# If $k < l$ show that smooth function on $\mathbb{R}^k$ considered as a subset of $\mathbb{R}^l$ are the same as usual.

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 $$\mathbb{R}^l$$

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 $F\colon \mathbb R^l\to\mathbb R^m$, $(x_1,\dots,x_l)\mapsto f(x_1,\dots,x_k)$.
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.