Mod-two duality for real projective space

The sphere is compact by [F8], so $P$ is compact: pull an open cover back under the continuous surjection $q$ and project a finite subcover. The map $q$ is open because $q^{-1}q(V)=V\cup (-V)$ for open $V\subset S^n$. Distinct orbits $\{x,-x\}$ and $\{y,-y\}$ have positive minimum distance $d$, the minimum of finitely many positive distances. The unions of radius-$d\mathrm{/}3$ balls about the points in the two orbits are disjoint, open and antipodally invariant. Their quotient images are disjoint open neighborhoods. Hence $P$ is Hausdorff. For each $0\le i\le n$, the open subset $U_i=\{[x]:x_i\mathrm{\ne }0\}$ has the affine chart $[x]\mapsto (x_j\mathrm{/}{x}_i{)}_{j\mathrm{\ne }i}$ onto ${\mathbb{R}}^n$. Ratios are continuous and invariant on the sphere preimage, so descend continuously through the open quotient restriction. The inverse sends $u$ to the orbit of $(u_0,\dots ,1,\dots ,u_n)\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, with the $1$ in coordinate $i$. These are continuous inverse maps. For $n=0$ this is the one-point chart and its singleton basis. For $n>0$ the pullbacks of rational balls in these finitely many charts form a countable basis: any open neighborhood intersects one of the covering charts, whose rational-ball basis refines it. Thus [F1] and [F9] make $P$ a boundaryless $n$-manifold.

This manifold is nonempty and connected. For $n=0$ both sphere points form one orbit, so $P$ is a point. For $n\ge 1$, any two sphere points $x,y$ not antipodal are joined by the path $((1-t)x+ty)\mathrm{/}\mathrm{\mid }(1-t)x+ty\mathrm{\mid }$, whose denominator cannot vanish unless $y=-x$. If they are antipodal, choose a unit vector $v$ perpendicular to $x$ and concatenate such paths through $v$. Such a vector exists explicitly: take a standard basis vector not parallel to $x$, subtract its projection on $x$ and normalize; at least one of the $n+1\ge 2$ standard vectors is not parallel. These paths project to paths between all projective points, proving path connectedness and hence connectedness by [F12]. No family of path choices is needed. Apply [F2] to the manifold in step 1.1: at every stalk take its unique nonzero mod-two value, which is continuous in every local trivialization. By [F11], compactness then gives its canonical fundamental class $[P{]}_2$. These steps require no AC.

By [F4], the integral cellular groups have one oriented generator in each degree $0,\dots ,n$, with positive-degree differentials alternating $0,2$. Reduction of chain coefficients modulo two commutes with boundary, relative quotient, and the connecting-map formula taking a relative cycle to its boundary. By [F10] it sends each integral characteristic-disk generator to the coefficient-one generator of the corresponding ${\mathbb{F}}_2$ cellular group. Thus the cellular differentials modulo two are the reductions of $0,2$, both zero. Therefore each cellular homology group in degrees $0,\dots ,n$ is ${\mathbb{F}}_2$, and the groups in all other degrees are zero. By [F5] these are the singular homology groups. In top degree $[P{]}_2$ is their nonzero generator because by step 2.1 its point restriction is nonzero.

Let $n>0$ be even. The top integral cellular differential in [F4] is multiplication by $2:\mathbb{Z}\to \mathbb{Z}$, which has zero kernel, and there is no cell above it. Thus [F5] gives $H_n(P;\mathbb{Z})=0$. If $P$ were integrally orientable, [F7], applied using compactness and connectedness from steps 1.1–2.1, would make its top restriction image the whole local group $\mathbb{Z}$, contradicting its zero domain. This proves integral nonorientability without assuming it from a picture of transition signs. The mod-two orientation and class of step 2.1 still exist. At $n=0$, $P$ is a point, so the multiplication-by-two argument has no positive differential to apply to; it is integrally orientable.

Assume [F6]. The compact boundaryless manifold and orientation in steps 1.1–2.1 satisfy exactly the hypotheses of [F3]. It gives the stated cap isomorphism, since compact support is ordinary cohomology on compact $P$. Step 3.1 then computes its domain as well as its codomain: both are ${\mathbb{F}}_2$ for $0\le p\le n$ and zero otherwise. Over ${\mathbb{F}}_2$ the isomorphism between one-dimensional spaces sends the unique nonzero class to the unique nonzero class, so this is a concrete complementary-degree pairing without any sign choice. In particular $1\cap [P{]}_2=[P{]}_2$ and the nonzero top cohomology class caps to the point generator of $H_0$.

The assertions include $n=1$, with zero cellular differential in degrees zero and one and cap interchanging the two nonzero degrees. Empty spaces and zero coefficient rings are not inputs here; the nonempty quotient and the specific field ${\mathbb{F}}_2$ were fixed. Degrees outside $0,\dots ,n$ are zero on both sides by step 4.1, including negative degrees. Collapsed or repeated boundary points in the attaching maps are included in [F4]'s quotient construction, and singular degeneracies remain in the coefficient comparison of step 3.1. The paths in step 2.1 have their prescribed endpoints, with the non-antipodal denominator check preventing a singularity. Only step 4.1 invokes AC, precisely the countable chart-neighborhood and local UCT free-module/projection/lift choices of [F3]; compactness, canonical mod-two orientation, fundamental-class existence and the integral nonorientability calculation above are choice-free.