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In the Lee‘s book there is a proposition stating:

If $M$ is a smooth $n$-manifold with or without boundary, and $M$ can be covered by a single smooth chart, then $TM$ is diffeomorphic to $M\times \mathbb{R}^n$.

This covers simple examples when $M$ is an open subset of $\mathbb R^n$. However, closed ball and closed annulus in $\mathbb R^n$ are not contained in the above statement.

My question is do we have (for any $n$)

  • $TB_r \cong B_r\times \mathbb R^n$
  • $TC(r_1,r_2) \cong C(r_1,r_2)\times \mathbb R^n$,

where $B_r:=\{x\in\mathbb R^n :|x|\leq r\}$ is closed ball, and $C(r_1,r_2):=\{x\in\mathbb R^n : r_1\leq|x|\leq r_2\}$ closed annulus.

By the stereographic projection, the problem with closed ball should be equivalent to one with closed hemisphere.

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Look at the hypotheses: the closed ball IS a smooth $n$-manifold with boundary, covered by a single chart. The answer in each case you asked about is "yes". In fact, the map you get is not just a diffeomorphism, but a bundle map. Once you see this,

The annular case follows from the ball case, since if you have a map $$ F: TB(r_2) \to B_r^2 \times R^n $$ such that for $v \in TB(r_2)$, we have $$ \pi(v) = \pi_1 (F(v)) $$ where $\pi$ is the projection to the base space on the left, and $\pi_1$ is projection to the first factor on the right (i.e., it's a bundle map), you can simply restrict the domain of $F$ to $TC(r_1, r_2) \subset TB(r_2))$ to get a trivialization of $TC$.

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    $\begingroup$ One thing is still not clear to me. I know that the closed ball is a smooth $n$-manifold, but I still missing why it can be covered by a single chart. $\endgroup$ Mar 15 '15 at 12:19
  • $\begingroup$ I think of definition that every point needs to have a neighbourhood diffeomorphic to a (relatively) open subset in $\mathbb H^n:=\mathbb R^n\cap\{x_n\geq 0\}$, so I know how to do it using $n$ charts. $\endgroup$ Mar 15 '15 at 12:27
  • $\begingroup$ The single chart is the inclusion map: you send $x \in B$ to $x \in \mathbb R^n$. Look at the north pole $(0, \ldots 0, 0,1)$. The artic (points $(x_1, x_2, \ldots)$ with $x_n > 1/2$, say), is diffeomorphic to a relatively open subset of $H^n$ via the map $x \mapsto x + u(x)$, where $u(x_1, x_2, \ldots) = -1 + \sqrt{1 - (x^2_1 + x^2_3 + x^2_{n-1})}$. Under this map, the surface of the artic maps to the $x_n = 0$ plane, while the rest of the arctic maps to a relatively open subset of $\mathbb R^n \cap \{ x_1 \le 0\}$. You can remap that to $H^n$ by negation. :) $\endgroup$ Mar 15 '15 at 12:47

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