# Loomis and Sternberg: Tangent Space to a manifold, using equivalence classes; help justifying one step of an argument

I am currently reading through the section in Loomis and Sternberg's Advanced Calculus on Tangent Spaces, but I'm having trouble justifying one step of the argument (shown below).

Here's the definitions and notations used by them. Let $$M$$ be a differentiable manifold (here they model their manifolds on a banach space $$V$$ rather than some $$\mathbb{R}^n$$). Let $$x \in M$$, and let $$\varphi : I \to M$$ be a differentiable map where $$I$$ is an interval in $$\mathbb{R}$$ containing $$0$$, and $$\varphi(0) = x.$$ Then, they define an operator $$D_{\varphi}: C^{\infty}(M) \to \mathbb{R}$$ by $$D_{\varphi}(f) = (f \circ \varphi)'(0)$$. Next, they define an equivalence relation on all the curves passing through $$x$$ by $$\varphi \sim \psi$$ if and only if $$D_{\varphi} = D_{\psi}$$, and call an equivalence class of curves, $$\xi$$ to be a tangent vector at $$x$$.

So far so good. Next, they go to a chart $$(W, \alpha)$$, and the underlined section below is what I don't fully understand. I get that $$\varphi \sim \psi$$ if and only if for every $$f \in C^{\infty}(M)$$, $$d(f \circ \alpha^{-1})_{\alpha(x)}((\alpha \circ \varphi)'(0)) = d(f \circ \alpha^{-1})_{\alpha(x)}((\alpha \circ \psi)'(0))$$.
But I don't see how to conclude from here that the two vectors $$(\alpha \circ \varphi)'(0)$$ and $$(\alpha \circ \psi)'(0)$$ are equal.

I'm guessing it has something to do with the fact that the two derivatives are equal for every $$f$$; if we somehow choose an $$f$$ such that the differential $$d(f \circ \alpha^{-1})_{\alpha(x)} : V \to \mathbb{R}$$ is injective, then its kernel is $$\{0\}$$, and thus equality follows. However I doubt this is always possible, since in general $$dim(V) > 1$$, so a linear map from $$V$$ to $$\mathbb{R}$$ cannot be injective.

Any help justifying this step is much appreciated.

The differential does not have to be injective. Note that for any vector space $$W$$, $$\phi(v) = 0$$ for any linear functional $$\phi$$ on $$W$$ implies that $$v = 0$$ necessarily. Therefore, since $$\mathrm{d}F$$ is always a linear functional regardless of the choice of $$f$$, you simply need to show that you can obtain any linear functional on $$V$$ with an appropriate choice of $$f$$.