Mod-two cohomology rings of complex projective spaces

The standard filtration gives one oriented cell in each even dimension and no odd cells. [given, construct] Write $P^r={\mathbb{C}\mathbb{P}}^r$ as nonzero vectors in ${\mathbb{C}}^{r+1}$ modulo nonzero complex scaling. Attach a real $2r$-disk to $P^{r-1}$ by $w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\parallel w{\parallel }^2}]\mathrm{.}$ The boundary lands in $P^{r-1}$, while each line outside $P^{r-1}$ has a unique unit representative whose last coordinate is positive real, so the open disk maps homeomorphically to the complement. The attachment quotient is compact. Projective space is Hausdorff because a unit vector $z$ maps to the rank-one matrix $z{z}^{\ast }$, whose fibres are precisely scalar-phase orbits; the induced continuous bijection from the compact phase quotient to its matrix image is a homeomorphism. Hence the attachment map from its compact quotient to $P^r$ is a homeomorphism. Starting from $P^0=\ast$ gives compatible cells in dimensions $0,2,\dots ,2r$. Orient the $2r$-cell by the ordered real and imaginary coordinates of ${\mathbb{C}}^r$.

Integral homology is one copy of $\mathbb{Z}$ in each occupied even degree, naturally under skeletal inclusions. [F1, step 1.1] There are no cells in adjacent dimensions, so [F1] makes every cellular differential zero and compares the resulting groups with singular homology. The positive $2k$-cell in the standard $P^k$ gives the generator of $H_{2k}(P^n;\mathbb{Z})$ for $k\le n$. A standard inclusion is the identity on each cell it contains, so [F1] makes its homology map the identity on these generators. The union $P^{\mathrm{\infty }}$ has the same calculation in every fixed degree: one copy of $\mathbb{Z}$ in nonnegative even degrees and zero in odd degrees.

Integral and mod-two cohomology are additively determined, with natural skeletal restrictions. [F2, A1, step 2.1] In [F2], all Ext terms vanish because the preceding integral homology group in each even degree is zero and the preceding group in each odd degree is free. Evaluation therefore gives $H^{2k}(P^n;\mathbb{Z})=\mathbb{Z}$ and $H^{2k}(P^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le k\le n$, with all odd groups zero; the same holds in every degree for $P^{\mathrm{\infty }}$. Naturality and step 2.1 make restriction to $P^n$ an isomorphism through degree $2n$. Let $u_n\in H^2(P^n;\mathbb{Z})$ be the class evaluating as $+1$ on the positive $P^1$ cell when $n\ge 1$, and put $u_0=0$. Restrictions preserve these normalized classes.

Reduction modulo two of the normalized integral class is the unique finite degree-two generator. [F2, F7, step 3.1] For $n\ge 1$, choose an integral cocycle representing $u_n$ and reduce its values modulo two. By [F7], this commutes with coboundary and is independent of the cocycle representative; it defines $c_n$. Coefficient naturality of evaluation in [F2] makes $c_n$ evaluate as $1$ on the mod-two reduction of the positive $P^1$ cell, so it is nonzero and hence is the unique class in degree two. The front/back formula in [F7] shows that reduction commutes with cup products. Put $c_0=0$. Restriction preserves every $c_n$ because it preserves $u_n$ and commutes with coefficient reduction.

Complementary coordinate projective subspaces admit the required explicit retractions. [F4, step 4.1] Fix $i,j\ge 1$ with $i+j=n$. Let $E=P^i$ use coordinates $z_0,\dots ,z_i$, let $F=P^j$ use $z_i,\dots ,z_n$, and let $p=E\cap F=[e_i]$. Put $V=P^n\setminus F$ and $W=P^n\setminus E$. Scaling $z_i,\dots ,z_n$ to zero retracts $V$ and $E\setminus \{p\}$ onto the same coordinate $P^{i-1}$; the first $i$ coordinates cannot all vanish on either domain. Symmetrically, $W$ and $F\setminus \{p\}$ retract onto $P^{j-1}$. Scaling only $z_i$ to zero retracts $P^n\setminus \{p\}$ onto a coordinate $P^{n-1}$. Each formula commutes with complex scaling, never produces the zero vector on its stated domain, fixes the target, and is continuous in affine coordinates, so [F4] applies.

The complementary classes lift uniquely to relative generators and the top local-to-global map is an isomorphism. [F3, F4, step 3.1, step 5.1] Step 5.1 and step 3.1 give $H^{2i-1}(V;{\mathbb{F}}_2)=H^{2i}(V;{\mathbb{F}}_2)=0$. Exactness in [F3] therefore makes $H^{2i}(P^n,V)\to H^{2i}(P^n)$ an isomorphism. The analogous map for $(E,E\setminus \{p\})$ is an isomorphism, and the natural pair square together with the absolute restriction isomorphism of step 3.1 identifies these two relative groups. The same holds for $F,W$ in degree $2j$. Finally the punctured-space retraction gives vanishing in degrees $2n-1$ and $2n$, so $H^{2n}(P^n,P^n\setminus \{p\})\to H^{2n}(P^n)$ is an isomorphism.

The product of the unique classes in complementary positive even degrees is the nonzero top class. [F3, F4, F5, F6, A1, step 6.1] In the affine chart $z_i\mathrm{\ne }0$, ordered ratios identify a neighborhood of $p$ with ${\mathbb{C}}^i\times {\mathbb{C}}^j$. They identify $E,F$ with the two coordinate planes, $V$ with $({\mathbb{C}}^i\setminus 0)\times {\mathbb{C}}^j$, and $W$ with ${\mathbb{C}}^i\times ({\mathbb{C}}^j\setminus 0)$. Excision in [F4], followed by contraction of the unused coordinate, identifies the nonzero relative classes from step 6.1 with the coefficient-one local generators. Their relative product is nonzero by [F5]. The open complements satisfy $V\cup W=P^n\setminus \{p\}$, so [F6] transports the product to the top relative group and then through step 6.1 to a nonzero absolute product. Since that top group is one-dimensional by step 3.1, this is its unique nonzero class.

Every finite projective space has the asserted truncated polynomial ring. [F6, step 3.1, step 4.1, step 7.1] For $n=0$, only the unit remains and $c_0=0$. For $n=1$, $c_1$ is nonzero and $c_1^2=0$ above dimension two. Inductively suppose the claim holds for $P^{n-1}$. Step 4.1 and [F6] make $c_n^k$ restrict to the nonzero $c_{n-1}^k$ for $k<n$. Step 7.1 with $i=n-1,j=1$ then makes $c_n^n=c_n^{n-1}{c}_n$ the nonzero top class. All higher powers vanish above dimension $2n$. With the additive calculation of step 3.1, polynomial evaluation is onto and its kernel is exactly $(c_n^{n+1})$.

Finite-skeleton detection gives the infinite polynomial ring. [F2, F6, A1, step 2.1, step 3.1, step 8.1] Let $c$ be the unique nonzero element of $H^2(P^{\mathrm{\infty }};{\mathbb{F}}_2)$. Restriction to every $P^n$ with $n\ge 1$ is an isomorphism in degree two, so it sends $c$ to $c_n$. By [F6], $c^k$ restricts to $c_n^k$, which is nonzero when $n\ge k$ by step 8.1. Hence $c^k$ is the unique nonzero class in degree $2k$ from step 3.1. Polynomial evaluation is onto degreewise and injective because a polynomial has finitely many homogeneous terms of distinct degrees.

Every boundary, degeneracy, and choice case is explicit. [F1, F2, F3, F4, F5, F6, F7, A1, step 1.1, step 2.1, step 3.1, step 4.1, step 5.1, step 6.1, step 7.1, step 8.1, step 9.1] The cases $n=0,1$, the unit power $k=0$, the top power $k=n$, and the first vanishing power $k=n+1$ were separated. There is no top degree for $P^{\mathrm{\infty }}$, but every fixed power is detected on a finite skeleton. The spaces are nonempty; zero classes and all odd groups are zero. The local product uses $i,j\ge 1$, so no zero-dimensional factor is smuggled into [F5]. The homotopies in step 5.1 specify both endpoints and their nonzero domains. The singular cochain definitions in [F7] retain degenerate simplices. Coefficient reduction is a specified map and introduces no choice. AC is used exactly through UCT [F2] and the local product [F5]; the finite coordinate constructions add none. No biconditional or converse is asserted. ∎