# Are all classes Stiefel-Whitney classes?

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 $$BO(m)$$?

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, $$BO(1) = K(\mathbb F_2,1)$$.)
• Sorry, why does this show that no nonzero classes can be pulled back from some suspension of some $BO(m)$? – John Greenwood May 25 '20 at 5:06