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I am trying to understand the following: Let $\gamma: \mathbb{R} \supset I \to M$ be a smooth curve (M is a smooth manifold). Then the push forward $\gamma_{*}(\partial t)$ of $\partial t \in T_{t_0}\mathbb{R}$ is the tangent vector $\gamma'(t_0) \in T_{\gamma(t_0)}M$, where $\gamma'(t_0) \sim t \mapsto \gamma(t-t_0)$.

We defined the pushforward of a function $F:N \to M$ at the point $p$ as $F_{*}([\gamma]) = [F \circ \gamma]$, where $l \in [\gamma]$ if $\gamma(0)=l(0)=p$ and $\gamma_{\alpha}'(0) = l_{\alpha}'(0)$ for any (thus every) given chart $\phi_{\alpha}$ and $l_{\alpha}= \phi_{\alpha} \circ l$.

Now if I follow this definition, then $\gamma_{*}(\partial t) = \gamma_{*}([\partial t]) = [\gamma \circ \partial t]$. This should be easy, but I have no idea how to go on. Can anyone give me a hint?

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  • $\begingroup$ Hint : What is a curve whose tangent vector is $\partial_t$ ? $\endgroup$ Commented May 8, 2022 at 21:24
  • $\begingroup$ Oh, is it meant to be $\gamma_{*}(\partial t) = \gamma_{*}[t] = [\gamma(t)]=\gamma'(t)$, and the notation is to write the curves in $[]$ and the derivatives of the curves without $[]$? So $[t]=\partial t$? $\endgroup$ Commented May 8, 2022 at 21:55
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    $\begingroup$ Does this answer your question? Clarifying notation of derivative in differential geometry $\endgroup$
    – Kurt G.
    Commented May 9, 2022 at 0:36
  • $\begingroup$ @ Kurt G. Your link explains the theory in common, but it is not explained why the push forward maps $\partial t$ to $\gamma'$ there, either. What I was missing was that $\partial t$ is the tangent vector of the equivalence class $t \mapsto t + t_0$, which was explained in the answer of SolubleFish. $\endgroup$ Commented May 9, 2022 at 10:39

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From your notations, I am assuming that you are defining a tangent vector as an equivalence classes of curves. With this definition, $\partial_t \in T_{t_0}\mathbb R$ is the equivalence class of $t\mapsto t+ t_0$.

Given a smooth map $f:N\to M$ and a tangent vector $X = [\gamma] \in T_p N$, the pushforward is defined as the equivalence class of the composite $F\circ \gamma$ ie : $$f_*[\gamma] = [f\circ\gamma]$$

In our case, we have $\gamma:\mathbb R\to M$ and : $$\gamma_*\partial t = \gamma_*[t\mapsto t+t_0] =[t\mapsto \gamma(t+t_0)] = \gamma'(t_0)$$

NB : I believe there is a sign error in your post.

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  • $\begingroup$ Thanks for your clear answer! The $-$ must but be a typo then. $\endgroup$ Commented May 9, 2022 at 10:41

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