Two set of axioms for Stiefel-Whitney classes

Question

Let $E \to X$ be a vector bundle. We can associate to $E$ several invariants: among them are the Stiefel-Whitney classes $w_i(E) \in H^i(X;\mathbb{Z}_2)$. These classes may be defined using the axioms:
0. $w_0(E)=1$ and $w_i(E) \in H^i(X;\mathbb{Z}_2)$.
1. $w(f^*E)=f^*w(E)$ for continuous maps $f$ (here $w=1+w_1+w_2+\cdots$ is the total class)
2. $w(E \oplus F)=w(E) \cup w(F)$
3. $w_1(\gamma_1) \neq 0$ for the tautological bundle $\gamma_1$ over $\mathbb{R}P^{\infty}=BO(1)$

In particular, we can rephrase these axioms to recognize the first Stiefel Whitney class:
1'. $w_1(f^*E)=f^*w_1(E)$ for continuous maps $f$
2'. $w_1(E \oplus F)=w_1(E) + w_1(F)$
3'. $w_1(\gamma_1) \neq 0$ for a tautological bundle $\gamma_1$ over $\mathbb{R}\mathbb{P}^{\infty}=BO(1)$.

However, in Lawson-Michelsohn's book "Spin Geometry" it is stated that in order for a cohomology class $v_1$ to equal $w_1$ we only need to check:
1''. $v_1(f^*E)=f^*v_1(E)$ for continuous maps $f$
2''. $v_1(\gamma_n) \neq 0$ for the tautological bundle $\gamma_n$ over $BO(n)$ for every natural number $n$.

How do we prove that these two sets of axioms are equivalent (and therefore characterize $w_1$)?

Concerning higher Stiefel-Whitney classes, does a set of axioms like 1''-2'' define $w_k$ for each $k$?

Answer 1

The naturality condition 1'' reduces the question of whether $v_1=w_1$ to the special case of the tautological bundles $\gamma_n$. In this case both $v_1$ and $w_1$ lie in $H^1(BO(n);{\mathbb Z}_2)$. This group is just ${\mathbb Z}_2$, as shown in the book of Milnor and Stasheff for example, or just from the fact that $\pi_1BO(n)=\pi_0O(n)={\mathbb Z}_2$. So if $v_1(\gamma_n)$ is nonzero it must equal the nonzero class $w_1(\gamma_n)$.

This argument does not work for $w_2$ since $H^2(BO(n);{\mathbb Z}_2)= {\mathbb Z}_2 \times {\mathbb Z}_2 $ with basis $w_2$ and $w_1^2$, assuming $n>1$. Thus to characterize $w_2$ one needs to know more than just that $w_2(\gamma_n)$ is nonzero since $w_1^2$ also has this property, as does $w_1^2+w_2$. On the other hand, if one considers only oriented vector bundles then $H^2(BSO(n);{\mathbb Z}_2)={\mathbb Z}_2$ when $n>2$ so in this situation $w_2$ is characterized by naturality and being nonzero for the tautological bundle over $BSO(n)$.

Answer 2

This follows from two standard facts:

1. The map $$f:\underbrace{BO(1)\times\cdots\times BO(1)}_{n}\to BO(n)$$ which classifies $\gamma_1\times \cdots\times \gamma_1$ induces an injection on mod 2 cohomology $$\mathbb{Z}/2[w_1,w_2,\ldots , w_n] \subseteq \mathbb{Z}/2[x_1,x_2,\ldots , x_n].$$ The generators on the left are the universal SW classes, and on the right are the polynomial generators of $H^*(BO(1))=H^*(\mathbb{R}P^\infty)=\mathbb{Z}/2[x]$ in each factor, each of which has degree one. The image is the subring of symmetric polynomials.
2. The splitting principle states that for any vector bundle $E\to X$ there exists a map $f:Y\to X$ which is injective on cohomology and such that $f^*(E)\cong L_1\oplus\cdots\oplus L_n$, a direct sum of line bundles over $Y$.

For example, the splitting principle allows to reduce 2' to the case when $E$ and $F$ are line bundles, where it follows from the fact that $f$ classifies line bundles and sends $w_1$ to $x_1 +\cdots + x_n$.