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I am working through Tu's "An Introduction to Manifolds" (I saw someone else here is looking at this book too! ) and I just want to make sure I am

**really**understanding everything. I am looking at the proof that the map from the tangent space at a point p to the set of all derivations at this p is an isomorphism.

**I will state the proof in the book below and include my questions in red.**

$\phi : T_p(\Bbb{R}^n) \rightarrow \textit{D}_p(\Bbb{R}^n)$

$v \mapsto D_v = \sum v^i \frac{\partial}{\partial x^i} \rvert_p$

**Theorem:**The linear map $\phi$ defined above is an isomorphism of vector spaces.

**Proof:**

To prove injectivity, suppose $D_v = 0$ for $v \in T_p(\Bbb{R}^n)$. Applying $D_v$ to the coordinate function $x^j$ gives

$0 = D_v(x^j) = \sum_{i} v^i \frac{\partial}{\partial x^i} \rvert_p = \sum_{i} v^i \delta_i^j = v^j$

What is $x^j$? An arbitrary $x = (x^1, ..., x^n)$ we are applying it to?

Hence, $v = 0$ and $\phi$ is injective.

I have proven for myself that a linear map $f$ is one to one if and only if $ker(f) = {0}$, so I am fine with this conclusion.

To prove surjectivity, let $D$ be a derivation at $p$ and let $(f, V)$ be a representative of a germ in $C_p^{infty}$. Making $V$ smaller if necessary, we may assume that $V$ is an open ball, hence star-shaped. By Taylor's theorem with remainder, there are functions $g_i(x)$ in a neighborhood of $p$ such that

$f(x) = f(p) + \sum (x^i - p^i) g_i(x)$, where $g_i(p) = \frac{\partial f}{\partial x^i}(p)$

Applying $D$ to both sides and noting that $D(f(p)) = 0$ and $D(p^i) = 0$, we get by the Leibniz rule,

$D(f(p)) = 0$ and $D(p^i) = 0$ because $f$ evaluated at a point $p$ is constant and each $p^i$ is a constant?

$Df(x) = \sum (Dx^i) g_i(p) + \sum (p^i - p^i) Dg_i(x) = \sum (Dx^i) \frac{\partial f}{\partial x^i}(p)$

Could I get a little explanation of getting from here

$f(x) = f(p) + \sum (x^i - p^i) g_i(x)$

to here

$Df(x) = \sum (Dx^i) g_i(p) + \sum (p^i - p^i) Dg_i(x)$ ?

I understand that $f(p)$ disappears because it is equal to $0$ when we apply $D$.

Are we applying the Leibniz rule to the functions $g_i(x)$ and $(x^i - p^i)$? If that is what's happening then I do understand.

This proves that $D=D_v$ for $v = <Dx^1, ... , Dx^n>$.