TITLE: What is the motivation behind the Thom space construction?
QUESTION [9 upvotes]: The Thom-Space of a vector bundle $\pi: V \to M$ is defined as the quotient of the total space of the associated disk bundle $D(V) \to M$ by the total space of the associated sphere bundle $S(V)$, so
$$
\text{Th}(V) = D(V)/S(V)
$$
what is the original motivation behind this construction?

REPLY [3 votes]: I brought up the connection between the homotopy groups of Thom spaces and cobordism groups in the comments, so I'll expand on it in this answer.  What I'm about to describe is the Pontryagin-Thom construction, which is covered in many other places, but here is a quick summary.  
Suppose I have a vector bundle $V$ of rank $n$ over $M$, and suppose I have a map $f: S^{n+k} \to \operatorname{Th}(V)$ that is transverse to the zero section $Z \subset \operatorname{Th}(V)$.  Then we consider $X := f^{-1}(Z)$, which is a $k$-dimensional manifold that is equipped with 

an embedding $X \hookrightarrow \mathbb{R}^{n+k} \subset S^{n+k}$, 
a map $f|_X: X \to M$, and 
a framing of the normal bundle $Df: \nu \to f^* V$.

Moreover, varying $f$ within its homotopy class yields manifolds $X'$ that are cobordant to $X$.  In fact, we have

Theorem (Pontryagin-Thom).  There is a bijection between $\pi_{n+k} \operatorname{Th}(V)$ and $k$-dimensional manifolds equipped with the above data, up to cobordism.  

To actually get the cobordism group $\mathfrak{N}_*$, we have to modify the construction so as to "forget" the extra data.  This can be done by

Passing to stable homotopy groups.  
Taking $V$ to be the universal rank $n$ bundle over $M = BO(n)$.  Denote the Thom space in this case by $MO(n)$.  

Once we do this, we arrive at the isomorphism $\mathfrak{N}_* \cong \operatorname{colim}_k \pi_{n+k} MO(k)$.  
This is the beginning of many different stories in geometric topology, stable homotopy theory, surgery theory, ....