$F$-related vector fields I'm having a bit of difficulty understanding $F$-related vector fields. I think I understand conceptually what's going on (taking a vector field on a manifold $M$ and getting a smooth vector field on $N$ by applying the differential of a map), but I'm having trouble applying the concept.
For example, here's a problem (8.14 in Lee's Smooth Manifolds textbook)
 Let $M,N$ be smooth manifolds with or without boundary and let $f:M\to N$ be a smooth map. Define $F:M\to M\times N$ by $F(x)=(x,f(x))$. Show that for each vector field $X$ on $M$, there is a smooth vector field $Y$ on $M\times N$ which is $F$ related to $X$.
Now, I see that $F$ is the graph of $f$, and so $F(M)$ is closed in $M\times N$, and there's a proposition in Lee which says that if we have a smooth vector field on a closed subset, we can extend it to a smooth vector field on the whole manifold. However, I'm really not sure where to go. I see that $F(M)\subset M\times N$ is an embedded manifold, so we have slice charts, but I'm struggling to see how to construct this $Y$.
 A: Let $\Gamma=F(M)$. As mentioned, $\Gamma$ is closed in $M\times N$. So it suffices to show that the rough vector field on $Y$ along $\Gamma$, defined by the equation $$Y_{(p,f(p))}=dF_p(X_p),$$
is a smooth vector field along $\Gamma$(, that is, each point $(p,f(p))\in\Gamma$ has a neighborhood on which $Y$ admits a smooth extension). 
Let $p\in M$  be arbitrary, and let $(U,\varphi)$ and $(V,\psi)$ be smooth charts about $x$ and $f(x)$, respectively, such that $f(U)\subset V$. Let $X^i$ denote the component functions of $X$ in $U$:$$X=X^i\frac{\partial}{\partial x^i}\text{ on } U.$$
Note that $X^i:U\to \mathbb{R}$ is smooth because $X$ is smooth. Using these functions, we define $\tilde{Y}:U\times V\to T(M\times N)$ by$$\tilde{Y}_{(p,q)}=X^i(p)\frac{\partial}{\partial x^i}|_{(p,q)}+ \frac{\partial \hat{f}^i}{\partial x^j}(\varphi (p))\frac{\partial}{\partial y^i}|_{(p,q)},$$
where $\hat{f}=\psi\circ f\circ \varphi^{-1}:\varphi(U)\to\psi(V)$ is the coordinate representation of $f$. Because the component functions of $\tilde{Y}$ are smooth, $\tilde{Y}$ is smooth, too. Clearly $\tilde{Y}$ and $Y$ agrees on their common domain. Thus $\tilde{Y}$ is a desired extension.
