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Can you find mistake in my computation of $H^{k}(S^{n})$.

Sphere is disjoint union of two spaces: $$S^{n} = \mathbb{R}^{n}\sqcup\mathbb{R^{0}},$$ so $$H^{k}(S^n) = H^{k}(\mathbb{R}^{n})\oplus H^{k}(\mathbb{R^{0}}).$$ In particular $$H^{0}(S^{n}) = \mathbb{R}\oplus\mathbb{R}=\mathbb{R}^{2}$$ and $$H^{k}(S^{n}) = 0,~~~k>0.$$

Where is mistake? Thanks a lot!

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You can't space with different dimensions! – checkmath Mar 10 '12 at 1:01
If $S^1$ were really the disjoint union of $R$ and a point $\bullet$ then the identity $S^1 \to S^1$ wouldn't be continuous. ($S^1$ is connected so its image must be connected, but a disjoint union (of non-empty things) is disconnected) – Jacob Bell Mar 10 '12 at 1:16
@donkeykong, well... if $S^1$ were the disjoint union of $\mathbb R$ and a point, then the identity $S^1\to S^1$ would be continuous because all identity maps of all topological spaces are continuous! Your argument is wrong in that if $S^1$ were the disjoint union of $\mathbb R$ and a point then $S^1$ would not be connected! – Mariano Suárez-Alvarez Mar 10 '12 at 3:35
@mariano: yeah, yeah, you got me. I was just (wrongly) putting pressure on the fact that $S^1$ is indeed connected (I should've just said that). thanks for pointing it out. – Jacob Bell Mar 10 '12 at 12:07

3 Answers 3

up vote 5 down vote accepted

You are wrong: $S^n$ is not the disjoint union $\mathbb R^n \sqcup \mathbb R^0$ - topologically.

Although $S^n$ is $\mathbb R^n$ with one point at infinity, the topology of this point at infinity is very different from that of $\mathbb R^0$.

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for example, the sphere is connected, but the disjoint union is not. – ziggurism Mar 10 '12 at 1:37

This should be a comment but is too long.

If your reasoning were correct, we could also do the following: write $S^1$ as the "disjoint union" of two open intervals and two points (by cutting out the north and south poles, for example) Then your idea would show that $H^0(S^1)=\mathbb R^4$. And you can cut it in more pieces...

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Dear Mariano, $H^0$ rather than $H^1$? Cheers, – Matt E Mar 10 '12 at 4:56
Indeed! Thanks for the catch :) – Mariano Suárez-Alvarez Mar 10 '12 at 4:58

de Rham cohomology is not additive, unless you have a partition of the manifold $M$ by two open subsets $U$ and $V$, then the Mayer-Vietoris exact sequence implies that $H^0(M)=H^0(U)\oplus H^0(V)$. Here a point $\{N\}$ is not open in $S^n$. However, the Euler characteristic is additive and $$\chi(S^n)=\chi(\mathbb{R}^n)+\chi(\{N\})=(-1)^n+1.$$

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