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I want to calculate de Rham cohomologies of $S^2$ with $g$ Mobius strips.

  1. Let's cut $g$ Mobius strips from $S^2$ by disks and then pull off disks to the border. So after that we have $S^2$ with $g$ holes and $g$ Mobius strips separately.
  2. The Mobius strip can be deformation retracted to the $S^1$ so it has the same cohomologies with it.
  3. $S^2$ with one hole is homotopy equivalent to $D^2$ so $S^2$ with $g$ holes is homotopy equivalent to $D^2$ with $g-1$ holes. Let's draw a circle around each hole and we get the wedge sum of $g-1$ copies of $S^1$. So we have $g$ Mobius strips and wedge sum of $g-1 ~S^1$.
  4. I think the best way to calculate de Rham cohomologies is to use the Mayer-Vietoris exact sequence which is $$\ldots \rightarrow H^k(U \cup V) \rightarrow H^k(U) \oplus H^k(V) \rightarrow H^k(U \cap V) \rightarrow H^{k+1}(U \cup V) \rightarrow \ldots$$ where $U$ and $V$ are open subsets such that $X = A \cup B$.

I don't understand how to calculate cohomolgies of wedge sum of $g-1$ copies of $S^1$. $H^0(X) \simeq \mathbb{R}$ because it's path-connected.

Let $U$ be the chosen point of wedge sum and $V$ be the $S^1$ without this point. Then in Mayer-Vietoris exact sequence we have $$H^0(A \cup B) \simeq \mathbb{R} \rightarrow \ H^0(U) \oplus H^0(V) \simeq \mathbb{R} \oplus \mathbb{R} \rightarrow H^0(U \cap V) \simeq 0 \rightarrow H^1(U \cup V) \simeq 0 \rightarrow \ldots$$

Is that true? If not how to calculate it and sum with known cohomologies of $g$ Mobius strips?

P.S. Sorry for my English -- I think there are a lof of mistakes.

Hasek
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  • You are talking about the de Rham cohomology, but many of hour spaces (e.g. wedge sums) are not manifolds. What do you mean? – Thomas Rot Dec 17 '15 at 14:00

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