= Solution
\b[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 $(x_1,\ldots,x_k,1)$, 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 map
$$
S^{k-1}\longrightarrow \mathbb{RP}^{k-1},\qquad x\longmapsto[x],
$$
which 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 $0\leq k\leq n$ 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,
$$
\boxed{d_k=1+(-1)^k=
\begin{cases}2,&k\text{ even},\\0,&k\text{ odd}.\end{cases}}
$$
For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.
\b[The mod-two cup products.] Modulo two every cellular differential vanishes, so <cellular cohomology> gives a one-dimensional group in each degree $0,\ldots,n$. 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 $a^j$, $0\leq j\leq n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne0$. Dimension makes $a^{n+1}=0$. The <mod-two cohomology ring of real projective space> is
$$
\boxed{H^*(\mathbb{RP}^n;\mathbb F_2)
\cong\mathbb F_2[a]/(a^{n+1}),\qquad |a|=1.}
$$
For $n=0$ this simply means the cohomology of a point.
\b[The product with integral coefficients.] The final product's coefficients are unstated; take $\mathbb Z$ as the default. Dualizing the <cellular chain complex> above gives
$$
\begin{aligned}
&H^*(\mathbb{RP}^2;\mathbb Z):\quad H^0=\mathbb Z,\ H^2=\mathbb Z/2,\\
&H^*(\mathbb{RP}^3;\mathbb Z):\quad H^0=\mathbb Z,\ H^2=\mathbb Z/2,\ H^3=\mathbb Z,\\
&H^*(\mathbb{RP}^4;\mathbb Z):\quad H^0=\mathbb Z,\ H^2=H^4=\mathbb Z/2,
\end{aligned}
$$
with all unlisted groups zero. The integral <Künneth theorem> has tensor terms with $i+j=q$ and <Tor functor> terms with $i+j=q+1$:
$$
0\to\bigoplus_{i+j=q}H^i(A;\mathbb Z)\otimes H^j(B;\mathbb Z)
\to H^q(A\times B;\mathbb Z)
\to\bigoplus_{i+j=q+1}\operatorname{Tor}^{\mathbb Z}_1(H^i(A;\mathbb Z),H^j(B;\mathbb Z))
\to0.
$$
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 is
$$
F_{A\times B}=F_AF_B,\qquad
T_{A\times B}=F_AT_B+T_AF_B+(1+t^{-1})T_AT_B.
$$
The last two factors record respectively the tensor contribution in summed degree and the <Tor functor> contribution one degree lower. The three factors have
$$
(F_2,T_2)=(1,t^2),\quad
(F_3,T_3)=(1+t^3,t^2),\quad
(F_4,T_4)=(1,t^2+t^4).
$$
The first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third gives
$$
F=1+t^3,\qquad
T=3t^2+3t^3+5t^4+6t^5+5t^6+4t^7+2t^8+t^9.
$$
Consequently the <integral cohomology of a product of finite real projective spaces> in this case is
$$
\boxed{
H^q(\mathbb{RP}^2\times\mathbb{RP}^3\times\mathbb{RP}^4;\mathbb Z)
\cong
\begin{cases}
\mathbb Z,&q=0,\\
0,&q=1,\\
(\mathbb Z/2)^3,&q=2,\\
\mathbb Z\oplus(\mathbb Z/2)^3,&q=3,\\
(\mathbb Z/2)^5,&q=4,\\
(\mathbb Z/2)^6,&q=5,\\
(\mathbb Z/2)^5,&q=6,\\
(\mathbb Z/2)^4,&q=7,\\
(\mathbb Z/2)^2,&q=8,\\
\mathbb Z/2,&q=9,\\
0,&\text{otherwise}.
\end{cases}}
$$
If the intended coefficients were instead $\mathbb F_2$, the <Künneth theorem> over a field gives the dimension polynomial
$$
(1+t+t^2)(1+t+t^2+t^3)(1+t+t^2+t^3+t^4).
$$
Thus the mod-two groups in degrees zero through nine are respectively
$$
\boxed{\bigl(\dim_{\mathbb F_2}H^q\bigr)_{q=0}^{9}=(1,3,6,9,11,11,9,6,3,1),}
$$
and all other degrees vanish.
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