# Is the differential between tangent bundles $F_*: TN \to TM$ smooth?

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:

1. If $$F$$ were smooth but not a smooth embedding, then is $$F_*$$ no longer necessarily smooth?

2. What if $$F$$ were smooth and injective?

3. What if $$F$$ were a smooth non-injective local diffeo?

4. What if $$F$$ were a smooth non-injective immersion but not local diffeo?

5. What if $$F$$ were a smooth injective immersion but not a topological embedding (My understanding is smooth embedding = smooth injective immersion + topological embedding)?

6. 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 $$F_*$$?

• Any derivative of a $C^{\infty}$ function is $C^{\infty}$ pretty much by definition. Using this you can deduce that locally the differential is smooth, and hence globally. – leibnewtz Jun 20 at 13:41
• @leibnewtz Thanks! Your second sentence refers to guide question (6)? – Selene Auckland Jun 20 at 13:48
• @leibnewtz Wait so to clarify, even if $F$ were not an embedding, $F_*$ is still smooth? – Selene Auckland Jun 20 at 13:49
• Yup. The map $F_*$ is always smooth as long as $F$ is – leibnewtz Jun 20 at 14:02
• @leibnewtz Thanks! Also, good intuition in your first sentence in your first comment. – Selene Auckland Jun 20 at 14:06

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 $$F_*$$.
• I was expecting just some link or reference to a textbook since I didn't really show effort into understanding how $TM$ and $TN$ are smooth manifolds in the first place. Thanks! – Selene Auckland Jun 20 at 13:53
• Careful — the linear transformation varies smoothly in the $x$ variables; your notation suggests that it is a fixed linear map. – Ted Shifrin Jun 20 at 17:14