Part two of exercise 11-C of Milnor-Stasheff book Can someone help with part two of exercise 11-C, showing that the euler class is two times a generator of $H^n(S^n)$, I think it is wrong because the Thom isomorphism should send generators to generators but it make sense too because of the relation $\langle e(\tau(S^n)),[S^n]\rangle=\chi(S^n)=2$.
 A: The Thom isomorphism for a real vector bundle $\xi: T(\xi) \rightarrow X$ sends the generator 1 of $H^0(X)$ to the generator u, the Thom class, of $H^n (T(\xi),T(\xi)-X)$. The Thom class restricts to the Euler class. The restriction is not necessarily an isomorphism. As you said, the Euler class evaluates on the fundamental class to the Euler characteristic, so for even dimensional spheres it must be 2 times a generator since the generator evaluates to 1 or -1 on the fundamental class.
A geometric way to see the 2 coefficient is as follows:
The Thom class lives in the Thom space of $\xi$. Let's call the Thom space of the tangent bundle of $S^{2n}$ T. The Thom class $u$ is dual to the generator of the compactification of the fiber over some point $x$. We wish to show that including the fundamental class of $S^{2n}$ gives us twice the dual of $u$.
Pick a vector field on $S^{2n}$ with one singularity. Using Hurewicz we may think of our homology class as a homotopy class. Outside some small disk around this singularity, we may take the fundamental class and stretch it in the direction of the fiber according to the vector field. This leaves us with most of the fundamental class lying in the point at infinity of $T$ and a disk around the singularity which has not been moved, along with some interpolation between these two. Collapse the disk into the singularity while pushing the interpolating area up the fiber to the point at infinity. We now have a class inside the compactification of one of the fibers. Because the index of our singularity was forced to be 2, we have that the resulting class is twice the dual of the Thom class.
