# Computing degree of attaching maps in $RP^2\times RP^2$

As a simple application of the Kunneth exact sequence $$0\to\oplus_{p=0}^n H_pX\otimes H_{n-p}Y\to H_n(X\times Y)\to\oplus_{p=0}^{n-1} \operatorname{Tor}(H_pX,H_{n-1-p}Y)\to0$$ one computes the cellular homology of $X=\mathbb{R}P^2\times \mathbb{R}P^2$ over $\mathbb{Z}$ from the homology of $\mathbb{R}P^2,H_0=\mathbb{Z},H_1=\mathbb{Z}_2,H_2=0$ as $$H_0(X)=\mathbb{Z},H_1(X)=\mathbb{Z}_2\oplus\mathbb{Z}_2, H_2(X)=\mathbb{Z}_2, H_3(X)=\mathbb{Z}_2,H_4(X)=0$$ The motivating question was this: what's a geometric explanation for the nontrivial third homology, which is the only contribution from the Tor term in the Kunneth sequence?

So, I had a look at the cellular structure of $X$. Call by $c_0,c_1,c_2$ the cells of $\mathbb{R}P^2$ in increasing order of dimension. Then $X$ has one 0-cell $c_0^2,$ two 1-cells $c_0c_1$ and $c_1c_0$, three 2-cells $c_1^2,c_2c_0,c_0c_2,$ two 3-cells $c_2c_1, c_1c_2,$ and one 4-cell $c_2^2$. What I'd like to do is understand how to fill in the degrees of the maps in the cellular chain of $X$: $$\begin{matrix} &&c_2c_1&\stackrel{\left(\begin{matrix}m\\n\\0\end{matrix}\right)}{\to}&c_2c_0& \stackrel{\left(\begin{matrix} 2\\0\end{matrix}\right)}{\to}&c_1c_0&\stackrel{0}{\to}\\ &&&&\oplus&&&&\\ c_2^2&\stackrel{\left(\begin{matrix}t\\u\end{matrix}\right)}{\to}&\oplus&&c_1c_1& \stackrel{\left(\begin{matrix}0\\0\end{matrix}\right)}{\to}&\oplus&&c_0^2\\ &&&&\oplus&&&&\\&&c_1c_2&\stackrel{\left(\begin{matrix}0\\r\\s\end{matrix}\right)}{\to}&c_0c_2& \stackrel{\left(\begin{matrix} 2\\0\end{matrix}\right)}{\to}&c_0c_1&\stackrel{0}{\to} \end{matrix}$$ Here the cells are standing for the basis elements of the free $\mathbb{Z}$-modules forming each dimension of the chain complex (sorry for the grotesque diagram!)

By symmetry, obviously either $t=u$ or $t=-u$. In the same way either $n=r$ or $n=-r$, and it seems we have $m=s$. Nothing gets covered more than twice, so I'm pretty sure all variables in sight are $\pm 2$ and $0$. $m=s=0$ is going to have to hold to get a big enough kernel to realize the $H_3=\mathbb{Z}_2$, although I hate to work backwards that way. Then I'd get what I came for out of $t,u,r,n$ all of absolute value $2$ and either $t=u$ and $n=-r$ or $t=-u$ and $n=r$.

So, how does one see in dimension higher than $1$ what these degrees ought to be? Thanks!

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