How do I show that $df_p$ is a linear map?

Question

Let $n \in \mathbb{N}$ and $\phi \neq U \subseteq \mathbb{R}^n$. For any $p \in U$, one has the following isomorphism: $$T_p(U) \to \mathbb{R}^n, [\gamma] \mapsto \gamma'(0), \forall [\gamma] \in T_p(U)$$ We denote the preimages of the standard basis vectors $e_i$ of $\mathbb{R}^n$ under this isomorphism by $\left.\frac{\partial}{\partial x_i}\right\vert_p \forall\text{ } 1\leq i\leq n$. Suppose $m \in \mathbb{N}$ and $V \subseteq \mathbb{R}^m$ is open and $f: U\to V$. The $i^{th}$ coordinate of $f$ is denoted by $f_i\text{ }\forall\text{ } 1\leq i\leq m$. Assume that $\frac{\partial f_i}{\partial x_j}$ is continuous for all $i,j$.

A. Show that, $\forall p \in U$, $df_p$ is a linear map: $$df_p:T_p(U) \to T_{f(p)}(V), [\gamma] \mapsto [f \circ \gamma], \forall \gamma \in T_p(U)$$ B. For any $p \in U$, find the matrix of $df_p$ relative to the ordered bases $\{\left.\frac{\partial}{\partial x_1}\right\vert_p, .., \left.\frac{\partial}{\partial x_n}\right\vert_p \}$ and $\{\left.\frac{\partial}{\partial x_1}\right\vert_{f(p)},..,\left.\frac{\partial}{\partial x_m}\right\vert_{f(p)} \}$

That was long! Here's what I did and where I am stuck:

1. So I understand that $T_p(U)$ is the set of all vectors starting at $p \in U$, which are tangent to $U$ at $p$. Is that right? I haven't done topology or analysis rigorously yet, and this is supposed to be an exercise in linear maps.
2. It makes intuitive sense that $f$ maps the entire subset $U \to V$ and $p \mapsto f(p)$, so if a vector is a tangent to $U$ at $p$, similar properties must hold after passing through the map $f$ as well. How do I prove this rigorously though?
3. I'm not sure I really understand what's happening here: $T_p(U) \to T_{f(p)}(V)$. Could someone elaborate? Is it the same as saying that "$df_p$ is a map from the set of vectors tangent to $U$ at $p \in U$ to the set of vectors tangent to $V$ at $f(p) \in V$"?
4. I'm pretty unsure whether or not I understand $T_p(U)$ correctly, as I mentioned earlier, so I'll just point out that it sounds weird to think about vectors tangent to a subset of $\mathbb{R}^n$ at a point $p$. Does this make sense? If yes, could someone give more details? If not, correct me, please?
5. Seems like $f$ isn't necessarily linear. Why is that okay?
6. Once all the above clarifications are made, how should I proceed to show parts (A) and (B) defined above? I just don't know where to start! I think I should be able to find the matrix in (B) once I understand $df_p$ better while solving (A), so I'd appreciate any help!

To sum it up, please help me understand what $df_p$ is, and is doing! (also why is it linear?) Thank you!

Answer 1

So I understand that $T_p U$ is the set of all vectors starting at $p\in U$, which are tangent to $U$ at $p$. Is that right? I haven't done topology or analysis rigorously yet, and this is supposed to be an exercise in linear maps.

You are right, but this is only the interpretation of what it seems to be your definition of the tangent space. It appears that your definition is the following: $$T_p U = \{\gamma\colon (-\varepsilon, \varepsilon)\to U; \gamma\,(0) = p\text{ and } \gamma \in C^1\}/{\sim}$$ Where $\alpha\sim\beta $ iff $\alpha'(0) = \beta'(0)$ (here it should be a little different, but since you are in $\mathbb R^n$ this is ok). That is why there is this isomorphism between $T_p(U)$ and $\mathbb R^n$. The idea behind the construction is: Identify the tangent space at a point with the space of all possible velocities of curves at that point. For an open set in $\mathbb R^n$ it may not seem the best definition, but if try to generalize this concept for a sphere, for example, it does make more sense.

It makes intuitive sense that $f$ maps the entire subset $U\to V$ and $p\mapsto f(p)$, so if a vector is a tangent to $U$ at $p$, similar properties must hold after passing through the map $f$ as well. How do I prove this rigorously though?

Here, once more, the problem you had was the definition of the tangent space. The item A of the problem tells you exactly how the differential of $f$ acts on a tangent vector, we have that $df(p)[\gamma] = [f\circ \gamma]\in T_{f(p)} V$, since $f\circ\gamma$ is a curve on $V$.

I'm not sure I really understand what's happening here: $T_p U\to T_{f(p)} V$. Could someone elaborate? Is it the same as saying that "$df(p)$ is a map from the set of vectors tangent to $U$ at $p\in U$ to the set of vectors tangent to $V$ at $f(p)\in V$"?

Yes, this is what the differential is doing. It takes a curve (which you should think as a particle with a given tangent velocity at p) and it give us the composition of $f$ and the curve (which you should think as the resulting velocity of the particle after you apply $f$)

I'm pretty unsure whether or not I understand $T_p U$ correctly, as I mentioned earlier, so I'll just point out that it sounds weird to think about vectors tangent to a subset of $\mathbb R^n$ at a point $p$. Does this make sense? If yes, could someone give more details? If not, correct me, please?

It is a little bit odd when you do it to an open set of $\mathbb R^n$, but you should think of this more like a definition the makes sense in more general setting (like manifolds for instance), and this particular case is just for you to get used to the definitions, in a familiar context.

Seems like $f$ isn't necessarily linear. Why is that okay?

When you think as velocity of a curve, it is usually not linear right? You don't need the linearity in $f$, you just need the linearity when f acts in the velocities vectors

For part A, I think you only have to check again what the definitions are of the tangent space, and it's operations.

For part B you should have in mind what those $\dfrac{\partial}{\partial x_i}$ are, in fact they are just the equivalence classes of the curves $\alpha_i(t)= p +te_i$

Answer 2

Your intuition for 3 is correct. Notice that $f: U \rightarrow V$ sends $p \mapsto f(p)$ and this is why the new base point is $f(p)$. I will however add that it is not just a set, it is a real vector space. In fact, as your opening line remarks, that this is just isomorphic to $\mathbb{R}^n$ for some $n \in \mathbb{N}$.

A nice (not as trivial) example is $f : S^2 \rightarrow S^2$, the map which sends the unit sphere to itself by rotating 180 degrees vertically. Consider the tangent plane to $S^2$ at the north pole, $N$. This is exactly $T_N(S^2) \simeq \mathbb{R}^2$. In fact, the tangent plane to any point in $S^2$ is isomorphic to $\mathbb{R}^2$. Now, for any map $f$ between manifolds with base point fixed (in our case $N \mapsto f(N) = S$, the south pole), there is an induced push-forward map between their respective tangent spaces at said base points. Namely, in our case $df_N : T_N(S^2) \rightarrow T_S(S^2) $ which is nothing else but an isomorphism of $\mathbb{R}^2 \to \mathbb{R}^2$.

For 4, just take your manifold to be $U$. This makes perfect sense as now the tangent plane to $\mathbb{R}^n$ is essentially itself.

For 6, the linearity of $df_p$ means $\mathbb{R}$-linear. So you must show that $df_p(av + w) = a df_p(v) + df_p(w)$ for any $a \in \mathbb{R}$ and any $v,w \in T_p(U)$.