the Thom isomorphism
/ TOM /
Once you have built the Thom space of a vector bundle, you would like to compute its cohomology in terms of the base, which you presumably understand. The Thom isomorphism does exactly this: it says the cohomology of the Thom space is a perfect copy of the cohomology of the base, just shifted up by the rank of the bundle. Adding a rank-n bundle behaves cohomologically like multiplying by an n-dimensional sphere.
Precisely, let E be an oriented (or, for Z/2 coefficients, any) rank-n real vector bundle over X. Then there is a distinguished Thom class u in H^n(Th(E)) (equivalently in H^n of the pair (E, E-minus-zero-section)) that restricts to a generator on each fiber sphere. The Thom isomorphism is the map H^k(X) -> H^{k+n}(Th(E)) sending a class a to (pullback of a) cup u — cup with the Thom class — and the theorem is that this is an isomorphism for every k. So the Thom class u is the universal n-dimensional generator and cupping with it transports all of H^*(X) up by n degrees. For unoriented bundles the same holds with Z/2 coefficients.
This is the workhorse behind several big results. The Euler class of E is obtained by restricting the Thom class to the zero section: e(E) = (zero-section)^* u, which is why the Euler class is the primary obstruction to a nonzero section. The Gysin sequence, the definition of the Thom class as a relative orientation, and the Pontryagin-Thom construction in cobordism all rest on it. The honest precondition to remember: the integral Thom isomorphism needs an orientation of the bundle (a coherent choice of fiber-orientation, i.e. w_1(E) = 0); without orientation you must work mod 2, where w_n becomes the relevant class and the clean integral statement is unavailable.
Take the trivial rank-n bundle over a point, E = R^n. Its Thom space is S^n, and the Thom isomorphism reads H^k(point) = H^{k+n}(S^n): the single class in H^0(point) = Z maps to the fundamental class in H^n(S^n) = Z, which is exactly the Thom class. Cupping with it is just the suspension isomorphism in disguise.
Over a point the Thom isomorphism is the suspension isomorphism, with the Thom class as the sphere's fundamental class.
The integral Thom isomorphism requires the bundle to be oriented (w_1 = 0); for non-orientable bundles you must use Z/2 coefficients. The Euler class is the restriction of the Thom class to the zero section — that is the precise link, not a vague analogy.