Obstruction cocycle of Stiefel manifold I was reading about the interpretation of Stiefel Whitney classes as obstructions from Milnor-Stasheff's book and I got stuck at a step. The context is the following. Let $E \to B$ be a vector bundle of rank $n$ and let $V_k(\mathbb{R}^n)$ be the Stiefel manifold, i.e. the manifold whose points are $k$-tuples of linearly independent vectors in $\mathbb{R}^n$. The main theorem I want to understand is that the reduction mod 2 of the obstruction class $c_j(E)$ is equal to the Stiefel Whitney class $w_j(E) \in H^j(B; \mathbb{Z}/2)$. 
Let $\gamma^n \to  G_n(\mathbb{R}^{n+1}) \cong \mathbb{R}P^n$ be the canonical bundle over the Grassmannian of $n$-planes in $\mathbb{R}^{n+1}$  and  consider the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1})$ whose fibers are $V_1(\mathbb{R}^n)$. They use the following fact in their proof:
"The obstruction cocycle of the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1})$ clearly assigns to the $n$-cell of $\mathbb{R}P^n \cong G_n(\mathbb{R}^{n+1})$ a generator of the cyclic group $\pi_{n-1}(V_1(\mathbb{R}^n))=\pi_{n-1}(\mathbb{R}^n - 0) = \mathbb{Z}$"
I understand  why $\pi_{n-1}(V_1(\mathbb{R}^n))=\mathbb{Z}$. The n-th obstruction cocycle of the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1}) \cong \mathbb{R}P^n $ assigns to the n-cell of $\mathbb{R}P^n$ an element of $\pi_{n-1}(V_1 (\gamma^n))$. So if we compose with the map induced by restriction to the fiber we get an element of  $\pi_{n-1}(V_1(\mathbb{R}^n))$.  Why is this element a generator of such group? 
Thank you!
 A: For your bundle $V_1(\gamma^n) \to G_n(\mathbb R^{n+1})$, $V_1(\gamma^n)$ is the space $\{(L,v) : L \subset \mathbb R^{n+1}, v \in L, |v|=1, dim(L)=n \}$, and the map is "forgetting $v$".  
I'm not sure if Milnor and Stasheff do it this way, but you can make a direct computation to justify their claim, using something like Schubert calculus techniques.  Since $G_n(\mathbb R^{n+1}) \equiv G_1(\mathbb R^{n+1})$ by orthogonal complements, you can think of the bundle as 
$$\{(L,v) : L \subset \mathbb R^{n+1}, dim(L)=1, v \perp L, |v|=1\}$$
Given a 1-dimensional subspace $L$ of $\mathbb R^{n+1}$, as long as it's not the $x$-axis, in the unit sphere orthogonal complement there is a unique vector $v$ such that it's x-component is minimal (most negative).  Choose that vector.  With some patience, you can think of this as a map from $D^n$ to this bundle, the idea is to think of $D^n$ as the $x=1$ plane in $\mathbb R^{n+1}$ suitably compactified with a point at infinity, and "blown up" along the x-axis.  For every line $L$, you can ask where it intersects this disc (provided the line isn't the x-axis), and take the $v$ with minimal $x$-coordinate.  
It's a fairly straightforward geometry exercise to show that for $L$ intersecting $x=1$ near the x-axis, it's a degree 1 map of spheres. 
Does that sort of make sense?   I imagine Milnor has a slicker argument but it's totally possible to "manhandle" this problem into submission. 
