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When I thought of this question, I was sure it must have been asked before on this site, but I could't find anything. Maybe my search skills are lacking, or maybe the question is obvious and it's my math skills that are lacking. Anyway, here it goes.

For a $CW$-complex $X$ let $sw^*X$ be the subring of $H^*(X,\mathbb{F}_2)$ generated by all classes which are Stiefel-Whitney classes of some vector bundle over $X$. It is not hard to see that $sw$ is a proper subfunctor of mod 2 homology. For example (and this might be overkill) if you take the right dimensional sphere $S^n$, then by Bott periodicity, $KO(S^n)=0$, so $sw^*S^n=0$.

Now let $SW^*X$ be the subring generated by all classes which are either Stiefel-Whitney classes of some vector bundle over $X$, or suspensions or desuspensions of such classes.

$\textbf{Edit}$: Perhaps it wasn't clear from context, but I want $SW^*$ to be a functor, so I force it to be closed under pullbacks. For that reason I am puzzled by Nicholas Kuhn's suggested answer below. Also, we know in retrospect that $H\mathbb{F}_2^*X$ is a summand in $MO^*X$, and that thing is sort of tautologically built out of characteristic classes...

Is $SW^*X=H^*(X,\mathbb{F}_2)$?

I suppose the question is equivalent to something like: does the identity map of $K(\mathbb{F}_2,n)$ factor, stably, through some $BO(m)$?

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A 1968 paper in Topology by Anderson and Hodgkin shows that $KO^*(K(\mathbb F_2, n)) = 0$ if $n \geq 2$. This implies that if $n \geq 2$, then no nonzero classes in $H^*(K(\mathbb F_2,n);\mathbb F_2)$ are SW classes. (And of course, $BO(1) = K(\mathbb F_2,1)$.)

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    $\begingroup$ Sorry, why does this show that no nonzero classes can be pulled back from some suspension of some $BO(m)$? $\endgroup$
    – John Greenwood
    Commented May 25, 2020 at 5:06
  • $\begingroup$ John, correct me if i'm wrong, but this follows from (i) F_2-cohomology of K(F_2,n) is polynomial algebra and (ii) multiplication in deg>0 cohomology of any suspension is trivial. But I don't see by now any trivial reason for the desuspension statement $\endgroup$
    – Bad English
    Commented May 25, 2020 at 15:50
  • $\begingroup$ @John, upd: for desuspension of the form \Sigma^N K(n)\to BO(\infty) it follows from Nicholas answer because after adjunction we obtain a map and as a class in KO^*(K(n)) he is equal to zero. Is there are a way to see why suspension on both sides is zero on cohomology? $\endgroup$
    – Bad English
    Commented May 25, 2020 at 22:43

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