# Smooth map on the square pyramid

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 ?

• 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. Commented Jun 24, 2021 at 9:21