=== Definition 2 implies Definition 1 === Let $\lambda \in \R$ and $f, g \in C^\infty \left({V, \R}\right)$.  {{begin-eqn}} {{eqn | l = X_m \left({f + \lambda g} \right)        | r = \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({f + \lambda g}\right) \circ \gamma \left({\tau}\right)        | c = Definition 2 }} {{eqn | r = \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({f \circ \gamma \left({\tau}\right)}\right) + \lambda \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({g \circ \gamma} \left({\tau}\right) \right)  }} {{eqn | r = X_m \left({f}\right) + \lambda \, X_m \left({g}\right)  }} {{end-eqn}} Thus $X_m$ is linear.  {{begin-eqn}} {{eqn | l = X_m \left({f g} \right)        | r = \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \left({f g}\right) \circ \gamma \left({\tau}\right)        | c = Definition 2 }} {{eqn | r = \frac {\mathrm d} {\mathrm d \tau} {\restriction_0}  f \circ \gamma \left({\tau}\right) \, g \circ \gamma \left({0}\right) + f \circ \gamma \left({0}\right) \, \frac {\mathrm d} {\mathrm d \tau} {\restriction_0}  g \circ \gamma  \left({\tau}\right)       | c = Product Rule }} {{eqn | r = X_m \left({f}\right) \, g \left({m}\right) + f \left({m}\right) \, X_m \left({g}\right)       | c = Definition 2, $\gamma \left({0}\right) = m$ }} {{end-eqn}} Hence $X_m$ satisfies the ''Leibniz law''.  Thus $X_m$ satisfies Definition 1. {{qed|lemma}} === Definition 1 implies Definition 2 === === Lemma 1 ===  Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1.  Let $V$ be an open neighborhood of $M$.  Let $f \in C^\infty \left({V, \R}\right)$ be constant.  Then $X_m \left({f}\right) = 0$. === Proof of Lemma 1 === Let $f \left({m}\right) = 0$.  Then, by constancy, $f = 0$ on $V$. Hence, by linearity, $X_m \left({0}\right) = 0$.  Let $f \left({m}\right) \ne 0$.  {{begin-eqn}} {{eqn | l = X_m \left({f}\right)       | r = X_m \left({1 \, f}\right) }} {{eqn | r = X_m \left({1}\right) \, f \left({m}\right) + X_m \left({f}\right)       | c = Leibniz law }} {{eqn | l = \iff       | o = }} {{eqn | l = X_m \left({1}\right)        | r = 0        | c = $f \left({m}\right) \ne 0$ }} {{end-eqn}} $f$ is constant, {{iff}} $\exists \lambda \in \R : f \left({V}\right) = \left\{ {\lambda} \right\}$ {{iff}} $f = \lambda$. {{explain|$\exists$ in the above? Should it not be $\forall$?}} {{begin-eqn}} {{eqn | l = \lambda X_m \left({1}\right)        | r = X_m  \left({\lambda}\right)        | c = Definition of Linear Mapping }} {{eqn | r = X_m \left({f}\right)        | c = as $f = \lambda$ }} {{eqn | r = 0       | c = as $X_m  \left({1}\right) = 0$ }} {{end-eqn}} {{qed|lemma}} Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1. Denote $n := \dim M$.  Let $f \in C^\infty \left({V, \R}\right)$. Let $(U, \kappa)$ be a chart with $\kappa \left( {m} \right) = 0$.  Let $\kappa^i$ be the $i$th coordinate function of the chart $(U, \kappa)$.  By, Taylor's Theorem/n Variables : {{begin-eqn}} {{eqn | l = f       | r = f \circ \kappa^{-1} \circ \kappa       | c =  }} {{eqn | r = \left({f \circ \kappa^{-1} }\right) \left({\kappa}\right)       | c =  }} {{eqn | r = \left({f \circ \kappa^{-1} }\right) \left({0}\right) + \sum_{i \mathop = 1}^n \frac {\partial} {\left({f \circ \kappa^{-1} }\right)} {\partial{\kappa^i} } \left({0}\right) \, \kappa^i + \mathcal O_2 \left({\kappa}\right)       | c =  }} {{end-eqn}} Observe that $\displaystyle \left({f \circ \kappa^{-1} }\right) \left({0}\right) = \left({f \circ \kappa^{-1} }\right) \left( {\kappa \left({m}\right)}\right) = f \left({m}\right) $ is a constant mapping on $V$. {{explain|The above needs more than observation.}} Define $X^i := X_m \left({\kappa ^i}\right)$. Then by linearity: :$\displaystyle X_m \left({f}\right) = X_m \left({f \left({m}\right)}\right) + \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({0}\right) \ X^i + X_m \left({\mathcal O_2 \left({\kappa}\right) }\right)$ === Lemma 2 ===  :$X_m \left({\mathcal O_2 \left({\kappa}\right)}\right) = 0$ === Proof of Lemma 2 === By Taylor's Theorem/n Variables, for each summand of $\mathcal O _2 \left({\kappa}\right)$ there exists $i \in \left\{ {1, \ldots, n} \right\}$ and an $h \in C^\infty \left({V, \R}\right)$ with $h \left({m}\right) = 0$ such that the summand is $\kappa^i h$. {{begin-eqn}} {{eqn | l = X_m \left({\kappa^i h}\right)       | r = X_m \left({\kappa^i}\right) h \left({m}\right) + \kappa^i \left({m}\right) X_m \left({h}\right)       | c = Definition 1 (Leibniz law) }} {{eqn | r = X_m \left({\kappa^i}\right) 0 + 0 X_m \left({h}\right)       | c = as $h \left({m}\right) = 0$, $\kappa^i \left({m}\right) = 0$ }} {{eqn | r = 0 }} {{end-eqn}} Thus the sum $\mathcal O_2 \left({\kappa}\right)$ vanishes.  {{qed|lemma}} By Lemma 1 :  :$\displaystyle X_m \left({f \left({m}\right)} \right) = 0$ By Lemma 2 :  :$\displaystyle X_m \left({\mathcal O_2 \left({\kappa }\right)}\right) = 0$ Hence: :$\displaystyle X_m \left({f}\right) = \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1}}\right)} {\partial \kappa^i} \left({0}\right) \ X^i$ Let $\left\{{e_i}\right\}$ be a basis of $\R^n$ such that: :$\displaystyle \kappa = \sum_{i \mathop = 1}^n \kappa^i e_i$ {{Disambiguate|Definition:Basis}} Choose a smooth curve $\gamma: I \to M$ with $0 \in I \subseteq \R$ such that $\gamma \left({0}\right) = m$ and: :$\dfrac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right) := X^i$ Then: {{begin-eqn}} {{eqn | l = X_m \left({f}\right)       | r = \sum_{i \mathop = 1}^n \frac{\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right)        | c = as $\kappa \left({m}\right) = 0$ }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({0}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right)        | c = as $m = \gamma \left({0}\right)$ }} {{eqn | r = \sum_{i \mathop = 1}^n \left. {\frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({\tau}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({\tau}\right)} \right \rvert_{\tau \mathop = 0} }} {{eqn | r = \intlimits {\frac {\mathrm d \left({f \circ \kappa^{-1} \circ \kappa \circ \gamma}\right)} {\mathrm d \tau} \left({\tau}\right)} {\tau \mathop = 0} {}       | c = Chain Rule for Derivatives }} {{eqn | r = \intlimits {\frac {\mathrm d \left({f \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right)} {\tau \mathop = 0} {}       | c = as $f \circ \kappa^{-1} \circ \kappa = f$ }} {{eqn | r = \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, f \circ \gamma \left({\tau}\right)  }} {{end-eqn}} Hence $X_m$ is a tangent vector according to Definition 2.  This proves the assertion.  {{qed}}