[proofplan]
We first check that differentiating a smooth function along a curve depends only on the curve's tangent class, so $D_v$ is well-defined. The one-variable linearity and product rules then give the derivation property. To prove that every derivation arises this way, we use a coordinate chart near $p$, extend the coordinate functions to global smooth functions by multiplying with a bump function, and show that a derivation is completely determined by its values on these extended coordinates. This identifies derivations with coordinate velocity vectors, and the two constructions are inverse to each other.
[/proofplan]
[step:Show that differentiating along a curve depends only on the tangent class]
Let $v=[\gamma]\in T_pM$, where
\begin{align*}
\gamma: (-\varepsilon,\varepsilon) &\to M
\end{align*}
is smooth and $\gamma(0)=p$. Let $(U,\varphi)$ be a smooth chart about $p$, with
\begin{align*}
\varphi: U &\to \varphi(U)\subseteq \mathbb{R}^n, &
\varphi(q)&=(x_1(q),\dots,x_n(q)).
\end{align*}
Choose $\delta\in(0,\varepsilon)$ such that $\gamma((-\delta,\delta))\subset U$. For $f\in C^\infty(M)$, define
\begin{align*}
F: \varphi(U) &\to \mathbb{R} \\
y &\mapsto f(\varphi^{-1}(y)).
\end{align*}
Then
\begin{align*}
f(\gamma(t))=F((\varphi\circ\gamma)(t))
\end{align*}
for $t\in(-\delta,\delta)$, so the ordinary chain rule gives
\begin{align*}
\frac{d}{dt}\Big|_{t=0} f(\gamma(t))
=
\sum_{i=1}^{n}
\frac{\partial F}{\partial y_i}(\varphi(p))
\frac{d}{dt}\Big|_{t=0} x_i(\gamma(t)).
\end{align*}
Thus the derivative depends only on the coordinate velocity
\begin{align*}
\left(
\frac{d}{dt}\Big|_{t=0} x_1(\gamma(t)),\dots,
\frac{d}{dt}\Big|_{t=0} x_n(\gamma(t))
\right)\in\mathbb{R}^n.
\end{align*}
Since equivalent curves have the same coordinate velocity, $D_v(f)$ is independent of the representative $\gamma$.
[guided]
We must first justify that the displayed formula defines a function of the tangent vector $v=[\gamma]$, not of the particular curve chosen to represent it. Fix a chart $(U,\varphi)$ around $p$, and write its coordinate functions as $x_i:U\to\mathbb{R}$ for $1\le i\le n$. Since $\gamma(0)=p$ and $U$ is open, there exists $\delta\in(0,\varepsilon)$ such that $\gamma((-\delta,\delta))\subset U$.
For a smooth function $f\in C^\infty(M)$, define the coordinate representative
\begin{align*}
F: \varphi(U) &\to \mathbb{R} \\
y &\mapsto f(\varphi^{-1}(y)).
\end{align*}
Then $F$ is smooth, and for $t\in(-\delta,\delta)$ we have
\begin{align*}
f(\gamma(t))=F((\varphi\circ\gamma)(t)).
\end{align*}
Applying the ordinary chain rule to the smooth curve $\varphi\circ\gamma:(-\delta,\delta)\to\mathbb{R}^n$ gives
\begin{align*}
\frac{d}{dt}\Big|_{t=0} f(\gamma(t))
=
\sum_{i=1}^{n}
\frac{\partial F}{\partial y_i}(\varphi(p))
\frac{d}{dt}\Big|_{t=0} x_i(\gamma(t)).
\end{align*}
The right-hand side uses only $f$, the chart, and the coordinate velocity of $\gamma$ at $0$. Therefore if another curve $\sigma$ represents the same tangent vector as $\gamma$, then the coordinate velocities of $\gamma$ and $\sigma$ agree, and the same formula gives
\begin{align*}
\frac{d}{dt}\Big|_{t=0} f(\gamma(t))
=
\frac{d}{dt}\Big|_{t=0} f(\sigma(t)).
\end{align*}
So $D_v(f)$ is well-defined.
[/guided]
[/step]
[step:Verify that $D_v$ is a derivation at $p$]
Let $a,b\in\mathbb{R}$ and $f,g\in C^\infty(M)$. By the one-variable linearity of derivatives,
\begin{align*}
D_v(af+bg)
&=
\frac{d}{dt}\Big|_{t=0} \bigl(af(\gamma(t))+bg(\gamma(t))\bigr) \\
&=
aD_v(f)+bD_v(g).
\end{align*}
Thus $D_v$ is $\mathbb{R}$-linear. By the one-variable product rule,
\begin{align*}
D_v(fg)
&=
\frac{d}{dt}\Big|_{t=0} \bigl(f(\gamma(t))g(\gamma(t))\bigr) \\
&=
f(\gamma(0))D_v(g)+g(\gamma(0))D_v(f) \\
&=
f(p)D_v(g)+g(p)D_v(f).
\end{align*}
Hence $D_v\in\operatorname{Der}_p(M)$.
[/step]
[step:Compute the coordinate formula for the derivation determined by a curve]
With the chart $(U,\varphi)$ fixed, let $v=[\gamma]\in T_pM$ and define the coordinate velocity components
\begin{align*}
v_i:=\frac{d}{dt}\Big|_{t=0} x_i(\gamma(t))
\end{align*}
for $1\le i\le n$. For $f\in C^\infty(M)$ and
\begin{align*}
F: \varphi(U) &\to \mathbb{R} \\
y &\mapsto f(\varphi^{-1}(y)),
\end{align*}
the chain rule computation from the first step gives
\begin{align*}
D_v(f)
=
\sum_{i=1}^{n} v_i \frac{\partial F}{\partial y_i}(\varphi(p)).
\end{align*}
Thus, in the chart $(U,\varphi)$, $D_v$ acts as the directional derivative with coordinate vector $(v_1,\dots,v_n)$.
[guided]
This step records the concrete coordinate expression for $D_v$. The coordinate velocity of the curve representative $\gamma$ is the vector
\begin{align*}
(v_1,\dots,v_n)
:=
\left(
\frac{d}{dt}\Big|_{t=0} x_1(\gamma(t)),\dots,
\frac{d}{dt}\Big|_{t=0} x_n(\gamma(t))
\right)\in\mathbb{R}^n.
\end{align*}
For $f\in C^\infty(M)$, pass to the coordinate representative
\begin{align*}
F: \varphi(U) &\to \mathbb{R} \\
y &\mapsto f(\varphi^{-1}(y)).
\end{align*}
The chain rule gives
\begin{align*}
D_v(f)
=
\frac{d}{dt}\Big|_{t=0} F((\varphi\circ\gamma)(t))
=
\sum_{i=1}^{n} v_i \frac{\partial F}{\partial y_i}(\varphi(p)).
\end{align*}
So the derivation obtained from a curve is exactly the coordinate directional derivative whose coefficients are the coordinate velocity components of the curve.
[/guided]
[/step]
[step:Show that derivations depend only on germs at $p$]
Let $D\in\operatorname{Der}_p(M)$. First, $D(1)=0$, because
\begin{align*}
D(1)=D(1\cdot 1)=1(p)D(1)+1(p)D(1)=2D(1).
\end{align*}
Hence $D(c)=0$ for every constant function $c:M\to\mathbb{R}$ by linearity.
Now suppose $h\in C^\infty(M)$ vanishes on an open neighbourhood $W\subset M$ of $p$. Choose a smooth function
\begin{align*}
\eta:M&\to\mathbb{R}
\end{align*}
such that $\eta(p)=1$ and $\operatorname{supp}\eta\subset W$. Since $h\eta=0$ on $M$, the Leibniz rule gives
\begin{align*}
0
=
D(h\eta)
=
h(p)D(\eta)+\eta(p)D(h)
=
D(h).
\end{align*}
Therefore $D(h)=0$ whenever $h$ vanishes in a neighbourhood of $p$. Consequently, if $f,g\in C^\infty(M)$ agree on a neighbourhood of $p$, then $D(f)=D(g)$.
[guided]
A derivation at $p$ should only measure first-order behaviour at $p$, so it must ignore any change made away from $p$. We prove this from the Leibniz rule.
First consider the constant function $1:M\to\mathbb{R}$. Since $D$ is a derivation,
\begin{align*}
D(1)=D(1\cdot 1)=1(p)D(1)+1(p)D(1)=2D(1).
\end{align*}
Subtracting $D(1)$ from both sides gives $D(1)=0$. By linearity, every constant function $c:M\to\mathbb{R}$ satisfies
\begin{align*}
D(c)=cD(1)=0.
\end{align*}
Now let $h\in C^\infty(M)$ vanish on an open neighbourhood $W$ of $p$. Choose a smooth function $\eta:M\to\mathbb{R}$ with $\eta(p)=1$ and $\operatorname{supp}\eta\subset W$. Then $h\eta=0$ everywhere on $M$, since $\eta$ is supported where $h$ is zero. Applying $D$ and using the Leibniz rule,
\begin{align*}
0
=
D(h\eta)
=
h(p)D(\eta)+\eta(p)D(h).
\end{align*}
Because $h(p)=0$ and $\eta(p)=1$, this becomes $D(h)=0$.
Finally, if $f,g\in C^\infty(M)$ agree on a neighbourhood of $p$, then $h:=f-g$ vanishes on a neighbourhood of $p$, so
\begin{align*}
D(f)-D(g)=D(f-g)=D(h)=0.
\end{align*}
Thus $D(f)=D(g)$.
[/guided]
[/step]
[step:Recover a coordinate vector from an arbitrary derivation]
Let $D\in\operatorname{Der}_p(M)$. Choose a smooth function
\begin{align*}
\rho:M&\to\mathbb{R}
\end{align*}
such that $\rho=1$ on a neighbourhood of $p$ and $\operatorname{supp}\rho\subset U$. For $1\le i\le n$, define
\begin{align*}
\widetilde{x}_i:M&\to\mathbb{R}
\end{align*}
by $\widetilde{x}_i=\rho x_i$ on $U$ and $\widetilde{x}_i=0$ on $M\setminus\operatorname{supp}\rho$. This is smooth because $\operatorname{supp}\rho\subset U$, and $\widetilde{x}_i=x_i$ near $p$.
Define [real numbers](/page/Real%20Numbers)
\begin{align*}
a_i:=D(\widetilde{x}_i)
\end{align*}
for $1\le i\le n$. These numbers are independent of the chosen cutoff and extension, because any two such extensions agree with $x_i$ on a neighbourhood of $p$, and derivations depend only on germs at $p$.
Let $f\in C^\infty(M)$, and define
\begin{align*}
F:\varphi(U)&\to\mathbb{R} \\
y&\mapsto f(\varphi^{-1}(y)).
\end{align*}
Set $y_0:=\varphi(p)\in\mathbb{R}^n$. On a neighbourhood of $y_0$, the smooth function $F$ has the local decomposition
\begin{align*}
F(y)=F(y_0)+\sum_{i=1}^{n}(y_i-y_{0,i})G_i(y),
\end{align*}
where each
\begin{align*}
G_i:\varphi(U)&\to\mathbb{R}
\end{align*}
is smooth near $y_0$ and satisfies
\begin{align*}
G_i(y_0)=\frac{\partial F}{\partial y_i}(y_0).
\end{align*}
Translating this identity back to $M$ and using the germ property, we obtain
\begin{align*}
D(f)
&=
D\left(f(p)+\sum_{i=1}^{n}(\widetilde{x}_i-\widetilde{x}_i(p))\,\widetilde{G}_i\right) \\
&=
\sum_{i=1}^{n}
D\bigl((\widetilde{x}_i-\widetilde{x}_i(p))\,\widetilde{G}_i\bigr),
\end{align*}
where each $\widetilde{G}_i\in C^\infty(M)$ is any smooth extension agreeing near $p$ with $G_i\circ\varphi$. Applying the Leibniz rule and using $\widetilde{x}_i(p)-\widetilde{x}_i(p)=0$ gives
\begin{align*}
D(f)
&=
\sum_{i=1}^{n}
\left(
\widetilde{G}_i(p)D(\widetilde{x}_i-\widetilde{x}_i(p))
+
(\widetilde{x}_i(p)-\widetilde{x}_i(p))D(\widetilde{G}_i)
\right) \\
&=
\sum_{i=1}^{n}
\frac{\partial F}{\partial y_i}(y_0)a_i.
\end{align*}
Thus every derivation is determined by the coordinate vector $(a_1,\dots,a_n)$.
[guided]
We now start with an abstract derivation $D$ and extract the coordinate vector that should represent it. The coordinate functions $x_i$ are only defined on $U$, while $D$ acts on global smooth functions on $M$. To feed coordinate functions into $D$, we first extend them smoothly to global functions.
Choose a smooth cutoff function
\begin{align*}
\rho:M&\to\mathbb{R}
\end{align*}
such that $\rho=1$ on a neighbourhood of $p$ and $\operatorname{supp}\rho\subset U$. For each coordinate function $x_i:U\to\mathbb{R}$, define
\begin{align*}
\widetilde{x}_i:M&\to\mathbb{R}
\end{align*}
by setting $\widetilde{x}_i=\rho x_i$ on $U$ and $\widetilde{x}_i=0$ outside $\operatorname{supp}\rho$. Since $\rho$ vanishes near the boundary of $U$, this extension is smooth on all of $M$. It agrees with $x_i$ near $p$ because $\rho=1$ near $p$.
Define
\begin{align*}
a_i:=D(\widetilde{x}_i)
\end{align*}
for $1\le i\le n$. These values are intrinsic to the germ of $x_i$ at $p$: if $\widehat{x}_i$ is another global smooth extension agreeing with $x_i$ near $p$, then $\widetilde{x}_i-\widehat{x}_i$ vanishes near $p$, and the germ property gives
\begin{align*}
D(\widetilde{x}_i)-D(\widehat{x}_i)=D(\widetilde{x}_i-\widehat{x}_i)=0.
\end{align*}
Now let $f\in C^\infty(M)$. Write its coordinate representative as
\begin{align*}
F:\varphi(U)&\to\mathbb{R} \\
y&\mapsto f(\varphi^{-1}(y)),
\end{align*}
and set $y_0:=\varphi(p)$. Near $y_0$, the first-order Taylor decomposition has the form
\begin{align*}
F(y)=F(y_0)+\sum_{i=1}^{n}(y_i-y_{0,i})G_i(y),
\end{align*}
where the functions $G_i$ are smooth near $y_0$ and satisfy
\begin{align*}
G_i(y_0)=\frac{\partial F}{\partial y_i}(y_0).
\end{align*}
This decomposition says that, up to first order at $p$, $f$ is determined by its coordinate partial derivatives.
Choose global smooth functions $\widetilde{G}_i:M\to\mathbb{R}$ agreeing near $p$ with $G_i\circ\varphi$. Since derivations depend only on germs at $p$, we may compute $D(f)$ using the local identity:
\begin{align*}
D(f)
&=
D\left(f(p)+\sum_{i=1}^{n}(\widetilde{x}_i-\widetilde{x}_i(p))\,\widetilde{G}_i\right).
\end{align*}
The constant term contributes zero, because derivations vanish on constants. Hence
\begin{align*}
D(f)
&=
\sum_{i=1}^{n}
D\bigl((\widetilde{x}_i-\widetilde{x}_i(p))\,\widetilde{G}_i\bigr).
\end{align*}
Applying the Leibniz rule to each summand gives
\begin{align*}
D(f)
&=
\sum_{i=1}^{n}
\left(
\widetilde{G}_i(p)D(\widetilde{x}_i-\widetilde{x}_i(p))
+
(\widetilde{x}_i(p)-\widetilde{x}_i(p))D(\widetilde{G}_i)
\right) \\
&=
\sum_{i=1}^{n}
\widetilde{G}_i(p)D(\widetilde{x}_i).
\end{align*}
Since $\widetilde{G}_i(p)=G_i(y_0)=\partial F/\partial y_i(y_0)$ and $D(\widetilde{x}_i)=a_i$, we obtain
\begin{align*}
D(f)
=
\sum_{i=1}^{n}
a_i\frac{\partial F}{\partial y_i}(y_0).
\end{align*}
Thus $D$ is completely determined by the coordinate vector $(a_1,\dots,a_n)$.
[/guided]
[/step]
[step:Construct a curve whose derivation is the given abstract derivation]
Let $D\in\operatorname{Der}_p(M)$, and let $(a_1,\dots,a_n)\in\mathbb{R}^n$ be the coordinate vector recovered in the previous step. Define a smooth curve
\begin{align*}
\gamma_D:(-\varepsilon,\varepsilon)&\to M \\
t&\mapsto \varphi^{-1}(y_0+t a),
\end{align*}
where $y_0:=\varphi(p)$, $a:=(a_1,\dots,a_n)$, and $\varepsilon>0$ is chosen so that $y_0+ta\in\varphi(U)$ for all $t\in(-\varepsilon,\varepsilon)$. Then $\gamma_D(0)=p$ and
\begin{align*}
\frac{d}{dt}\Big|_{t=0} x_i(\gamma_D(t))=a_i
\end{align*}
for each $1\le i\le n$.
For every $f\in C^\infty(M)$, the coordinate formula for $D_{\,[\gamma_D]}$ gives
\begin{align*}
D_{[\gamma_D]}(f)
=
\sum_{i=1}^{n}
a_i\frac{\partial F}{\partial y_i}(y_0),
\end{align*}
where $F=f\circ\varphi^{-1}$. The previous step gives the same formula for $D(f)$. Hence $D=D_{[\gamma_D]}$, so $\Theta_p$ is surjective.
[guided]
The vector $(a_1,\dots,a_n)$ extracted from $D$ should be realised as the velocity of an actual curve through $p$. In the coordinate chart, the simplest curve is the straight line with initial point $y_0:=\varphi(p)$ and velocity $a:=(a_1,\dots,a_n)$. Since $\varphi(U)$ is open in $\mathbb{R}^n$, there exists $\varepsilon>0$ such that
\begin{align*}
y_0+ta\in\varphi(U)
\end{align*}
for every $t\in(-\varepsilon,\varepsilon)$. Define
\begin{align*}
\gamma_D:(-\varepsilon,\varepsilon)&\to M \\
t&\mapsto \varphi^{-1}(y_0+t a).
\end{align*}
Then $\gamma_D$ is smooth and $\gamma_D(0)=p$. Moreover,
\begin{align*}
x_i(\gamma_D(t))=y_{0,i}+t a_i,
\end{align*}
so
\begin{align*}
\frac{d}{dt}\Big|_{t=0} x_i(\gamma_D(t))=a_i.
\end{align*}
For any $f\in C^\infty(M)$, let
\begin{align*}
F:\varphi(U)&\to\mathbb{R} \\
y&\mapsto f(\varphi^{-1}(y)).
\end{align*}
The coordinate formula for the curve derivation gives
\begin{align*}
D_{[\gamma_D]}(f)
=
\sum_{i=1}^{n}
a_i\frac{\partial F}{\partial y_i}(y_0).
\end{align*}
But the previous step proved that the abstract derivation $D$ satisfies exactly the same formula:
\begin{align*}
D(f)
=
\sum_{i=1}^{n}
a_i\frac{\partial F}{\partial y_i}(y_0).
\end{align*}
Therefore $D(f)=D_{[\gamma_D]}(f)$ for every $f\in C^\infty(M)$, and hence $D=D_{[\gamma_D]}$. This proves surjectivity of $\Theta_p$.
[/guided]
[/step]
[step:Prove injectivity and preserve the vector space structure]
The map $\Theta_p$ is linear because for tangent vectors represented in the chart by coordinate velocities $v=(v_1,\dots,v_n)$ and $w=(w_1,\dots,w_n)$, and for $\lambda,\mu\in\mathbb{R}$, the coordinate formula gives
\begin{align*}
\Theta_p(\lambda v+\mu w)(f)
&=
\sum_{i=1}^{n}(\lambda v_i+\mu w_i)\frac{\partial F}{\partial y_i}(y_0) \\
&=
\lambda\Theta_p(v)(f)+\mu\Theta_p(w)(f)
\end{align*}
for every $f\in C^\infty(M)$.
If $\Theta_p(v)=0$, let $v_i$ be the coordinate velocity components of $v$. Applying $\Theta_p(v)$ to the global coordinate extensions $\widetilde{x}_i$ from above gives
\begin{align*}
0=\Theta_p(v)(\widetilde{x}_i)=v_i
\end{align*}
for every $1\le i\le n$. Hence the coordinate velocity of $v$ is zero, so $v=0$ in $T_pM$. Thus $\Theta_p$ is injective.
Since $\Theta_p$ is linear, injective, and surjective, it is a [vector space](/page/Vector%20Space) isomorphism from $T_pM$ onto $\operatorname{Der}_p(M)$.
[/step]
