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This is a follow up of my previous MO question "Non orientable, closed manifold covered by two simply-connected charts." Nick L's nice answer shows that such manifolds actually exist, examples being provided by some non-orientable $S^n$-bundles over $S^1$, with $n \geq 2$.

In these examples, the two charts have the homotopy type of $S^n$. So, let me ask the following

Question. Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two contractible charts?

Note. A comment by Denis Nardin to the aforementioned question shows that, by the strong form of Seifert-van Kampen theorem, a manifold covered by two contractible charts $U$, $V$ has the same homotopy type of the suspension of $U \cap V$.

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    $\begingroup$ A closed manifold covered by two contractible charts has LS category 1, and hence is homeomorphic to a sphere. I don't know a reference but this fact is mentioned on the top of p.2 of arxiv.org/pdf/0706.1625.pdf. LS category is defined on p.7 (definition 3.1). $\endgroup$ – Igor Belegradek Apr 22 at 12:58
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No. Recall that the Lusternik-Schnirelmann category of a space $X$, denoted $\operatorname{cat}(X)$, is the minumum $k$ such that $X$ may be covered by open sets $U_0,U_1,\ldots, U_k$ such that each inclusion is null-homotopic. The standard lower bound for LS-category is the cup-length of reduced cohomology: if $R$ is a commutative ring, and $x_1,\ldots, x_k\in \tilde{H}^*(X;R)$ are cohomology classes whose cup product $x_1\cdot \cdots \cdot x_k\in \tilde{H}^*(X;R)$ is non-zero, then $\operatorname{cat}(X)\geq k$. The proof of this is a nice exercise in the naturality of relative cup products.

Now suppose $N$ is a closed non-orientable $n$-manifold which is covered by two contractible charts $U_0$ and $U_1$. It follows that $\operatorname{cat}(N)=1$ (it can't be $0$, because closed manifolds are never contractible), and all cup products in $\tilde{H}^*(N;\mathbb{Z}/2)$ are trivial.

Since $N$ is non-orientable, its first Stiefel-Whitney class $w_1(N)\in H^1(N;\mathbb{Z}/2)$ is non-zero. Poincaré duality gives a non-singular pairing $H^1(N;\mathbb{Z}/2)\times H^{n-1}(N;\mathbb{Z}/2)\to H^n(N;\mathbb{Z}/2)$. In particular $w_1(N)\cdot y\neq 0$ for some $y\in H^{n-1}(N;\mathbb{Z}/2)$. Contradiction.

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    $\begingroup$ Another way of seeing that the cup product is trivial is to notice that the manifold is homotopy equivalent to the suspension of $U_0\cap U_1$ by the van Kampen theorem. $\endgroup$ – Denis Nardin Apr 22 at 13:14

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