False proof that $\chi(A\times B)=\chi(A)+\chi(B)$

There is a well known formula for the Euler characteristic which says that if $A$ and $B$ are two topological spaces, $\chi(A\times B)=\chi(A)\cdot\chi(B)$. I know very little about algebraic topology, but I'm know some homological algebra, so I wanted to try to come up with some proof of this formula using the cohomology of derived functors. However, after some fiddling around with diagrams, I instead came up with the following apparent proof that $\chi(A\times B)=\chi(A)+\chi(B)$:

Let $A$ and $B$ be topological spaces and $F:Top\to Ab$ a left exact functor. Consider the long exact sequence induced in cohomology by the short exact sequence $$0\to A\to A\times B\to B\to 0$$ That is, $$0\to FA\to F(A\times B)\to FB\to R^1FA\to R^1F(A\times B)\to R^1FB\to R^2FA\to ...$$ Let $K^i$ represent the $i$th object in this sequence such that $K^1=FA$, and $d^i:K^{i}\to K^{i+1}$ the respective maps. At each place in this sequence, there is an associated short exact sequence $$0\to\ker d^{i-1}\to K^i\to \ker d^{i}\to 0.$$ By the rank theorem, we have $\text{r} K^i=\text{r} \ker d^{i-1}+\text{r} \ker d^i$, where $\text{r}$ is the rank operator shortened for ease of notation. Consider the rank of $F(A\times B)$. This is \begin{align} \text{r} F(A\times B)&=\text{r} \ker d^1+\text{r} \ker d^2\\ &=\text{r} FA+(\text{r} K^3 - \text{r}\ker d^3)\\ &=\text{r} FA+\text{r} FB - \text{r} R^1FA+\text{r} R^1F(A\times B)-\text{r} R^1FB+\text{r} R^2FA-\text{r} R^2F(A\times B)+... \end{align} One easily sees that by rearranging and moving the $R^kF(A\times B)$ terms to the left side, we get the equation $$\sum_{k=0}^{\infty} (-1)^j\text{r} R^kF(A\times B)=\sum_{k=0}^{\infty} (-1)^j\text{r} R^kFA+\sum_{k=0}^{\infty} (-1)^j\text{r} R^kFB.$$ But by definition these are simply the Euler characteristics of $A$, $B$, and $A\times B$ when the derived functors $R^kF$ are treated as cohomology functors $H^k$. Thus we arrive at the formula $\chi(A\times B)=\chi(A)+\chi(B)$.

Am I misunderstanding something here? My first thought was perhaps that the Euler characteristic product formula does not extend to derived cohomology, but apparently it is possible to formulate singular cohomology as a derived functor, and this proof would seem to apply to that setting as well, implying that the product formula does not hold even for singular cohomology, which is false. Thus I can only assume there is some issue with the proof, but I am not sure what the issue is. Any help would be greatly appreciated.

• You cannot derive a functor that has domain in $\mathsf {Top}$, you need an abelian category. The fact that Euler characteristic works multiplicatively is deduced from the EZ theorem for the cohomology of a product of spaces and the usual Künneth formulas for complexes. Over a field, we have $H(C\otimes C') =H(C)\otimes H(C')$, and the Euler characteristic of a tensor product of complexes is the product of the original Euler characteristics. This in turn follows from the Cauchy formulas for the product of two series, which is built up in the definition of the tensor product of two complexes. – Pedro Tamaroff Oct 8 '16 at 1:44
• Your proof rather shows that Euler characteristic relative to a fixed functor $F: A\to B$ between abelian categories and a rank function in $B$, where you can derive $F$ is additive respect to the direct sum of objects (under certain hypothesis). – Pedro Tamaroff Oct 8 '16 at 1:57