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Consider the map $f:\mathbb{R}\rightarrow T_{(x_1,x_2)}\mathbb{S}^1$ defined by $f(t)=t(x_2,-x_1)$. Is this map a diffeomorphism? I think it is; I am trying to prove that the tangent bundle of $\mathbb{S^1}$ is diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$. I have already proved that it is a homeomorphism and it is smooth; how do I get smoothness of $f^{-1}$? Thank you in advance!

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That's a diffeomorphism from the real number line to one tangent line to the circle, yes. In fact, it's linear and invertible, so certainly a diffeomorphism. But what you need is a diffeomorphism from $S^1 \times \mathbb R$ to $TS^1$; when you write that down, we can tell you how to invert it and prove the inverse smooth.

John Hughes
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  • But my argument is the following: I defined the map $\phi:S^1\times\mathbb{R}\rightarrow TS^1$ by $\phi(x,t)=((x_1,x_2),t(x_2,-x_1))$ and now I am using the fact that if $f$ and $g$ are diffeomorphisms then also $(f\times g)(x,y)=(f(x),g(y))$ is a diffeomorphism (which I have already proved). Then $\phi$ is a diffeomorphism, because $\phi=id_{S^1}\times f$ –  Oct 03 '16 at 11:51
  • I was only missing smoothness of $f^{-1}$ to conclude that $f$ was a diffeomorphism, and your comment gave me that :) so thank you! –  Oct 03 '16 at 11:52
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    Ah, good. Nice use of the "products of diffeos" idea in there. Glad to have helped. :) – John Hughes Oct 03 '16 at 13:48
  • How can it be deduced that $f$, $f^{-1}$ are smooth? You said that it is linear and invertible, so doe this imply smoothness directly? @John Hughes – Mat999 Apr 22 '22 at 12:17
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    A linear map is always smooth (its derivative is more or less itself!); if it's invertible, the inverse is also linear (easy to see if you write out matrices), hence also smooth. If the linear map's NOT invertible, then the inverse can't be smooth of course: it doesn't even exist. – John Hughes Apr 22 '22 at 12:40