Take $BO=\bigcup_{n\geq 1}BO(n)$ and $BTOP=\bigcup_{n\geq 1}BTOP(n)$ where $TOP(n)$ is the set of homeomorphisms of $\mathbb{R}^n$ which send $0$ to $0$ and let $\phi: BO \rightarrow BTOP$ be the map induced by the inclusion of $O(n) \rightarrow TOP(n)$.
I've already shown that $\phi_{*}: \pi_i(BO) \otimes \mathbb{Q} \rightarrow \pi_i(BTOP) \otimes \mathbb{Q} $ is injective for $i\geq 0$ and I want to show that this implies $\phi^*: H^i(BTOP;\mathbb{Q})\rightarrow H^i(BO;\mathbb{Q})$ surjective for $i\geq 0$.
I first tried to use the rational Huerwics and the fact that $H^*(BO;\mathbb{Q}) \cong \mathbb{Q}[p_1,p_2,\dots]$ where $p_i$ are the universal Pontryagin classes of the $EO \rightarrow BO$ bundle, but it seems that this is not enough. Then I thought about that $BO$, $BTOP$ are H-spaces and $\phi$ is a H-map and may work with Hopf-algebras, but the cohomology of $BTOP$ is not finally generated, at least I don't see why it should be, so $H^*(BTOP,\mathbb{Q})$ doesn't allow a Hopf-algebra structure.
I would be grateful for any advice or hint.


1 Answer 1


Write $kO$ for the connective $k$-theory, and $X$ for the connective delooping of $BTOP$. Then $H_*(BO;Q)$ and $H_*(BTOP,Q)$ are free commutative (in graded sense) generated by $\pi _*(kO,Q)\cong \pi _*(BO,Q)$ and $\pi _*(X,Q)\cong \pi _* (BTOP,Q)$ respectively. Thus the injection of the homotopy groups imply the injection of the homology. By dualizing ($H_*(BO,Q)$ is surely of finite type, and this suffices) you get the surjection of homotopy groups.


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