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I am reading a book on differential topology and the first question in it has me confused.

If $k < l$ we can consider $\mathbb{R}^k$ to be the subset $\{(a_1, \cdots,a_k, 0, \cdots, 0)\}$ in $\mathbb{R}^l$. Show that smooth functions on $\mathbb{R}^k$, considered as a subset of $\mathbb{R}^l$, are the same as usual.

What does it mean same as usual?

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Guillemin and Pollack, right? –  Neal May 13 '12 at 0:56

1 Answer 1

up vote 3 down vote accepted

Presumably, the text has presented a definition of what it means for a function whose domain is a subset $S\subseteq\mathbb{R}^l$ to be smooth. The question is asking you to show that, when $S=\mathbb{R}^k$ (viewed as a subset of $\mathbb{R}^l$), this definition is equivalent to the usual definition of what it means for a function to be smooth on $\mathbb{R}^k$ (whatever that is according to your book).

Of course, to show that the two definitions are equivalent is just to show that a function is smooth according to one definition if and only if it is smooth according to the other definition.

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