$f:M\to N$ smooth manifold map. $F(x)=(x,f(x))\in M\times N$. For each $X\in \mathfrak{X}(M)$ there's a F-related $Y\in \mathfrak{X}(M\times N)$. Let $M$ and $N$ be to manifolds and $f:M\to N$ smooth map. Define $F:M\to M\times N$ by $F(x)=(x,f(x))$. Show that for each $X\in \mathfrak{X}(M)$ there's a F-related $Y\in \mathfrak{X}(M\times N)$.
I can't see how that is true, as if $|N|>1$, then $F$ cannot be surjective. How can one decide what what vector to assign to the points outside the image of $F$?
 A: Take $p\in M$. Let $x_1,\dots,x_m$ be coordinates around $p$ and $y'_{1},\dots,y'_{n}$ around $f(p)$. Note that for the coordinate functions of $F$ we have that $$F^i=\left\{\begin{array}{ll}x_i,& \textrm{if }i\leq m\\ f^{i-m}, & \textrm{otherwise}. \end{array}\right.$$
Also, $$\frac{\partial x^j(p)}{\partial x_i}=\delta_i^j.$$
Set $y_i$ the coordinates around $(p,f(p))$ by $$y_i=\left\{\begin{array}{ll}x_i,& \textrm{if }i\leq m\\ y'_{i-m}, & m<i\leq n. \end{array}\right.$$
 Then, we have
$$F_*\left(\left.\frac{\partial}{\partial x_i}\right|_p\right)=\sum_j\frac{\partial F^j(p)}{\partial x_i}.\left.\frac{\partial}{\partial y_j}\right|_{(x,f(x))}=$$ $$\left.\frac{\partial }{\partial x_i}\right|_{(x,f(x))}+\sum_{j=1}^n\frac{f^j(p)}{\partial x_i}.\left.\frac{\partial}{\partial y'_j}\right|_{(x,f(x))}=$$
$$\left.\frac{\partial }{\partial x_i}\right|_{(x,f(x))}+f_*\left(\left.\frac{\partial}{\partial x_i}\right|_p\right).$$
Then, by linearity $$F_*(X)=X\oplus f_*(X).$$
So, $F_*(X)\in \mathfrak{X}(M\times N)$, as wanted.
