If $F$ is smooth then the pushforward $F_*$ is smooth Question: Suppose $F:M\to N$ is a smooth map. Show that $F_*:TM\to TN$ is a smooth map.
We need to find charts $(U,\phi)\subset TM$ and $(V,\psi)\subset TN$ such that $\phi\circ F_*\circ\psi^{-1}$ is smooth. Since for any vector $r$, $\psi^{-1}(r)$ gives a derivation $X_r$, $F_*(\psi^{-1}(r))=F_*(X_r)$. And for any $f\in C^{\infty}(M)$, $F_*(X_r(f))=X(f\circ F)$. But this doesn't lead anywhere. Please help.
 A: In short: write it out in coordinates, where you use $\phi$ and the associated tangent vectors (perhaps you call these $\partial/\partial \phi_k$ or something like that) to build a coordinate system on TM. 
Details:
If $\phi : U \subset M \rightarrow {\mathbb R}^m$ is a coordinate chart, let's let $k = \phi^{-1}$. Then for any $(x_1, x_2, \ldots, x_m)$ in the image of $\phi$, a path of the form $t \mapsto (x_1, x_2, \ldots, t + x_i, \ldots x_m)$, is transformed, by $k$, to a path in $M$. Let's call the tangent vector to this path, at $t = 0$, the vector $v_i(x_1, ..., x_m)$. Then the vectors $v_1(x_1, \ldots, x_m) \ldots v_m(x_1, \ldots, x_m)$ form a basis for $T_p(M)$, where $p = k(x_1, \ldots, x_m)$. And thus every element of $T_p(M)$ can be written as a combination $\sum c_i v_i$. This implicitly defines a function $c_i$ on $T_p(M)$, for each $p$. 
With this in hand, we can say that the pair
$$(\phi, c)$$
constitutes a chart in the manifold $TM$; similarly, we can build a chart $(\psi, b)$ in $TN$. 
In this pair of coordinate charts, what's the map $F_{*}$ look like? It looks like $(F, E)$, where $E$ is the derivative of $\phi \circ F \circ \psi^{-1}$, i.e., the function whose value at $x = (x_1, \ldots, x_m)$ is a linear transformation that's the best local approximation of $\phi \circ F \circ \psi^{-1}$ at $x$. 
Because $F$ is smooth, this map is also smooth, so you're done.
Let me make this more explicit. Suppose that $M = N = \mathbb R$, and that the coordinates on $M$ and $N$ are called $x$ and $y$ respectively. Then a typical element of $T_p(M)$ is a vector tangent to $R$, which is a multiple $c[1]_p$ of the unit vector $[1]_p$ that points in the direction of increasing $x$; a similar description holds for $T_q(N)$. So coordinates on $TM$ are $(x, c)$, by which I mean that the vector $c[1]_x$ in $T_x M$ is sent to the pair $(x, c) \in \mathbb R^2$. Clear? 
OK, suppose $f(x) = x^2$. Then the element of $TM$ whose coords are $(x, c)$ is sent, by $DF$, to an element of $TN$ whose first coordinate is $y = x^2$ and whose second coordinate is $2xc$. Clearly $(x, c) \mapsto (x^2, 2xc)$ is a smooth map. 
The general case is just a working out of this argument for higher dimensions. 
