Compute \(H_*(\RP^2 \cross \RP^2; \ZZ_2)\)
Solution
1. With coefficients in \(\mathbb F_2=\mathbb Z_2\),\[ H_i(\mathbb{RP}^2;\mathbb F_2)\cong \begin{cases} \mathbb F_2,&i=0,1,2,\\ 0,&\text{otherwise}. \end{cases} \]
The standard CW structure on \(\mathbb{RP}^2\) has one cell in each dimension \(0,1,2\). Its integral cellular boundary maps are respectively \(0\) and multiplication by \(2\); after tensoring with \(\mathbb F_2\), both boundary maps vanish.
2. Over the field \(\mathbb F_2\), the Künneth theorem gives\[ H_n(\mathbb{RP}^2\times\mathbb{RP}^2;\mathbb F_2) \cong\bigoplus_{i+j=n}H_i(\mathbb{RP}^2;\mathbb F_2)\otimes H_j(\mathbb{RP}^2;\mathbb F_2). \]
Over a field there are no Tor terms in the homological Künneth theorem.
3. Therefore\[ \boxed{ H_n(\mathbb{RP}^2\times\mathbb{RP}^2;\mathbb Z_2) \cong \begin{cases} \mathbb Z_2,&n=0,4,\\ (\mathbb Z_2)^2,&n=1,3,\\ (\mathbb Z_2)^3,&n=2,\\ 0,&\text{otherwise}. \end{cases}} \]
Count pairs \((i,j)\in\{0,1,2\}^2\) with \(i+j=n\). Their numbers are \(1,2,3,2,1\) for \(n=0,1,2,3,4\) respectively, and each tensor factor is one-dimensional over \(\mathbb F_2\).