Stiefel–Whitney class, obstructions and exact sequences 
*

*The first Stiefel–Whitney class $w_1$ is zero if and only if the bundle is orientable. In particular, a manifold $M$ is orientable if and only if $w_1(TM) = 0$.

*The first and second Stiefel–Whitney classes are zero, $w_1(TM) =w_2(TM) = 0$, if  and only if  the bundle admits a spin structure.

*The third integral Stiefel–Whitney class is zero if and only if the bundle admits a spin$^c$ structure.

Questions: Are there obvious ways to relate and encode the above obstructions in terms of group extension languages? For example, the failure of the pullback from a group $G$ to a group $G'$. And how do $w_1(TM)$, $w_1(TM)^2$, $w_2(TM)$ and the third integral Stiefel–Whitney class enter into the homomorphism map in the exact sequences?

My attempts:


*

*It looks to me that for $w_1(TM)^2$, is has something to do with the classifying the extensions:
$$
1 \to \mathbb{Z}_2  \to SO(3) \rtimes \mathbb{Z}_4  \to O(3) \to 1
$$ 
or for the odd $n$ (how about the even $n$)
$$
1 \to \mathbb{Z}_2  \to SO(n) \rtimes \mathbb{Z}_4  \to O(n) \to 1
$$ 
What is the extension and obstruction for classifying $w_1(TM)$?

*It looks to me that for $w_2(TM)$, is has something to do with the classifying the extensions:
$$
1 \to \mathbb{Z}_2  \to Spin(n)  \to SO(n) \to 1
$$ 

*It looks to me that for the integral $\tilde w_3(TM)$, is has something to do with the classifying the extensions:
$$
1 \to \mathbb{Z}_2  \to Spin^c(n)  \to SO(n)\times U(1) \to 1
$$ 
which we may view (?) the 
$$
Spin^c(n) =\frac{Spin(n)\times U(1)}{\mathbb{Z}_2}.
$$

 A: You might be thinking about the Whitehead tower of the orthogonal group.  Given a manifold $M$, its tangent bundle classifies a map $\tau_M: M \to BO$.  If we are given a group homomorphism $G \to O$, we can ask whether the structure group of  the tangent bundle reduces to $G$.  For example, asking for a $SO$-structure is the same as requiring $M$ to be orientable.  In terms of classifying maps, this means a lift of the tangent classifier along $BG \to BO$:
$$\begin{array}{ccc}
& & BG \\
& \nearrow & \downarrow \\
M & \xrightarrow[\tau_M]{} & BO
\end{array}$$
In the case $G = SO$, we have that $BSO$ is the fiber of a map $w_1: BO \to K(\mathbb{Z}/2,1)$.  So the existence of a lift $M \to BSO$ is equivalent to asking that the composite $w_1(M): M \to BO \to K(\mathbb{Z}/2,1)$ is nullhomotopic.  This means that $M$ is orientable iff $w_1(M) = 0$.  
We can continue up the Whitehead tower for $BO$.  Suppose $M$ is orientable.  To lift the tangent classifier to $B\mathrm{Spin}$ and endow $M$ with a spin structure is to ask that $w_2(M): M \to BSO \to K(\mathbb{Z}/2,2)$ is zero.
$$\begin{array}{ccccc}
& & \downarrow \\
& & B\mathrm{Spin} & \xrightarrow{\frac{p_1}{2}} & K(\mathbb{Z},4) \\
& & \downarrow \\
& & BSO & \xrightarrow{w_2} & K(\mathbb{Z}/2,2) \\ 
& & \downarrow \\
M & \xrightarrow[\tau_M]{} & BO & \xrightarrow{w_1} & K(\mathbb{Z}/2,1)
\end{array}$$
In this diagram, each "L"-shaped part (e.g., $B\mathrm{Spin} \to BSO \to K(\mathbb{Z}/2,2)$) is a fiber sequence that we use to rephrase the problem of reducing to a structure group in terms of cohomology classes.  These fiber sequences are incarnations of the group extensions you've described, e.g.:
$$1 \to SO \to O \xrightarrow{\det} \mathbb{Z}/2 \to 1$$
$$1 \to \mathbb{Z}/2 \to \mathrm{Spin} \to SO \to 1.$$
