# Image of Thom Class under Sequence of Maps?

So I've been trying to do problems in Milnor & Stasheff's Characteristic Classes as a quick review, not having done anything with them in a while. However, I'm stuck on some parts in attempting Problem 9-C and I was hoping to get some help.

Let $E=TS^n$ be the tangent bundle of $S^n$, and let $A=\{(p,-p):p\in S^n\}\subset S^n\times S^n$ be the anti-diagonal. We know that $E\approx S^n\times S^n-A$ by stereographic projection, and by excision and homotopy that $$H^*(E, E_0)\approx H^*(S^n\times S^n, S^n\times S^n-\text{ diagonal})\approx H^*(S^n\times S^n, A)\subset H^*(S^n\times S^n).$$

Problem: Let $n$ be even. Show that the Euler class $e(E)=\phi^{-1}(u\cup u)$ is $2\in H^n(S^n;\mathbb{Z})$, where $\phi:H^i(S^n)\to H^{i+n}(E,E_0)$ is the Thom isomorphism.

Attempt: To prove this, first we want to see what happens to the Thom class $u$ of $E$ under this sequence of maps starting from $H^*(E,E_0)$ to $H^*(S^n\times S^n)$, where the last map is the map from relative to absolute cohomology. Then we can what happens to $u\cup u$, and then the Thom isomorphism will give us the rest.

However, I'm having trouble with what the image of $u$ is under this sequence of maps. I can see that $u$ is mapped to an element in $\mathbb{Z}=H^n(S^n\times S^n,A)$ (most probably the generator $1$), which must then be mapped to an element in $\mathbb{Z}\oplus\mathbb{Z}=H^n(S^n\times S^n)$ (most probably the generator $(1,1)$). But I'm not exactly sure how to explain this rigorously. I appreciate any help!