This question arose during my Differential Geometry course. Possibly there is an obvious answer, but I do not see it, and I could not find it in the literature. The same question was [asked](https://math.stackexchange.com/questions/4109257/non-orientable-closed-manifold-covered-by-exactly-two-charts) yesterday on MSE, but it did not get much attention.



> 
> **Question.** Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two *simply-connected*
> charts? If so, what is a reference?
> 
> 
> 


Of course, the intersection of the two charts must be disconnected. As noted in the linked MSE question, the open Möbius band and the closed Möbius band give examples in the open case and in the non-empty boundary case, respectively. Moreover, the Klein bottle is covered by two charts, both homeomorphic to cylinders, but they are not simply-connected.



Non orientable, closed manifold covered by two simply-connected charts
Take a non-orientable $S^n$ bundle over $S^1$ with $n \geq 2$ (\*), (sometimes called generalised Klein bottles) then covering $S^1$ by two intervals and taking preimages should work.


(\*) Let $\tau : S^n \rightarrow S^{n}$ be a non-orientable diffeomorphism of $S^n$ given by $(x\_{1},\ldots,x\_{n+1}) \mapsto (x\_1,\ldots,-x\_{n+1}) $. Then such a bundle is given by taking the quotient of $[0,1] \times S^{n}$ by the equivelance closure of the relation $$(0,p) \sim (1,\tau(p)). $$


Such bundles are discussed in the following paper <https://academic.oup.com/plms/article-abstract/s3-4/1/196/1497906?redirectedFrom=PDF>


The non-orientable $S^2$-bundle over $S^1$ appear's in Hatchers 3-manifold notes, since it plays a role in the prime decomposition theorem for non orientable $3$-manifolds. See page 8 of
[https://pi.math.cornell.edu/~hatcher/3M/3Mdoublepage.pdf](https://pi.math.cornell.edu/%7Ehatcher/3M/3Mdoublepage.pdf)


