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Is the torus the union of two connected, simply-connected open sets? A routine computation with the Mayer-Vietoris sequence shows that if so, then their intersection must have exactly three components.

Also, exactly one of the components must have $H_1=\mathbb{Z}$; the other two must be homologically trivial. (That's assuming that $H_2(X)=0$ for any proper open subset $X$ of the torus, which seems obvious.)

Michael Weiss
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  • Perhaps considering the universal covering $\mathbb{R}^2$ of the torus would help? Namely the set $[0,2\pi)\times [0,2\pi)$. Can we find two open subsets of $[0,2\pi)\times[0,2\pi)$ in the subspace topology which cover the square and will not allow us to create a nontrivial loop after projection? –  Jan 30 '14 at 21:39
  • I feel like the answer will be Seifert-van Kampen somehow. Unfortunately, the intersection of simply-connected sets is not necessarily simply-connected. –  Jan 30 '14 at 21:52

2 Answers2

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Edit: The answer is still negative. What you have to use is the Lusternik-Shnirelmann category $cat(X)$:

Definition. $cat(X)$ for a topological space $X$ is the least number of contractible open sets needed to cover $X$.

It is known that $cat(T^n)=n+1$, see here. Thus, you cannot cover 2-torus with two simply-connected open sets (since such sets are contractible).

Moishe Kohan
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Suppose the torus is the union of simply connected open sets $A,B$. We have that those sets are manifolds which are not compact, and hence have trivial $H_2$. Since they are simply-connected, we infer that they are acyclic. But if $X=U \cup V$ with $U,V$ open acyclic sets, then $\alpha \cup \beta=0$ (cup product) for all $\alpha,\beta \in H^*(X)$ of positive degree*. However, we can generate $H^2(T^2) \neq 0$ with some $\alpha,\beta \in H^1(T^2)$ via cup product.

*You can see this in Bredon, page 332.

Aloizio Macedo
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