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\marginnote[-5em]{Using Einstein's notation, this could be written as \begin{equation}\frac{\partial (g^i\circ f)}{\partial x^j}(x) = \frac{\partial g^i}{\partial y^r}(f(x)) \frac{\partial f^r}{\partial x^j}(x).\end{equation}}
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Theorem~\ref{thm:chainrule} has some very deep consequences.
\textit{\small Hint: is $Df(x)$ an invertible matrix? If so, what is its inverse?} %$D(f^{-1})(f(x))$.
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\end{exercise}
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Since differentiability is a \emph{local} property and topological manifolds are \emph{locally} like euclidean spaces, it seems reasonable to expect that we can lift the definitions directly from $\R^n$.
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If we are given a continuous map between two topological manifolds, we can locally view it as a continuous map between two Euclidean spaces.
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Since differentiability is a \emph{local} property and topological manifolds are \emph{locally} like euclidean spaces, it seems reasonable to expect that we can lift the definitions directly from $\R^n$ using the charts to obtain functions between euclidean spaces:
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for example, if we are given a continuous map between two topological manifolds, we can locally view it as a continuous map between two Euclidean spaces.
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Generalizing this further, we could conceivably say that our original map is differentiable if the local map is.
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\newthought{As usual, the devil is in the details}: a topological manifold is only homeomorphic to a Euclidean space, and a different choice of homeomorphism might affect whether the local map is differentiable or not.
Let $N$ denote the north pole $(0,\ldots,0,1)\in\bS^n\subset\R^{n+1}$ and let $S$ denote the south pole $(0,\ldots,0,-1)$.
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Define the \emph{stereographic projections} $\sigma:\bS^n\setminus\{N\}\to\R^n$ by
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\begin{marginfigure}
@@ -588,7 +593,7 @@ \section{Smooth maps and differentiability}
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Before considering the general definition of a differentiable map, let's look at the simpler example of differentiable functions $f:M\to\R$ between a smooth manifold $M$ and $\R$.
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\begin{definition}
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A function $f:M\to\R$ from a smooth manifold $M$ of dimension $n$ to $\R$ is \emph{smooth}, or \emph{of class $C^\infty$}, if for any chart $(\varphi, V)$of$M$ the map $f\circ\varphi^{-1}:\varphi(V)\subset\R^n \to\R$ is smooth as a euclidean function.
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A function $f:M\to\R$ from a smooth manifold $M$ of dimension $n$ to $\R$ is \emph{smooth}, or \emph{of class $C^\infty$}, if for any smooth chart $(\varphi, V)$for$M$ the map $f\circ\varphi^{-1}:\varphi(V)\subset\R^n \to\R$ is smooth as a euclidean function on the open subset $\varphi(V)\subset\R^n$.
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\begin{marginfigure}
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\includegraphics{1_5-diff-fun-v2.pdf}
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\label{fig:diff-fun}
@@ -617,13 +622,13 @@ \section{Smooth maps and differentiability}
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\begin{enumerate}[(i)]
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\item$f\in C^\infty(M)$;
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\item$M$ has an atlas $\cA$ such that for every chart $(U, \varphi)\in\cA$, $f\circ\varphi^{-1} : \R^n\supset\varphi(U)\to\R$ is $C^\infty$;
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\item for every chart $(V,\psi)$on$M$, the function $f\circ\psi^{-1} : \R^n\supset\psi(U)\to\R$.
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\item for every point $p\in M$, there exists a smooth chart $(V,\psi)$for$M$ such that $p\in V$ and the function $f\circ\psi^{-1} : \R^n\supset\psi(V)\to\R$ is $C^\infty$ on the open subset $\psi(V)\subset\R^n$.
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\end{enumerate}
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\end{proposition}
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\begin{exercise}
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Prove the proposition.\\
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\textit{\small Hint: go cyclic, for example show $(ii)\Rightarrow(i)$, $(i)\Rightarrow(iii)$, $(iii)\Rightarrow(ii)$.}
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\textit{\small Hint: go cyclic, for example show $(i)\Rightarrow(ii)$, $(ii)\Rightarrow(iii)$, $(iii)\Rightarrow(i)$.}
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\end{exercise}
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At this point, the generalization of smooth functions to smooth maps between manifolds should not come as a surprise.
@@ -651,15 +656,19 @@ \section{Smooth maps and differentiability}
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\label{fig:1.3-differentiable_maps}
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\end{figure}
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A first observation about our definition of smooth maps is that as one would hope, smooth imply continuity.
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For a very simple and familiar example, consider the real valued function $f(x,y)= x^2+y^2$ defined on $\R^2$.
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In polar coordinates on $U=\{(x,y)\in\R^2\mid x>0\}$, $f$ has the coordinate representation $\hat f (\rho, \theta) = \rho^2$.
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Very often, where there is no ambiguity, we will simply identify $f$ and $\hat f$ and just write ``in the local coordinates $(\rho,\theta)$ on $U$, $f(\rho,\theta) = \rho^2$.''
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A first observation about our definition of smooth maps is that as one would hope, smoothness implies continuity.
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\begin{exercise}
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Show that every smooth map is continuous.
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\end{exercise}
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\begin{definition}
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A \emph{diffeomorphism} $F$ between two smooth manifolds $M_1$ and $M_2$ is a bijective map such that $F\in C^\infty(M_1, M_2)$ and $F^{-1}\in C^\infty(M_2, M_1)$.
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%
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Two smooth manifolds $M_1$ and $M_2$ are called \emph{diffeomporphic} if there exists a diffeomorphism $F:M_1\to M_2$ between them.
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\end{definition}
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\end{proposition}
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\end{exercise}
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\begin{exercise}
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\begin{exercise}[\textit{[homework 1]}]
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Prove that $\R^2\setminus\{(0,0)\}$ is a two-dimensional manifold and construct a diffeomorphism from this manifold to the circular cylinder
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\begin{equation}
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C := \{ (x,y,z)\in\R^3 \mid x^2+y^2 = 1\}\subset\R^3.
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Why is $\varphi_1$ not appearing in $\partial\cC$?
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\end{exercise}
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\begin{exercise}
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\begin{exercise}[\textit{[homework 1]}]
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Let $M = D_1\subset\R^n$ be the $n$-dimensional closed unit ball from Example~\ref{ex:uball}.
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Show that $M$ is a topological manifold with boundary in which each point of $\mathring M = \bS^{n-1}$ is a boundary point and each point in $\{x\in\R^n\mid\|x\|<1\}$ is an interior point.
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Give a smooth structure to $M$ such that every smooth interior chart is a smooth chart for the standard smooth structure on $\mathring M$.\\
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\textit{\small Hint: consider the map $\pi\circ\sigma^{-1}:\R^n\to\R^n$ where $\sigma:\bS^n\to\R^n$ is the stereographic projection from Exercise~\ref{ex:stereo} and $\pi:\R^{n+1}\to\R^n$ is a projection that omits one of the first $n$ coordinates.}
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\begin{enumerate}[(a)]
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\itemShow that $M$ is a topological manifold with boundary in which each point of $\mathring M = \bS^{n-1}$ is a boundary point and each point in $\{x\in\R^n\mid\|x\|<1\}$ is an interior point.
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\item Give a smooth structure to $M$ such that every smooth interior chart is a smooth chart for the standard smooth structure on $\mathring M$.\\
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\textit{\small Hint: consider the map $\pi\circ\sigma^{-1}:\R^n\to\R^n$ where $\sigma:\bS^n\to\R^n$ is the stereographic projection from Exercise~\ref{ex:stereo} and $\pi:\R^{n+1}\to\R^n$is a projection that omits one of the first $n$ coordinates.}
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