I have a question about the proof of Proposition 3.13 from John Lee's Introduction to Smooth Manifolds.
We give $V$ the natural smooth structure given by the following Example.
Now the proposition is as follows:
It says we can use the same argument as in the proof of Proposition 3.2, to show that $D_v|_a$ is an isomorphism. I can see that the exact same argument applies to show injectivity. However, I do not understand how we can show surjectivity, as in the proof below, this requires using Taylor's theorem. However, Taylor's theorem does not apply here as $f$ is a smooth function from a vector space $V$ that may not be the Euclidean space. So how can we adapt the proof for a general finite-dimensional vector space $V$ here?
Maybe this question is a bit old but I've managed to find a way.
I'll use the same notations as in the OP. Let $(V,E^{-1})$ be the chart and $E^{-1} = (x^1, \ldots ,x^n)$ the coordinate maps. Your choice of the vector $v \in V$ is the right one, indeed take $v = w(x^i) \, E_i$ where $(E_1, \ldots ,E_n)$ is the basis associated with $E$ and $w \in T_aV$ is given.
Now, for any $f \in C^\infty(V)$ holds $ w(f) = (d(E^{-1})_a (w))(f \circ E)$ because $E \in C^\infty$ and we applied the definition of differential. You can now use Taylor's theorem, as in the proposition 3.2 of Lee's book, on $f \circ E \in C^\infty(\mathbb{R}^n)$ at the point $E^{-1}(a)$ to get
$$ (f \circ E)(x) = f(a) + \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (e^i(x)-x^i(a)) \, + \sum_{i,j}... $$
where $e^i$ are the coordinate maps of the standard basis of $\mathbb{R}^n$ (to not get confused with $x^i$).
So we have \begin{align} w(f) &= \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (d(E^{-1})_a (w))(e^i) = \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, w(x^i) =\\ &= \frac{\partial (f \circ E)}{\partial e^i}|_{E^{-1}(a)} \, (E^{-1}(v))^i = \frac{d}{dt}|_{t=0} (f \circ E)(E^{-1}(a)+tE^{-1}(v)) = \\ &= \frac{d}{dt}|_{t=0} f(a + tv) = D_v|_a f \end{align}
Notice that we exploited the linearity of $E^{-1}$ to use the chain rule for maps $\mathbb{R} \rightarrow \mathbb{R}^n \rightarrow \mathbb{R}$, after we expressed $f = (f \circ E) \circ E^{-1}$.