I am trying to solve the following exercise (which is an exercise in Milnor-Stasheff's book). It basically says the following:
Let $ M =S^n $ be the $n$-sphere and let $TM$ be its tangent bundle, and let $A =\left\lbrace (v, -v) : v \in S^n \right\rbrace \subset S^n\times S^n$ be the anti-diagonal. I need to show that the total space $TM$ is homeomorphic to $ S^n\times S^n-A$ , and I need to use excision and homotopy to show that $H^*(TM, TM_0)\cong H^*(S^n\times S^n, S^n\times S^n-\text{ diagonal})\cong H^*(S^n\times S^n, A)\subset H^*(S^n\times S^n).$
The hint for the first part was to use stereographic projection which basically tells me that $\mathbb{R}^n$ is homeomorphic to $S^n$... How can I continue from here?
I also need some help for the second part please (I didn't quite get where/how to use excision) ?
THANK YOU!
I write an answer summarizing the discussions in the comments.
Point 1) the homeomorphism.
For any $v\in S^n$, let $\pi_v:S^n\setminus\{-v\}\to T_vS^n$ be the stereographic projection with pole $-v$ to the tangent space $T_vS^n$. (Note that in $\mathbb R^{n+1}$ the tangent space at $v$ is its orthogonal.)
The map $(v,u)\mapsto (v,\pi_v(u))$ gives the desired homeo.
Point 2) the isomorphisms of homology.
$TM_0$ denotes the complement of the zero section (see the quoted book, beginning of chapter 8).
The zero section corresponds to the diagonal under our homeomorphism.
Thus, excision gives $H^*(S^n\times S^n,S^n\times S^n-diagonal)=H^*(S^n\times S^n -A, S^n\times S^n-diagonal -A)=H^*(TM,TM-zero section)=H^*(TM,TM_0)$.
The last isomorphisms follows from the fact that $S^n\times S^n-diagonal$ retracts to $A$