TITLE: The first Stiefel-Whitney class is zero if and only if the bundle is orientable
QUESTION [8 upvotes]: Ralph Cohen's notes on the topology of fiber bundles pp.84 (theorem 3.3) says that it follows immediately from the definition of the first Stiefel-Whitney class of real vector bundles (pp.83)
\begin{equation}
  (BSO(n) \to BO(n)) \in \text{Prin}_{O(1)}(BO(n)) \cong [BO(n),BO(1)] \cong H^1(BO(n);\mathbb{Z}_2) \implies w_1(\eta) = f_{\eta}^*(w_1) \in H^1(X;\mathbb{Z}_2)
\end{equation}
and that a bundle has an $SO(n)$ structure if and only if it is orientable.  But I am still stuck with how to apply the orientability to the definition.  Could someone please point that out for me?

REPLY [8 votes]: Consider the short exact sequence of groups (note that I use $O(1) \cong \mathbb{Z}_2$) $$SO(n) \rightarrow O(n) \xrightarrow{\det} \mathbb{Z}_2.$$  This induces an exact sequence $$[X, BSO(n)] \rightarrow [X, BO(n)] \xrightarrow{(B\det)_*} [X, B\mathbb{Z}_2].$$
The real rank $n$ bundle $\eta$ is classified by the map $f_\eta \in [X, BO(n)]$.  We want to show that it lifts to an element in $[X, BSO(n)]$.  Then it will be a $SO(n)$-bundle and hence orientable.  By exactness, to show that it lifts is to show that $(B\det)_* (f_\eta) = 0$.  
Consider the commutative square
$$\begin{array}{ccc}
[BO(n),BO(n)] & \rightarrow & [BO(n), B\mathbb{Z}_2] \cong H^1(BO(n);\mathbb{Z}_2) \\
\downarrow & & \downarrow \\
[X, BO(n)] & \rightarrow & [X, B\mathbb{Z}_2] \cong H^1(X;\mathbb{Z}_2)
\end{array}$$
The vertical maps are $f_\eta^*$ and the horizontal maps are $(B\det)_*$.  Trace the image of the identity $\operatorname{id}_{BO(n)}$ through the square.  By definition of the Stiefel-Whitney class from Cohen's notes, $(B\det)_*(\operatorname{id}_{BO(n)}) = w_1 \in H^1(BO(n);\mathbb{Z}_2)$, so going right and then down gives $$f_\eta^*(B\det)_*(\operatorname{id}_{BO(n)}) = f_\eta^*(w_1) =: w_1(\eta).$$ On the other hand, going down and then right gives $$(B\det)_*f_\eta^*(\operatorname{id}_{BO(n)}) = (B\det)_* (f_\eta).$$
So $\eta$ is orientable iff $(B\det)_*(f_\eta) = w_1(\eta) = 0$.