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Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.