Consider the Homotopy cardinality (or $\infty$-groupoid cardinality) $\chi(X):=\sum_{[x]\in\pi_0(X)}\prod_{i\geq0}|\pi_i(X,x)|^{(-1)^{i+1}}$ associated to a space $X$. Suppose we have a fibre bundle $F\to E\stackrel{p}{\to}B$ such that for each $\chi(F),\chi(E),\chi(B)<\infty$. If we make that assumption that each space have just finitely many components with each component having finite homotopy groups that vanish from a certain point, then the cardinalities are certainly finite.

I am trying to calculate $\chi(E)$ in terms of $\chi(B)$ and $\chi(F)$. Is it true that $\chi(E)=\chi(F)\chi(B)$ in these circumstances? I have tried to show it using the long exact sequence of the fibration: For each $[x]\in\pi_0(E)$ the long exact sequence yields $\frac{|\pi_{2i}(E,x)|}{|\pi_{2i-1}(E,x)|}= \frac{|\pi_{2i}(F,x)|| \pi_{2i}(B,p(x)) |}{|\pi_{2i-1}(F,x)|| \pi_{2i-1}(B,p(x)) |}$, which takes a little bit of work decomposing the long exact sequence into short ones, but it is not particularly hard. Hence $\prod_{i\geq1}\frac{|\pi_{2i}(E,x)|}{|\pi_{2i-1}(E,x)|}= \prod_{i\geq1} \frac{|\pi_{2i}(F,x)|| \pi_{2i}(B,p(x)) |}{|\pi_{2i-1}(F,x)|| \pi_{2i-1}(B,p(x)) |}$ and if we sum over $[x]\in\pi_0(E)$ we obtain $$\chi(E)= \sum_{[x]\in\pi_0(E)}\prod_{i\geq1} \frac{|\pi_{2i}(F,x)|| \pi_{2i}(B,p(x)) |}{|\pi_{2i-1}(F,x)|| \pi_{2i-1}(B,p(x)) |} $$ and it appears that we are quite close to finishing an argument. However, I donâ€™t see any slick way to split the rhs into the product $\chi(B)\chi(F)$.

My question is: If it is possible to split the rhs into that product, can someone tell me how to do it?

If it however it s not the case that $\chi(E) = \chi(F)\chi(B)$, can someone tell me how to express $\chi(E)$ in terms of $\chi(F)$ and $\chi(B)$?

Thank you very much.