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I have a question related to this one: link.

Let $f: \mathbb{R}^3 \to \mathbb{R}$ be a smooth map.

Now, consider the restriction $f'$ of $f$ to the square pyramid (denoted $\Delta$ here). As $f$ is differentiable on $\mathbb{R}^3$, I would like to say something about the differentiability of $f'$, but there is something I don't understand:

1- Because the square pyramid is not a smooth manifold with corners (so, is not a smooth manifold with or without boundary), technically, we cannot say that $f'$ is smooth at a given point of $\Delta$, right ?

2- But the point is that $f$ is smooth at any point of $\mathbb{R}^3$, so in particular, at any point of $\Delta$. Thus, I would like to say that $f'$ is also smooth and that its tangent map at any point is the one of $f$.

Where is my mistake ?

Thanks for your help !

deeppinkwater
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  • Possibility relevant: the Wikipedia article on smoothness proposes (right at the end) a definition of the notion of a smooth function between arbitrary subsets of smooth manifolds. I don't know how well-known or widely-used that definition is. – Rob Arthan Jun 24 '21 at 00:50
  • Thank you for your answer @RobArthan. There is something "weird" with that definition (in my opinion). It's like "general theory" (i.e. differential geometry) would say "we cannot talk about smoothness on $\Delta$" and this definition would say the opposite... it seems pretty strange to me that the "general theory" does not cover a particular case as this one. – deeppinkwater Jun 24 '21 at 09:21

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