This is a follow up of my previous MO question "[Non orientable, closed manifold covered by two simply-connected charts](https://mathoverflow.net/questions/390739/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?
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**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 $U \cap V$.



Non orientable, closed manifold covered by two contractible charts
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.


