Why does $D\pi^i(p)(v)=v^i$?

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There's an equality that I don't understand from Spivak' book "Calculus on Manifolds" (p. 89).

We define $$ df(p)(v_p)=Df(p)(v) $$ Let us consider in particular the 1-forms $d\pi^i$. [...] $$ dx^i(p)(v_p)=d\pi^i(p)(v_p)=D\pi^i(p)(v)=v^i$$

($\pi^i$ is the $i$'s component of $x=(x^1,...,x^n)$).

Why does $D\pi^i(p)(v)=v^i$?

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$\pi^i\colon\Bbb R^n\to\Bbb R$ is projection onto the $i$th coordinate. As such it is a linear map, and the derivative of a linear map (at any point) is itself. Of course, $\pi^i(v) = v^i$, by definition.