Definition of Tangent Map is compatible with the definition of a Tangent Vector

Question

A tangent vector at a point p $\in$ M can be defined as an equivalence class of curves $\gamma:I\rightarrow M$ by the equivalence relation \begin{align} \gamma_1 \sim \gamma_2 \Leftrightarrow \frac{d}{dt}(\phi \circ\gamma_1 )|_{t=0} = \frac{d}{dt}(\phi \circ\gamma_2 )|_{t=0} \end{align} Where $(U,\phi)$ is any chart containing $p$.
In my lecture notes we defined the tangent map of a map $f:M\rightarrow N$ between two smooth manifolds as follows:
The tangent map $T_pf:T_pM\rightarrow T_{f(p)}N$ is a map between the corresponding tangent spaces, i.e. it maps a tangent vector $v\in T_pM$ to a tangent vector in $T_{f(p)}N$ by: \begin{align} T_pf(v)=[f\circ\beta] \end{align} where $\beta: I\rightarrow M$ is a curve with $\beta(0)=p$ and $v=[\beta]$.
I understand this definition but what I dont see is how this is compatible with the definition of a tangent vector above, i.e. why $\gamma_1\sim \gamma_2$ implies that $f\circ\gamma_1\sim f\circ\gamma_2$ and vice versa.

Answer 1

The equivalence $f\circ \gamma_1\sim f\circ \gamma_2$ need not imply $\gamma_1\sim \gamma_2$. If $f$ is constant, $T_pf(v)$ is the zero vector for any $v$, so $f\circ \gamma_1\sim f\circ \gamma_2$, whether or not $\gamma_1\sim \gamma_2$.

Smoothness of $f$ implies that if $(U,\phi)$ is a chart containing $p$ and if $(V,\psi)$ is a chart containing $f(p)$, then $\psi\circ f\circ \phi^{-1}$ is smooth (as function from a subset of $\mathbb{R}^m$ to $\mathbb{R}^n$) on $\phi(U\cap f^{-1}(V))$.

Suppose $\gamma_1\sim \gamma_2$ for smooth curves $\gamma_1,\gamma_2:I\to M$ with $\gamma_1(0)=\gamma_2(0)=p$. Fix a chart $(V,\psi)$ containing $f(p)$. We need to show that $\frac{d}{dt}(\psi\circ f\circ \gamma_1)|_{t=0}=\frac{d}{dt}(\psi\circ f\circ \gamma_2)|_{t=0}$. Fix any chart $(U,\phi)$ containing $p$ and, by translating, assume $\phi(p)=0$ (for readability). By replacing $I$ with a subinterval if necessary, we can assume $\gamma_1(I),\gamma_2(I)\subset U\cap f^{-1}(V)$.

Define $g=\psi\circ f\circ \phi{-1}:\phi(U\cap f^{-1}(V))\to V$ and $h_i=\phi\circ \gamma_i:I\to \phi(U\cap f^{-1}(V))$ for $i=1,2$. Then if $n=\dim N$ and $g=(g_1,\ldots,g_n)$, \begin{align*} \frac{d}{dt}(\psi\circ f \circ \gamma_i)|_{t=0} & =\frac{d}{dt}(\psi\circ f\circ \gamma_i\circ \phi^{-1}\circ \phi\circ \gamma_i)|_{t=0}=\frac{d}{dt}(g\circ h_i)|_{t=0} \\ & = \Bigl(\frac{d [g_j\circ h_i]}{dt}\Bigr|_{t=0}\Bigr)_{j=1}^n = \Bigl(\text{grad}(g_j)\cdot \frac{d}{dt}(h_i)|_{t=0}\Bigr)_{j=1}^n\end{align*} The only dependence here on $\gamma_i$ is through $\frac{d}{dt}(h_i)|_{t=0}=\frac{d}{dt}(\phi\circ \gamma_i)|_{t=0}$, which is equal for $i=1,2$.