Jae Choon Cha
POSTECH
The remaining part of this semester will focus on differentiable techniques in topology. We begin with the notion of smooth manifolds, which comes with tangent spaces.
Definition of smooth manifolds
To define smooth manifolds, we need to use smooth maps. Briefly speaking, "smooth" means that all partial derivatives exist. The easiest case to formulate this is when the domain and codomain of the concerned map are open subsets in Euclidean spaces, as in the following definition.
Definition. Suppose $U$ and $V$ are open subsets in $\R^n$ and $\R^m$ respectively. For a map $f\colon U\to V$, let $f_i$ be the $i$th coordinate function ($i=1,\ldots,m$). That is,
We say that $f$ is smooth if all partial derivatives of arbitrary order
exists and are continuous for each $f_i$.
More generally, suppose $X\subset \R^n$ and $Y\subset \R^m$ are just subsets, not necessarily open. In this case, a map $f\colon X\to Y$ is called smooth if the following holds: for every $x\in X$, there is an open neighborhood $W$ of $x$ in $\R^n$ and a smooth map $g\colon U\to \R^m$ such that $g_ {W\cap X} = f_ {W\cap X}$. (Here, the smoothness of $g$ is defined as above, since the domain $U$ is an open subset of $\R^n$.)
Note that a composition of smooth maps is smooth.
A map $f\colon X\to Y$ is called a diffeomorphism if $f$ is a smooth map which has a smooth inverse $f^{1}\colon Y\to X$.
Definition. A subspace $M\subset \R^k$ is a smooth manifold of dimension $n$ if for every $x\in M$, there is a neighborhood $V$ of $x$ in $M$ and a diffeomorphism $f\colon V\to U$ onto an open subset $U$ in $\R^n$.
The above $f\colon V\to U$ is called a coordinate system. Its inverse $f^{1}\colon U\to V$ is called a parametrization.
Example: $S^n$ is a smooth manifold of dimension $n$.
Recall the definition $S^n=\{x\in \R^{n+1}\mid x_1^2+\cdots+x_n^2=1\}$. Define
by $\phi(x_1,\ldots,x_n) = (x_1,\ldots,x_n,\sqrt{1(x_1^2+\cdots+x_n^2)})$. This is smooth, and has smooth inverse $\phi^{1}\colon U\to \R^n$ which is defined by $\phi^{1}(x_1,\ldots,x_n,x_{n+1}) = (x_1,\ldots,x_n)$. (Verify that this is smooth, using the above definition!) Therefore $\phi$ is a diffeomorphism, and thus $\phi$ is a parametrization. Exchanging the roles of the coordinates $x_i$ and $x_{n+1}$, we obtain more parametrizations, and the images of these $n+1$ parametrizations cover $S^n$. It follows that $S^n$ is a smooth manifold of dimension $n$.
Derivatives of maps on Euclidean spaces
Recall that the first order partial derivatives of a multivariable vectorvalued function form a linear map, which we call the derivative. More precisely, for a smooth map $f\colon U\to V$ with $U\subset \R^k$ and $V\subset \R^\ell$ open, and for each point $x\in U$, the map $df_x\colon \R^k \to \R^\ell$ defined by
is a linear map, which is represented by the matrix
The following are some basic properties:

The chain rule holds: $d(g\circ f)_ x = dg_ {f(x)} \circ df_x$.

For the identity map $I\colon U\to U$, $dI_x$ is the identity (matrix) for each $x\in U$.

If $f\colon \R^k \to \R^\ell$ is a linear transformation, then $df_x = f$ for all $x\in \R^k$.
The following is an immediate consequence of the above properties (1), (2).
Theorem. Suppose $U\subset \R^k$ and $V\subset \R^\ell$ are open. If $f\colon U\to V$ is a diffeomorphism, then $df_x \colon \R^k \to \R^\ell$ is a linear isomorphism for every $x\in U$. In particular, $k=\ell$.
Proof. Let $f^{1}\colon V\to U$ be a smooth inverse of $f$. Let $x\in U$ and $y=f(x)$. Then, since $I=f^{1}f$ and $I=ff^{1}$, we obtain that $d(f^{1})_y \circ df_x$ and $df_x \circ d(f^{1})_y$ are equal to the identity, by applying (1) and (2) above. It follows that both $df_x$ and $d(f^{1})_y$ are isomorphisms. $\quad\square$
A partial converse of this is the wellknown inverse function theorem:
Inverse function theorem. For a smoooth map $f\colon U\to V$ with $U\subset \R^k$, $V\subset \R^\ell$ open, if $df_x$ is nonsingular, then there exist open neighborhoods $U'$ of $x$ in $U$ such that $V'=f(U')$ is an open neighborhood of $f(x)$, and $f_{U'} \colon U' \to V'$ is a diffeomorphism.
Tangent spaces and derivatives of smooth maps on manifolds
Note that the above discussion on the derivatives is limited to maps defined on an open neighborhood of the Euclidean space. To generalize this to the case of smooth maps on smooth manifolds, we first define tangent spaces.
Definition. Suppose $M\subset \R^k$ is a smooth manifold of dimension $n$, and $g\colon U\to M$ is a parametrization, where $U\subset \R^n$ is open. Let $x=g(p)$ for some $p\in U$. We call $TM_x = dg_p(\R^n)$ the tangent space of $M$ at $x$.
Note that $TM_x$ is a linear subspace of the vector space $\R^k$ when $M$ is contained in $\R^k$.
We list some basic properties:

$TM_x$ is determined by $x$, independent of the choice of the parametrization $g$ used in the definition. Indeed, if $h\colon V\to M$ is another parametrization with $h(q)=x$, then $U'=g^{1}(g(U)\cap h(V))$ and $V'=h^{1}(g(U)\cap h(V))$ are open subsets, and $h^{1}g_{U'}\colon U'\to V'$ is a diffeomorphism. So $(dh_q)^{1} dg_p$ is an isomorphism. It follows that $dh_q(\R^n) = dh_q((dh_q)^{1} dg_p(\R^n)) = dg_p(\R^n)$.

$TM_x$ is an $n$dimensional vector space if $M$ has dimension $n$. To prove this, take a parametrization $g\colon U\to M\subset \R^k$ with $g(p)=x$. Then by definition, $g^{1}$ is smooth. This implies that there is a smooth map $F\colon W\to \R^n$ which is defined on an open neighborhood $W$ of $x$ in $\R^k$ and satisfies $F_ {W\cap g(U)} = g^{1}_ {W\cap g(U)}$. Let $U' = g^{1}(g(U)\cap W)$. Then, $F\circ g_{U'}$ is equal to the identity on $U'$. It follows that $dF_x dg_p = I$, so $dg_p$ is 11. Therefore $TM_x = dg_p(\R^n)$ has dimension $n$.
Now, we generalize the notion of derivative to the case of smooth maps defined on manifolds, not only for those on open subsets of Euclidean spaces.
For a smooth map $f\colon M\to N$ with $M\subset \R^k$ and $N\subset \R^\ell$ smooth manifolds and for $x\in M$, define
by $df_x(v) = dF_x(v)$, where $F\colon W\to \R^\ell$ is a smooth map defined on an open neighborhood $W$ of $x$ in $\R^k$ satisfying $F_ {W\cap M} = f_{W\cap M}$. Note that such $F$ exists by the definition of the smoothness of $f$.
To show that $df_x$ is welldefined, we first claim that $df_x(v)\in TN_{f(x)}$. To show this, choose parametrizations $g\colon U\to M$ and $h\colon V\to N$ with $U\subset \R^m$ and $V\subset \R^n$ open, $g(p)=x$, $h(q)=f(x)$. We may assume $g(U)\subset W$ and $f(g(U))\subset h(V)$. Then, $Fi=f$, so $Fg=fg=h(h^{1}fg)$. It follows that $dF_x dg_p = dh_q d(h^{1}fg)_p$. Applying this to $\R^n$, we have $dF_x(TM_x) \subset TN_{f(x)}$.
Second, we claim that $df_x(v)$ is independent of the choice of $F$. Indeed, it follows from the above: we have $dF_x = dh_q d(h^{1}fg)_ p (dg_p)^{1}$ on $TM_x=dg_p(\R^n)$.
The above properties of derivatives generalize to this more general context:

Chain rule: if $M\xrightarrow{f} N \xrightarrow{g} P$ are smooth maps of manifolds, then $d(g\circ f)_x = dg_{f(x)}\circ df_x$.

For the identity map $I\colon M\to M$, $dI_x$ is the identity on $TM_x$. More generally, if $M\subset N$, then for the inclusion $i\colon M\to N$, $di_x$ is the inclusion $TM_x \to TN_x$.
Theorem. If $f\colon M\to N$ is a diffeomorphism of smooth manifolds, then $df_x\colon TM_x \to TN_{f(x)}$ is an isomorphism for all $x\in M$. Moreover, $M$ and $N$ has the same dimension.
Proof. The first statement is an immediate consequence of the above properties 1 and 2. For the second conclusion, recall that the dimension of a manifold is equal to the dimension of the tangent space. Since $TM_x \cong TN_{f(x)}$, $TM_x$ and $TN_{f(x)}$ have the same dimension, so $M$ and $N$ have the same dimension. $\quad\square$