Kunneth formula I am trying to understand the proof of the Kunneth formula, as described by Ravi Vakil's notes in 18.2.8 here:
http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf#page=475
I will follow Ravi's notation. Why is that the the tensor product of the Cech complexes of $X$ and $Y$ the same as the Cech complex of $X \times Y$ with respect to the product cover? More specifically, let $\mathcal U =  \{U_i \}_{i \in I}$ and $\mathcal V = \{ V_j \}_{j \in J}$ be open affine covers of $X$ and $Y$. Then $\mathcal U \times_k \mathcal V = \{ U_i \times_k V_j \}$ is an open affine cover of $X \times Y$. The $n$-cochains in Cech complex of $X \times Y$ are:
$$
C^n(\mathcal U \times_k \mathcal V, \mathcal F \boxtimes \mathcal G)
=
\prod_{ (i_0,j_0), \dots, (i_n, j_n) \in (I \times J)^{n+1} }
(\mathcal F \boxtimes \mathcal G) ( (U_{i_0} \times V_{j_0}) \cap \dots \cap (U_{i_n} \times V_{j_n}) )\\
=
\prod_{ (i_0,j_0), \dots, (i_n, j_n) \in (I \times J)^{n+1} }
(\mathcal F \boxtimes \mathcal G) ( U_{i_0 \dots i_n} \times V_{j_0 \dots i_n} )\\
=
\prod_{ (i_0,j_0), \dots, (i_n, j_n) \in (I \times J)^{n+1} }
\mathcal F(U_{i_0 \dots i_n}) \otimes_k \mathcal G(V_{j_0 \dots j_n})
$$
This looks to me like $ C^n(\mathcal U, \mathcal F) \otimes_k C^n(\mathcal V, \mathcal G)$. I don't see how this is supposed to be the degree $n$ part of $C^\bullet(\mathcal U, \mathcal F) \otimes C^\bullet(\mathcal V, \mathcal G)$, which is
$$
\bigoplus_{p+q=n} C^p(\mathcal U, \mathcal F) \otimes C^q(\mathcal V, \mathcal G).
$$
Thanks for you help.
 A: Let $X=Y=\Bbb P^1$ with the standard open covers $U_0,U_1$ on $X$ and $V_0,V_1$ on $Y$. Then our open cover for $X\times Y$ is $W_{ij} = U_i\times V_j$ for $i,j \in \{0,1\}$. Let's write down exactly what the Cech complex is:
Degree 0 single intersections: $W_{00},W_{01},W_{10},W_{11}$
Degree 1 double intersections is as follows: $W_{00}\cap W_{01}=U_0\times (V_0\cap V_1), W_{00}\cap W_{10}=(U_0\cap U_1) \times V_0, W_{00}\cap W_{11} = (U_0\cap U_1)\times(V_0\cap V_1), W_{01}\cap W_{10} = (U_0\cap U_1)\times (V_0\cap V_1), W_{01}\cap W_{11}= (U_0\cap U_1)\times V_1, W_{10}\cap W_{11} = U_1\times (V_0\cap V_1)$
Degree 2 triple intersections: $W_{00}\cap W_{01} \cap W_{10} = (U_0\cap U_1)\times (V_0\cap V_1), W_{00}\cap W_{01}\cap W_{11}=(U_0\cap U_1)\times(V_0\cap V_1), W_{00}\cap W_{10}\cap W_{11} = (U_0\cap U_1)\times (V_0\cap V_1), W_{01}\cap W_{10}\cap W_{11} = (U_0\cap U_1)\times(V_0\cap V_1)$
Degree 3 quadruple intersection: $W_{00}\cap W_{01}\cap W_{10}\cap W_{11} = (U_0\cap U_1)\times (V_0\cap V_1)$
By writing out the maps, we can see that some of the identical intersections don't contribute to the cohomology of this complex - that is, the complex is quasi-isomorphic to the complex given with the following intersections:
Degree 0 single intersections: $W_{00},W_{01},W_{10},W_{11}$
Degree 1 double intersections is as follows: $U_0\times (V_0\cap V_1), (U_0\cap U_1) \times V_0, (U_0\cap U_1)\times V_1, U_1\times (V_0\cap V_1)$
Degree 2 triple intersections: one copy of $(U_0\cap U_1)\times(V_0\cap V_1)$
Degree 3 quadruple intersection: empty
The Cech complex of the standard cover of $\Bbb P^1$ is given as follows:
Degree 0: $U_0,U_1$
Degree 1: $U_0\cap U_1$
Writing out the tensor product of two of these complexes, we see it has the exact form as the complex we found above, and they are indeed quasi-isomorphic. The key point here is that you can cancel away the intersections that appear in places "they shouldn't" to get exactly what you want.
