Homotopy cardinality of fiber bundles

Question

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 don’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.

Answer 1

First, it should be assumed that $B$ is connected, in order to make $F$ sufficiently well-defined. This is actually not too crucial since we can treat a fiber bundle over a disconnected base as a collection of bundles over each of its connected components, and apply this result individually to each.

Second, we use the definition of homotopy cardinality given by Baez-Dolan: $$\chi(X) = \sum_{[x] \in \pi_0 X} \prod_{i \geq 1} \left\lvert \pi_i(X, x) \right\rvert^{(-1)^i}.$$ Notice that your formula is slightly different.

Then, with this definition in hand, we have $\chi(E) = \chi(F) \chi(B)$ for any fiber bundle (in fact, any fibration) $F \to E \to B$. This result is also stated in Baez-Dolan, but let us flesh out the proof.

For now, suppose $E$ is furthermore connected. We suppress writing out the basepoint. Your argument with the homotopy long exact sequence shows that $$\begin{align} \chi(E) &= \prod_{i \geq 1} \left\lvert \pi_i(E) \right\rvert^{(-1)^i} \\ &= \left\lvert \pi_0(F) \right\rvert \prod_{i \geq 1} (\left\lvert \pi_i(F) \right\rvert^{(-1)^i} \left\lvert \pi_i(B) \right\rvert^{(-1)^i}) \\ &= \left\lvert \pi_0(F) \right\rvert \prod_{i \geq 1} \left\lvert \pi_i(F) \right\rvert^{(-1)^i} \cdot \underbrace{\prod_{i \geq 1} \left\lvert \pi_i(B) \right\rvert^{(-1)^i}}_{\chi(B)}. \end{align}$$

It remains to show that $$\chi(F) := \sum_{[x] \in \pi_0 F} \prod_{i \geq 1} \left\lvert \pi_i(F) \right\rvert^{(-1)^i} = \left\lvert \pi_0(F) \right\rvert \prod_{i \geq 1} \left\lvert \pi_i(F) \right\rvert^{(-1)^i}.$$ For this it suffices to show that each connected component of $F$ is homotopy equivalent to each other. This can be shown by using the fact that the monodromy action of $\pi_1(B)$ on $\pi_0(F)$ is transitive since $E$ is connected, so takes a distinguished connected component of $F$ to every other component homeomorphically. So this proves the multiplicative formula $\chi(E) = \chi(F) \chi(B)$ in the case that $E$ (and $B$) is connected.

For general $E$, we use the fact that $\chi(X_1 \sqcup X_2) = \chi(X_1) + \chi(X_2)$ by construction and induct on $\left\lvert \pi_0 E \right\rvert$. Suppose $E = E_1 \sqcup E_2$ is a disconnection of $E$. Then the fiber bundle is separated into two pieces: there exist $F_1$ and $F_2$ such that $F = F_1 \sqcup F_2$ and $F_1 \to E_1 \to B$ and $F_2 \to E_2 \to B$ are fiber bundles. Applying the induction hypothesis to these new bundles, we get: $$\begin{align} \chi(E) &= \chi(E_1) + \chi(E_2) \\ &= (\chi(F_1) \chi(B)) + (\chi(F_2) \chi(B)) \\ &= (\chi(F_1) + \chi(F_2) \chi(B) \\ &= \chi(F) \chi(B). \end{align}$$

This completes the argument.