I guess this question has more to do with notations then with concepts.
Let $I:=[0, 1]$ and $\alpha:I\longrightarrow M$ a path with values in a smooth manifolds $M$.
I'm reading a paper in which the author says to use the basis $\partial_t$ of $T_iI$ for doing a given computation (which is not relevant by now). How is $\partial_t$ defined?
Furtheremore, does $d\gamma_t(\partial_t)=\gamma^\prime(t)$ hold?
Generally, if $M$ is a smooth manifold (without boundary) then a tangent vector in $p\in M$ is a derivation of the algebra of germs $C^\infty_p(M)$.
Thanks.
You know that the tangent space $T_t[0,1]\cong \mathbb R$ is isomorphic (via $\Phi$, see below) to $\mathrm {Der}_t([0,1])$, the space of derivations at the point $t\in[0,1]$. Now define $\partial_t$ to be $\Phi(1)$, where $$ \Phi: T_t[0,1]\to\mathrm{Der}_t([0,1]), v\mapsto \Phi(v) $$ and $\Phi(v)(f):=df_t(v)$.
Now $\dot\gamma(t)$ is just a notation for the following: $T_t[0,1]\stackrel{d\gamma_t}{\to}T_{\gamma(t)}M, \partial_t\mapsto d\gamma_t(\partial_t)=:\dot\gamma(t)$ and you can think of this tangent vector again as a derivation via an isomorphism $$\widetilde\Phi:T_{\gamma(t)}M\stackrel{\cong}{\to}\mathrm{Der}_{\gamma(t)}(M).$$
Although if your manifold is $\mathbb R^n$, then $\dot\gamma(t)$ coincides with its usual meaning (just using the identity as a chart)...