TITLE: Is the differential between tangent bundles $F_*: TN \to TM$ smooth?
QUESTION [3 upvotes]: Let $F: N \to M$ be a smooth map of smooth manifolds (with dimensions). Let $F_{*,p}: T_pN \to T_{F(p)}M$ be the differential at $p \in N$. Let $F_*: TN \to TM$ be the map between tangent bundles given by $F_*(X_p)=F_{*,p}(X_p)$. This says $F_*$ is a smooth embedding if $F$ is a smooth embedding.
What are some sufficient conditions to say $F_*$ is smooth besides $F$ being a smooth embedding?

I'm not really interested in deducing $F_*$ to be a smooth embedding or topological embedding. I'm just hoping for smooth for 1.4 here.
Some guide questions:

If $F$ were smooth but not a smooth embedding, then is $F_*$ no longer necessarily smooth?
What if $F$ were smooth and injective?
What if $F$ were a smooth non-injective local diffeo?
What if $F$ were a smooth non-injective immersion but not local diffeo?
What if $F$ were a smooth injective immersion but not a topological embedding (My understanding is smooth embedding = smooth injective immersion + topological embedding)?
I think each $F_{*,p}$ is smooth as a map of manifolds, besides linear as a map of vector spaces. What does this mean for $F_*$?

REPLY [5 votes]: It is still smooth. If $F\in \mathscr{C}^\infty(M,N)$, then fixing $p\in M$ and local coordinates $(x^1,\ldots, x^n)$  centred at $p$ on a trivializing neighborhood $U\subseteq M$ for $TM\to $M, and fixing analogous coordinates on $(y^1,\ldots, y^m)$ centred at $F(p)$ on a neighborhood $V\subseteq N$ containing $F(U)$ trivializing  $TN\to N$, we can write down $F_*: TM\to TN$ in local coordinates as a map $TU\to TV$. 
In local coordinates, $F$ is given by an $m-$tuple of smooth functions, $y^i=F_i(x^1,\ldots, x^n)$ for $1\le i \le m$. And given the local trivialization condition we can view $TU\cong U\times \mathbb{R}^n$ and $TV\cong V\times \mathbb{R}^m$. Then $F_*:U\times \mathbb{R}^n\to V\times \mathbb{R}^m$ is $F\times L$ where 
$$L|_{\{x\}\times \mathbb{R}^n}=L_x:\{x\}\times\mathbb{R}^n\to \{F(x)\}\times\mathbb{R}^m$$
is a linear transformation and the transformations $L_x$ vary smoothly according to the choice of $x\in U$. So, denoting the variable in $U$ by $x$ and the variable in $\mathbb{R}^n$ by $y$, $F_*$ can be  viewed as a map $F_*(x,y)=(F(x),L_x(y))$. All the components  are smooth, and hence so is $F_*$.
{"set_name": "stack_exchange", "score": 3, "question_id": 3268623}