multiplication, not the existence of any pairing.