# Euler class of tangent bundle of the sphere

I am working through Milnor's Characteristic classes and am currently working problems on the topic of oriented bundles and euler class. I am having trouble computing the euler class of the tangent bundle of a sphere. (The actual question is in the third paragraph)

The first part which worked out fine for me was to show that the total space of the tangent bundle is equal to $S^{n} \times S^{n} \setminus A$ where $A$ is the anti-diagonal subset of the product space and furthermore that $H^{*}(E,E_{0};Z) \cong H^{*}(S^{n} \times S^{n},A;Z)$ (Here $(E,E_{0})$ is the pair (total space, total space - zero section), as used in the book). This follows by excision and arguing that $S^{n}\times S^{n} \setminus (Diagonal) \sim A$.

The part I am having trouble with is showing that, in the $n$ even case, the euler class corresponds to twice a generator of $H^{n}(S^{n};Z)$. It seems obvious to me that the fundamental class ,$u$, is a generator of $Z = H^{n}(E,E_{0})$. Next, since $-\cup u$ is an isomorphism, $u \cup u$ is a generator of $Z = H^{2n}(E,E_{0})$ and finally, since the Thom isomorphism is in fact an isomorphism this seems to suggest that the euler class should in fact be a generator of $H^{n}(S^{n};Z)$.

For one thing, I haven't used the fact that $n$ is even, though I don't see where that would be relevant. I also am not 100% sure about my computation of the fundamental class or the cohomology groups. For reference I will state my computation of the cohomology below.

To start, $H^{i}(S^{n}\times S^{n};Z) = (Z, i = 0);(Z \oplus Z, i = n);(Z, i = 2n)$ as proved with the Kunneth formula. Next, using the long exact sequence of a triple $(S^{n}\times S^{n},A,\not{0})$, we get that $H^{i}(S^{n} \times S^{n},A;Z) = (Z,i = 0,n,2n); 0\,else$.

So, all you have to do is pick a vector field on $S^n$ with isolated zeroes whose indices are easy to compute. The easiest is the vector field which has a "source" at the south pole and a "sink" at the north pole. The index of a source equals $+1$, and the index of a sink equals $(-1)^n$.
this is in Hatcher's Vector Bundles and K-Theory Book, Page 92. Alternatively you can use the fact that $\chi(M) = <e(M), [M]>$ to conclude that $e(M)$ must correspond to the cohomology class $2$. Hope this helps!
Also, your sphere must be even for otherwise you will have an odd-dimensional bundle which will produce a $2$-torsion in $H^n(S^n; \mathbb{Z}) = \mathbb{Z}$ which is clearly impossible