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Suppose that $M,N$ are smooth manifolds without boundary and $F: N \rightarrow M$ is a continuous map, then we know that $F$ is homotopic to a smooth map (Th6.26, Lee's Smooth Manifolds),i.e there is a continuous map $G: [0,1]\times N \rightarrow M$ such that $G(0, \cdot) = F$ and $G(1, \cdot)$ is smooth.
My question is : Can we choose $G$ so that $G$ is smooth in $(0,1]\times N$?

Why I came up with this small (and trivial?) question? : I has been learning Lee's book for fun by trying to prove each Theorems in his book by myself. Then for this Theorem, I obtained the existence of such $G$ so I wanted to check if it is really right to make sure that I understand correctly most of concepts related to Smooth Manifolds.

Thank you for your time.

Arctic Char
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Taro Tokyo
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    Yes. It is essentially contained in the answer here, you just have to modify it to work with manifolds. – Moishe Kohan Dec 02 '21 at 17:23
  • Thank you for your answer. – Taro Tokyo Dec 02 '21 at 20:02
  • If you have an argument for this, why not share it? This way you wouldn't require that much work on the part of the (possible) answerer. – Michał Miśkiewicz Dec 02 '21 at 22:39
  • @MichałMiśkiewicz Dear Michal, I thought about doing the same however, based on my experiences on MSE, people seem to do not like "proof-verification" type of question so I referred to not post my attempt. – Taro Tokyo Dec 03 '21 at 07:00
  • It’s hard to know what techniques are used if one does not have Lee’s text at hand. Locally, convolutions wirk just fine to answer your question. – Ted Shifrin Dec 05 '21 at 02:50

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