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Above is from p.13 of John Lee's Introduction to Smooth Manifolds. I am curious if this proposition also holds when $M$ is a topological manifold with boundary, thus making it into a smooth manifold with boundary. I am quite suspicious, but not certain. I question this because I think I have to apply the above proposition to the exercise presented below. I think the atlas mentioned in (a) of Exercise 1.44 below is a smooth atlas for $U$. So, by the direction "define a smooth structure on U", does it mean that I have to apply the proposition 1.17 to this exercise?

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I think I see your confusion. The way this proposition is phrased is a bit misleading. In part (a), the phrase "Every smooth atlas" is implicitly assuming that $M$ has a smooth atlas--otherwise part (a) doesn't apply. That is, all the topological manifolds in this proposition are secretly smooth manifolds.

To answer your second question, this proposition should be true for topological manifolds with boundary, but ${ \bf beware}$: Just as not all topological manifolds can be given a smooth structure, not all topological manifolds with boundary can be given the structure of a smooth manifold with boundary (Can you think of an example?).

To do the exercise, you will need to invoke the (manifold with boundary version of) the proposition to say that the suggested atlas determines a unique "manifold with boundary" structure on $U$. Note that you still need to show that the suggested atlas is actually an atlas in order for the proposition to apply.

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