TITLE: Intuition behind Thom class
QUESTION [48 upvotes]: The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences.
What is the "actual reason" behind the existence of Thom class? It seems strange that such an interesting class would exist just because some Mayer-Vietoris sequences routinely produce it.

REPLY [8 votes]: There is a nice formulation and interpretation of the Thom isomorphism in terms of sheaf theory, or "Grothendieck's six functors". The statement of the Thom isomorphism in this setting is that if $f \colon E \to X$ is a rank $n$ orientable vector bundle, then
$$ Rf_! \mathbf Z_E \cong \mathbf Z_X[-n].$$
In this form it is also trivial to prove. Note first that it's clear that $Rf_!\mathbf Z_E$ is concentrated in degree $n$, where it is a rank $1$ local system: the statement is local on $X$, so it is enough to prove this for a trivial bundle. So we should argue that this rank $1$ local system is trivial, i.e. that it has a global section. But a section of this local system is precisely an orientation of the vector bundle; again one argues locally, using that an orientation of the fiber $\mathbf R^n$ is the same thing as a generator of $H^n_c(\mathbf R^n,\mathbf Z)$. 
So why does this global statement on $X$ imply the usual cohomological form of the Thom isomorphism? Let $j \colon E \to \overline E$ be the inclusion into the fiberwise one-point compactification, and $\overline f \colon \overline E \to X$ the projection. For any space $Y$ we let $a^Y$ be the projection from $Y$ to a point. The relative cohomology group $H^\bullet(\overline E, \overline E \setminus E)$ can be computed as $R a^{\overline E}_\ast j_! \mathbf Z_E$ (this is how you compute relative cohomology in general in the six functors language). But
\begin{align*} Ra^{\overline E}_\ast j_! \mathbf Z_E & \simeq Ra^X_\ast R\overline f_\ast j_! \mathbf Z_E && \text{since $a^{\overline E} = a^X \circ \overline f$} \\
 & \simeq Ra^X_\ast R\overline f_! j_!\mathbf Z_E && \text{since $\overline f$ is proper} \\ 
& \simeq Ra^X_\ast Rf_!\mathbf Z_E && \text{since $f = \overline f \circ j$} \\
& \simeq Ra^X_\ast \mathbf Z_X[-n] && \text{by the Thom isomorphism}
\end{align*}
and the final equation manifestly computes $H^{\bullet+n}(X)$.