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I'll be quoting from the Wikipedia page on smoothness. Smooth function between manifolds are defined as follows:

If $F$ is a map from an $m$-manifold $M$ to an $n$-manifold $N$, then $F$ is smooth if, for every $p\in M$, there is a chart $(U, \varphi)$ in $M$ containing $p$ and a chart $(V, \psi)$ in $N$ containing $F(p)$ with $F(U) \subset V$, such that $\psi\circ F \circ \varphi^{-1}$ is smooth from $\varphi(U)$ to $\psi(V)$ as a function from $\mathbb R^m$ to $\mathbb R^n$.

They then define a notion of a smooth map between arbitrary subsets of manifolds:

If $f\colon X \to Y$ is a function whose domain and range are subsets of manifolds $X \subset M$ and $Y \subset N$, respectively, $f$ is said to be smooth if for all $x\in X$ there is an open set $U\subset M$ with $x\in U$ and a smooth function $F\colon U\to N$ such that $F(p) = f(p)$ for all $p \in U \cap X$.

The funny thing is, they never defined a smooth map on an open subset of a manifold, even though they use this notion in their definition of smooth maps between arbitrary subset of manifolds! So my question is, how are we supposed to define a smooth map on an open subset of a manifold? Are we supposed to give the open set the structure of a manifold by giving it the subspace topology and forming an atlas on it from the set of all charts in the initial atlas contained in the open subset?

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    $\begingroup$ Yes, your last paragraph is correct. $\endgroup$ – user98602 Feb 25 '16 at 1:34
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Let $M$ be a topological space and $U$ be a subset of $M$. If you want the set-theoretical inclusion map $\iota : U \longrightarrow M$ to be continuous, the subspace topology is the smallest one you need. In the case of the manifold and its open set, if you want $\iota$ to be a morphism in this category ($C^0$, $C^r$, $C^\infty$, complex, etc.), the extra condition you will find is that the atlas of $M$ and that of $U$ are "compatible". Note you cannot just collect all the members of the atlas which are contained in $U$ to form the atlas of $U$. Those charts may simply be too large. Take $S^n$ with only two charts as an example. The canonical way is to collect all the pairs of the form $(U \cap V, \varphi \big|_U)$ to form the atlas, where $(V, \varphi)$ is a chart of $M$.

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You are suggesting the right idea. Recall that if $f: U \to V$ is a smooth map between vector spaces then $f \in C^{\infty}(U,V)$. Now let $M^n,N^n$ be manifolds and suppose $f: M^n \to N^n$ is a smooth map. Here saying $f$ is $C^{\infty}$ doesn't make since in the general manifold setting. Instead we look at charts on $M,N$ say $(\phi, U)$ and $(\psi,V)$ where $U,V$ are open subsets of $M^n,N^n$ respectively.

Then we say $f$ is smooth if;

$$\phi^{-1} \circ f \circ \psi^{-1}: \phi(U) \subset \mathbb{R}^n \to \phi(V) \subset \mathbb{R}^n$$

Now we have a map between vector spaces and so our $C^{\infty}$ definition makes complete sense. In the case of takin $U$ to be an open subset of $M$ then the definition naturally holds, just for every $\{(\phi_i,U_i)\}$ in your original atlas on $M$ take the canonical atlas $\{(\phi_i, U_i \cap U)\}$. I believe this is what you were suggesting, I just wanted to formalize it just in case there was still some confusion.

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