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If $k < l$ we can consider $\mathbb{R}^k$ to be the subset $\{(a_1, ..., a_k, 0, ..., 0)\}$ 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.

There are two definitions given in Guillemin and Pollack for smoothness

Definition 1: A mapping $f$ of an open set $U \subseteq \mathbb{R}^n$ into $\mathbb{R}^m$ is called smooth if it has continuous partial derivatives of all orders

Definition 2: A map $f : X \to \mathbb{R}^m$ defined on an arbitrary subset $X \subseteq \mathbb{R}^n$ is smooth if for every $x \in X$ there exists an open set $U \subseteq \mathbb{R}^n$ containing $x$ and a smooth map (in the sense of definition 1) $F : U \to \mathbb{R}^m$ such that $F$ equals $f$ on $U \cap X$

Now I'm slightly unsure what the authors mean by "Show that smooth functions on $\mathbb{R}^k$ considered as a subset of $\mathbb{R}^l$ are the same as usual." but based on the answer here I guess that means I have to show that Definition 1 $\iff$ Definition 2 in this case

Also I'm sure the authors mean to consider the topological embedding of $\mathbb{R}^k$ into $\mathbb{R}^l$, which I'll denote by $\phi[\mathbb{R}^k]$

I managed to show Definition 1 $\implies$ Definition 2 in this case

Proof: Suppose that $f : \phi[\mathbb{R}^k] \to \mathbb{R}^m$ is smooth in the sense of Definition 1. Then define $\varphi : \mathbb{R}^l \to \phi[\mathbb{R}^k]$ by $\varphi(x_1, ..., x_n, .., x_l) = (x_1, .., x_n, 0, .., 0)$ and note that $\varphi$ is smooth (in the sense of definiton 1).

Then define $F : \mathbb{R}^l \to \mathbb{R}^m$ by $F = f \circ \varphi$. This map is smooth in the sense of definition 1 since $\mathbb{R}^l$ is an open subset of $\mathbb{R}^l$ and the composition of smooth maps is known to be smooth. Furthermore $F$ equals $f$ on $\mathbb{R}^l \cap \phi[\mathbb{R}^k] = \phi[\mathbb{R}^k]$. So definition 1 implies definition 2. $\square$

I'm not sure how to show that Definition 2 implies definition one. I could let $f : \phi[\mathbb{R}^k] \to \mathbb{R}^m$ a map and suppose that for each $x \in \phi[\mathbb{R}^k]$ there existed an open set $U \subseteq \mathbb{R}^l$ containing $x$ and a smooth map $F : U \to \mathbb{R}^m$ such that $F$ equals $f$ on $U \cap X$, but after that I'm unsure how to show that $f$ is smooth in the sense of definition 1


EDIT: I think there's a big error in my proof that Definition 1 $\implies$ Definition 2 because $\phi[\mathbb{R}^k]$ isn't even an open set of $\mathbb{R}^l$ so we can't suppose that $f$ is smooth in the sense of definition 1.

Similary I don't see how we can prove the converse because no matter what $\phi[\mathbb{R}^k]$ will never be open in $\mathbb{R}^l$

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What the exercise asks you to show is that any map $f\colon \mathbb{R}^k\to \mathbb{R}^m$ is smooth in the sense of Definition 1 (with $U = \mathbb R^k$) if and only if it is smooth in the sense of Definition 2 (where we consider $\mathbb R^k = \mathbb R^k\times 0\subset\mathbb R^l$).

Hint: It should be straightforward that if $f$ is smooth in terms of Definition 2, then it is also smooth according to Definition 1. For the converse implication consider $F\colon \mathbb R^l\to\mathbb R^m$, $(x_1,\dots,x_l)\mapsto f(x_1,\dots,x_k)$.

Hover the following to see some more details:

Suppose $f$ is smooth in Definition 2; that is, around every point $x = (x_1,\dots,x_k,0,\dots,0)\in\mathbb R^k\times 0\subset \mathbb R^l$ there exists an open neighbourhood $U\subset \mathbb R^l$ and a map $F\colon U\to\mathbb R^m$ which is smooth according to Definition 1 and such that $F|_{\mathbb R^k\cap U} = f|_{\mathbb R^k\cap U}$. Since all partial derivatives of $f|_{\mathbb R^k\cap U}$ are also partial derivatives of $F$, we get that $f|_{\mathbb R^k\cap U}$ is smooth in Definition 1 as well. In particular, $f$ is smooth at $x$. But $x$ was arbitrary, so that we get that $f$ is smooth in Definition 1.


Now suppose that $f$ is smooth in Definition 1 instead and consider the map $F$ defined in the hint. Its partial derivatives are trivial or those of $f$. In particular, $F$ is smooth in Definition 1 (with $U = \mathbb R^l$). But since $f = F|_{\mathbb R^k}$, this shows that $f$ is also smooth in Definition 2.

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