The tangent to a smooth map.

If $f \colon M \to N$ is a smooth map and $\gamma \colon (-\epsilon, \epsilon) \to M$ is a smooth path through $x$ then $f \circ \gamma$ is a smooth path in $N$ through $f(x)$. Moreover if we consider another path $\rho$ which is tangent to $\gamma$ then $f \circ \gamma$ and $f\circ \rho$ are tangent. To see this choose co-ordinates $(U, \psi)$ and $(V, \chi)$ with $f(U) \subset V$. Assume without loss of generality that $\gamma(-\epsilon, \epsilon)$ and $\rho(-\epsilon, \epsilon)$ are in $U$. Then we have

$$\chi\circ (f\circ\gamma)'(0) = d(\chi\circ f \circ \psi^{-1})(\psi(x)) (\psi\circ\gamma)'(0)$$

and

$$\chi\circ (f\circ\rho)'(0) = d(\chi\circ f \circ \psi^{-1})(\psi(x)) (\psi\circ\rho)'(0)$$

so that $(\psi\circ\rho)'(0) = (\psi\circ\gamma)'(0)$ implies that $\chi\circ (f\circ\rho)'(0) = \chi\circ (f\circ\gamma)'(0)$ and hence $f \circ \gamma$ and $f\circ \rho$ are tangent. So associated with $f$ there is a well-defined map from $T_xM$ to $T_{f(x)}N$ that sends $\gamma'(0)$ to $(f \circ \gamma)'(0)$. This map is denoted $T_xf$ and called the tangent to $f$ at $x$. So we have that

$$T_x(f)(\gamma'(0)) = (f\circ\gamma)'(0).$$

Notice that the tangent map satisfies

$$T_x(f) = d\chi(f(x))^{-1} \circ T_x(f) \circ d\psi(x).$$

so that, being a composition of three linear maps it is, itself linear. Moreover this formula also shows that with respect to the bases of $T_xM$ and $T_{f(x)}N$ given by the co-ordinate vector fields we have

$$T_x(M)(\frac{\partial\phantom{\psi}}{\psi}(x))= \sum_{i=1}^n \fr... ...^j \circ f}{\partial \psi^i}(x) \frac{\partial\phantom{\chi^j}}{\chi^j}(f(x)).$$

In other words it is given by the action of a matrix whose entries are the partial derivatives of the co-ordinate expression for $f$.

Example 4.5   The tangent space to $\mathbb{R}^n$ at $\psi(x)$ is just $\mathbb{R}^n$ again. The map $d\psi(x) \colon T_xM \to \mathbb{R}^n$ is just the map $T_x\psi \colon T_xM \to T_{\psi(x)} \mathbb{R}^n$.

Example 4.6   If $F \colon \mathbb{R}^n \to \mathbb{R}^m$ is a smooth map then, after identifying $T_x \mathbb{R}^n$ with $\mathbb{R}^n$ and $T_{F(x)} \mathbb{R}^m$ with $\mathbb{R}^m$ we see that the tangent map $T_x(F)$ is just the matrix of partial derivatives $dF(x)$.

The chain rule for smooth functions in $\mathbb{R}^n$ generalises to manifolds as follows.

Proposition 4.3 (Chain Rule)   Let $f \colon M \to N$ and $g \colon N \to P$ be smooth functions. Then the map $g \circ f \colon M \to P$ is smooth and $T_x(g \circ f) = T_{f(x)}(g) \circ T_x(f)$.