# Can every manifold be written as the union of two contractible open subspaces?

Let $$n \in \mathbb{N}$$ and $$X$$ be a connected $$n$$-manifold. Must there exist contractible open subsets $$A, B \subseteq X$$ with $$A \cup B = X$$?

This question arose when toying around with the Seifert-Van Kampen Theorem.

• Wouldn’t $X$ have to be simply connected? Commented May 3 at 20:52
• @TedShifrin Not necessarily; consider $S^1$. Commented May 3 at 20:52
• Fair enough. In higher dimensions, can be arrange for the intersection to be connected? Commented May 3 at 20:54
• @TedShifrin Not sure. My manifold-fu really isn't great, but perhaps it's possible to show that (in the smooth case, say) one can deformation retract one of the open sets onto a closed subset still intersecting the other open set, argue that the inclusion is a cofibration and reduce to the case that one of the sets is a disk (although at that point you're not far from showing that the whole question is equivalent to asking whether the punctured manifold is contractible, which would be quite nice). Commented May 3 at 21:15
• On the other hand, every connected $n$-manifold is the union of $n+1$ contractible open subsets. Commented May 4 at 4:50

No, although I can only offer something a bit more high-tech than van Kampen for an explanation: The $$\mathbb{Z}$$-algebra structure on the cohomology $$H^*(X)$$ of any space $$X$$ which is the union of two contractible subspaces is trivial, but there are plenty of manifolds where this isn't the case; the simplest example is perhaps $$T = \mathbb{S}^1 \times \mathbb{S}^1$$.

Edit: As Lee Mosher pointed out in the comments, your question is in fact asking whether every manifold has Lusternik-Schnirelmann category at most 2. But the Lusternik-Schnirelmann category of any space is bounded below by its cup length (i.e. the number of positive-degree factors in the longest non-trivial cup product in the cohomology ring) minus 1 as by my answer here, generalizing the argument above.

• The key word is Lusternik-Schnirelmann category... AND I left off an n. Commented May 3 at 21:42
• @LeeMosher Thank you for pointing this out. I somehow answered a question about this exact relationship between LS-category and cup products less that a month ago yet didn't even think about it here... Commented May 3 at 22:22

Let $$M$$ be a connected $$n$$-dimensional manifold. Then $$M$$ is covered by $$n+1$$ open contractible subsets. This is a special case of Theorem 1.7 in

O. Cornea, G. Lupton, J. Oprea, D. Tanré, Lusternik-Schnirelmann category. Mathematical Surveys and Monographs. 103. Providence, RI: American Mathematical Society (AMS). xvii, 330 p. (2003).

No, for instance let $$X$$ be the surface of genus $$g$$.

The Mayer-Vietoris sequence implies the existence of an exact sequence $$H_{2}(A)\oplus H_2(B) \to H_2(X) \to H_1(A\cap B) \to H_1(A)\oplus H_1(B) \to H_1(X) \to H_0(A \cap B)$$

Thus, $$H_2(X) = H_1(A\cap B)$$ and there exists an injective homomorphism from $$H_1(X)$$ to $$H_0(A\cap B)$$.

We now obtain a contradiction by considering whether the nonempty set $$A \cap B$$ is connected. If $$A\cap B$$ is connected then $$H_0(A \cap B)=\mathbb{Z}$$, so that there is no injective homomorphism from $$H_1(X) = \mathbb{Z}^{2g}$$ to $$\mathbb{Z}$$. If $$A \cap B$$ is not connected then the connected components $$X_1,X_2, \ldots, X_k$$ of $$A \cap B$$ are open manifolds of dimension 2, so that $$\pi_1(X_i)$$ is a free group of rank $$r_i$$. Thus, $$H_1(A \cap B) = \oplus_{i=1}^k \mathbb{Z}^{r_i}$$ is not isomorphic to $$H_2(X) = \mathbb{Z}$$.

• I think that there's a gap in your argument - why can't the $r_i$'s sum to $1$? In other words, a connected component $X_i$ can be a disc and have trivial first homology. Commented May 4 at 6:15
• You are correct, there is a gap in the proof. It is not clear if this gap can be fixed by a small modification in the argument.
– idk
Commented May 4 at 12:28