482 APPENDIX A. HOMOLOGICAL METHODS∗
Then Sn−1 is the intersection of Rn with Sn. Also U will be all of Sn except for the toppoint t while V will be all of Sn except the bottom point b. The line illustrates how Rn ishomeomorphic to U and similarly homeomorphic to V . Thus, from the picture, U ∩V ishomeomorphic to Rn \0. From Lemma A.3.11
Hm (U ∩V )≈ Hm (Rn \0)≈ Hm(Sn−1) .
Then from the Mayer Vietoris sequence above and letting X = Sn =U ∪V,
→0=Hm(U)
Hm (Rn)⊕0=Hm(V )
Hm (Rn)g∗→ Hm (Sn)
∆→Hm−1(U∩V )
Hm−1(Sn−1)
f∗→0
Hm−1 (Rn)⊕0
Hm−1 (Rn)g∗→ (1.17)
Proposition A.8.3 For m ≥ 1,Hm (Sm) = Z and if m ̸= n, then Hm (Sn) = 0. Also, for anyn≥ 1,H0 (Sn) = Z.
Proof: The last claim follows because Sn is path connected. The first claim was shownabove in case n = 1. So suppose the claim is true for n− 1. Consider n and the casewhere m = n. Then from 1.17, the left side is 0 because Rn is convex, so Im(g∗) =0 = ker(∆) which shows that ∆ is one to one. Also Hm−1 (Rn)⊕Hm−1 (Rn) = 0 and soker f∗ = Im(∆) = Hm−1
(Sn−1
)which shows that ∆ is an isomorphism. Hence by induction
Hn (Sn)≈ Hn−1(Sn−1
)≈ Z.
Next consider the case that m < n. By induction, Hm−1(Sn−1
)= 0 but ∆ in 1.17 is still
one to one. Hence Hm (Sn) = 0 since otherwise ∆ would fail to be one to one.Next consider the case that m > n. In this case Im(g∗) is still 0 and so ∆ is still one to
one. Again, by induction, we have Hm−1(Sn−1
)= 0 so again Hm (Sn) = 0 since otherwise
∆ would fail to be one to one. This proves the proposition. ■
A.9 Brouwer Fixed PointsCorollary A.9.1 Sn and Sm are not homeomorphic if n ̸= m.
Proof: If they were homeomorphic, they would have the same homology groups andthey don’t. ■
Note that if Rn and Rm for n ̸= m were homeomorphic, then, their one point compact-ifications would be homeomorphic and hence, using steriographic projection Sn and Sm
would also be homeomorphic which they are not. Steriographic projection involves addinga point at ∞ with the understanding that neighborhoods of this point are complements ofcompact sets so {xn} converges to ∞ means that limn→∞ |xn|= ∞ in the usual manner fromcalculus. The picture illustrating the idea is as follows. The point at ∞ maps to the top pointof the sphere.
•
•
•(⃗0,2)
(⃗0,1)•
p
θ(p)
Rn
Corollary A.9.2 If m ̸= n, then Rn is not homeomorphic to Rm.