关于满足曲线导数对应性质的光滑映射微分的唯一性证明问询

Hi folks,

I'm currently learning about differentials of smooth maps between manifolds. I've already used Theorem 1 (the coordinate-based characterization of the differential) to prove that the linear map described in Theorem 2 exists, but I want to solidify my understanding of why this map is unique. I've worked up a proof attempt on my own, but I'd love to hear if it's rigorous, plus any alternative approaches to the uniqueness proof.

Background Definitions & Theorems #

First, let's recap the key definitions and results I'm working with:

1. Tangent Vectors
   Given an $m$-dimensional manifold $M$ and $p \in M$, a tangent vector at $p$ is a function
   $$
   v:{\text{charts of } M \text { around } p} \rightarrow \mathbb{R}^m, \quad \chi \mapsto v^\chi,
   $$
   such that for any two charts $\chi, \chi'$ around $p$,
   $$
   v^{\chi{\prime}} = \mathrm d c_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi], \tag{1.2}
   $$
   where $c_{\chi, \chi^{\prime}} :=\chi' \circ \chi^{-1}$ is the coordinate transition map. The set of all such tangent vectors forms the vector space $T_p M$.

2. Smooth Curves Through $p$
   Define the set of smooth curves starting at $p$ as:
   $$
   \operatorname{Curve}_p (M) := {\gamma: (-\epsilon, \epsilon) \to M \text{ smooth such that } \epsilon>0, \gamma (0)=p}.
   $$
   For any $\gamma \in \operatorname{Curve}_p(M)$, the derivative $\frac{\mathrm d \gamma}{\mathrm d t} (0)$ is an element of $T_p M$ (this follows from the chain rule).

3. Theorem 1 (Coordinate-Based Differential)
   Given a smooth map $f: M \rightarrow N$ and $p \in M$, there exists a unique linear map
   $$
   \mathrm d f_p: T_p M \rightarrow T_{f(p)} N, \quad v \mapsto \mathrm d f_p(v)
   $$
   such that for any chart $\chi$ on $M$ around $p$ and chart $\chi'$ on $N$ around $f(p)$,
   $$
   \left( \mathrm d f_p(v)\right)^{\chi{\prime}} = \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi] \quad \forall v \in T_p M . \tag{1.3}
   $$
   Here, $f_{\chi, \chi'} := \chi' \circ f \circ \chi^{-1}$ is the coordinate representation of $f$.

4. Theorem 2 (Curve-Based Differential)
   Given a smooth map $f: M \rightarrow N$ and $p \in M$, there exists a unique linear map
   $$
   \mathrm d f_p: T_p M \rightarrow T_{f(p)} N
   $$
   such that
   $$
   \mathrm d f_p \left(\frac{\mathrm d \gamma}{\mathrm d t}(0)\right) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) \quad \forall \gamma \in \operatorname{Curve}_p (M). \tag{1.4}
   $$

My Existence Proof #

To show existence, I used the linear map $F = \mathrm d f_p$ from Theorem 1, and verified it satisfies equation (1.4):

Fix a chart $\chi$ on $M$ around $p$ and $\chi'$ on $N$ around $f(p)$. Using (1.3) and the chain rule:
$$
\begin{align}
\left [ F \left ( \frac{\mathrm d \gamma}{\mathrm d t}(0) \right ) \right ]^{\chi'} &= \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) \left [ \frac{\mathrm d \gamma}{\mathrm d t}(0) \right ]^\chi \
&= \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) \circ \mathrm d (\chi \circ \gamma)(0)[1] \
&= \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi(p)) \circ \mathrm d (\chi \circ \gamma)(0)[1] \
&= \mathrm d (\chi' \circ f \circ \gamma) (0) [1] \
&= \left ( \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) \right )^{\chi'}.
\end{align}
$$
Since a tangent vector is uniquely determined by its coordinate representations across all charts, this shows $F\left(\frac{\mathrm d \gamma}{\mathrm d t}(0)\right) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0)$, so $F$ satisfies (1.4).

My Attempt at Uniqueness #

Here's the uniqueness proof I came up with:

Suppose there are two linear maps $F_1, F_2: T_p M \to T_{f(p)} N$ both satisfying (1.4). We need to show $F_1(v) = F_2(v)$ for all $v \in T_p M$.

First, recall a key fact: every tangent vector $v \in T_p M$ can be written as the derivative of some curve $\gamma \in \operatorname{Curve}_p(M)$. For example, take a local chart $\chi$ around $p$, define $\gamma^\chi(t) = \chi(p) + t v^\chi$, then $\gamma = \chi^{-1} \circ \gamma^\chi$ is a smooth curve through $p$ with $\frac{\mathrm d \gamma}{\mathrm d t}(0) = v$.

For such a $\gamma$, equation (1.4) gives:
$$
F_1(v) = F_1\left(\frac{\mathrm d \gamma}{\mathrm d t}(0)\right) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0),
$$
$$
F_2(v) = F_2\left(\frac{\mathrm d \gamma}{\mathrm d t}(0)\right) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0).
$$
But to make this rigorous, we need to confirm that if two curves $\gamma, \lambda \in \operatorname{Curve}_p(M)$ have $\frac{\mathrm d \gamma}{\mathrm d t}(0) = \frac{\mathrm d \lambda}{\mathrm d t}(0)$, then $\frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (f \circ \lambda)}{\mathrm d t}(0)$.

To verify this, fix local charts $\chi$ (around $p$ on $M$) and $\chi'$ (around $f(p)$ on $N$). From $\frac{\mathrm d \gamma}{\mathrm d t}(0) = \frac{\mathrm d \lambda}{\mathrm d t}(0)$, we know $\frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (\chi \circ \lambda)}{\mathrm d t}(0)$.

Now compute the coordinate representation of $\frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0)$:
$$
\begin{align}
\frac{\mathrm d (\chi' \circ f \circ \gamma)}{\mathrm d t}(0) &= \mathrm d (\chi' \circ f \circ \gamma)(0)[1] \
&= \mathrm d (\chi' \circ f \circ \chi^{-1} \circ \chi \circ \gamma)(0)[1] \
&= \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi(p)) \circ \mathrm d (\chi \circ \gamma)(0)[1].
\end{align}
$$
By the same logic,
$$
\frac{\mathrm d (\chi' \circ f \circ \lambda)}{\mathrm d t}(0) = \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi(p)) \circ \mathrm d (\chi \circ \lambda)(0)[1].
$$
Since $\mathrm d (\chi \circ \gamma)(0)[1] = \frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (\chi \circ \lambda)}{\mathrm d t}(0) = \mathrm d (\chi \circ \lambda)(0)[1]$, the two expressions are equal. This means $\frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (f \circ \lambda)}{\mathrm d t}(0)$, so regardless of which curve we use to represent $v$, the value of $F(v)$ is the same.

Thus, $F_1(v) = F_2(v)$ for all $v \in T_p M$, so the linear map satisfying (1.4) is unique.

Questions for the Community #

1. Is my uniqueness proof rigorous, or are there gaps I'm missing?
2. Are there alternative ways to prove uniqueness without relying on the fact that every tangent vector is the derivative of a curve?

备注：内容来源于stack exchange，提问作者Akira