A question about characteristic classes I have a map $\phi:BO(1)^n\rightarrow BO(n)$ which is given by sending any $n$-tuple in $BO(1)^n$ to an $n$-plane through the origin.  Thus, this induces a group action on the symmetric group $S_n$ such that $BO(1)^n/{S_n}\cong BO(n)$.  There is an induced map on cohomology rings $\phi^*:H^*(BO(n);F_2)\rightarrow H^*(BO(1)^n;F_2)$.  I claim that $\phi^*:H^*(BO(n);F_2)\rightarrow (H^*(BO(1)^n;F_2))^{S_n}$ is an isomorphism, where $(H^*(BO(1)^n;F_2))^{S_n}$ denotes those fixed elements of $H^*(BO(1)^n;F_2)$ by the induced $S_n$ action.  Here $H^*(BO(1)^n;F_2)$ = $F_2[\alpha_1, ..., \alpha_n]$.  Also, $BO(1)\cong RP^\infty$ and $BO(n)\cong G_n$.
Any suggestions?  This is a claim in May A Concise Course and some online notes by him and others found here:  http://www.math.uchicago.edu/~may/CHAR/charclasses.pdf
page 35.
Thanks
 A: It's true that there's a natural map $BO(1)^n \to BO(n)$, which you should think of as taking the direct sum of $n$ real line bundles to get an $n$-dimensional real vector bundle, inducing a quotient map $BO(1)^n / S_n \to BO(n)$, although we need to be a bit careful about what we mean by that quotient: properly speaking it's a homotopy quotient. It is not true that this map is a homotopy equivalence, and May never makes a claim of this form. 
What is true is that the pullback map
$$H^{\bullet}(BO(n), \mathbb{F}_2) \to H^{\bullet}(BO(1)^n, \mathbb{F}_2)^{S_n}$$
is an isomorphism - this is one way to define the Stiefel-Whitney classes, as symmetric functions via the splitting principle - but this is a weaker claim. It's discussed, for example, around Theorem 3.9 in Hatcher's Vector Bundles and K-Theory (I'd put a link here but Hatcher's site appears to be down). 
In particular, the corresponding map in rational cohomology is very far from an isomorphism; $BO(1)$ has the rational cohomology of a point, but the rational cohomology of $BO(n)$ is given by the Pontryagin classes. Over $\mathbb{Q}$ the splitting principle takes a more complicated form involving maximal tori (in particular $O(1)^n$ is very far from being the maximal torus of $O(n)$). I don't know what the statements are for disconnected groups so let me restrict attention to $SO(n)$ instead. Then the maximal torus is $SO(2)^{\lfloor \frac{n}{2} \rfloor}$, and the Weyl group is not the symmetric group but a group related to the group of signed permutations. 
