Exercise 2.2

Question

Let $U$ be an open submanifold of $\mathbb{R}^n$ with its standard smooth manifold structure. Show that a function $f: U \rightarrow \mathbb{R}^k$ is smooth in the sense just defined if and only if it is smooth in the sense of ordinary calculus. Do the same for an open submanifold with boundary in $\mathbb{H}^n$ (see Exercise 1.44).

Elu Thingol · Accepted Answer

\paragraph{Manifolds without boundary.} By Example/Definition 1.22, the smooth structure of the Euclidean space $\mathbb{R}^n$ is the maximal atlas $\mathcal{A}$ generated by the single chart $(\mathbb{R}^n,\mathrm{Id})$. By Example/Definition 1.26, the smooth structure of $U$ is determined by the atlas
\begin{equation} \label{submanifold_atlas}
    \mathcal{A}_U = \left\{ (V,\psi) \in \mathcal{A} \mid V \subseteq U ight\}.
\end{equation}

Since $(U,\operatorname{Id}|_{U})$ is smoothly compatible with $(\mathbb{R}^n,\mathrm{Id}_{\mathbb{R}^n})$ and therefore is contained $\mathcal{A}$, it must also be contained in $\mathcal{A}_U$. 
For all $p \in U$, the global chart $(U,\mathrm{Id|}_{U})$ suffices when demonstrating that the coordinate representation
$$f \circ \mathrm{Id}|_U^{-1}: \operatorname{Id}(U) 	o \mathbb{R}$$
is smooth. But $f \circ \mathrm{Id}|_U^{-1}:\operatorname{Id}(U) 	o \mathbb{R}$ is the same thing as $f:U 	o \mathbb{R}$, and so both conditions for smoothness are equivalent.

\paragraph{Manifolds with boundary.} Recall that $\mathbb{H}^n$ is a smooth $n$-manifold with boundary and $U$ is an open subset of $M$. By \href{./exercise_1-44}{Exercise 1.44}, $U$ is a topological $n$-manifold with boundary and its smooth structure is determined by the atlas defined analogously to that in ef{submanifold_atlas}. The rest of the proof is practically the same.

Exercise 2.2

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Exercise 2.2

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