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.