# [SOLVED](An Introduction to Manifolds) Prove that the Map is an Isomorphism of Vector Spaces

#### joypav

##### Active member
Hi all! I wasn't sure where to post this... so feel free to move it.

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 $v = <Dx^1, ... , Dx^n>$.

#### GJA

##### Well-known member
MHB Math Scholar
Hi joypav ,

Hope you're enjoying Tu. As you noticed, I mentioned to another user that it's a book I like quite a bit.

What is $x^j$? An arbitrary $x = (x^1, ..., x^n)$ we are applying it to?
$x^{j}$ serves two related purposes: (i) it represents the $j^{\text{th}}$ Cartesian axis variable in $\mathbb{R}^{n}$ (e.g., in $\mathbb{R}^{2}$ we would usually write $x^{1} = x$ and $x^{2} = y$); (ii) $x^{j}$ is also used to represent the function whose value for a point $p\in\mathbb{R}^{n}$ is $x^{j}(p) = p^{j}$ (assuming $p=(p^{1}, \ldots, p^{n})$ is the Cartesian coordinate representation for $p$).

$D(f(p)) = 0$ and $D(p^i) = 0$ because $f$ evaluated at a point $p$ is constant and each $p^i$ is a constant?
This is correct. Both $f(p)$ and $p^{i}$ are real-valued constants, so applying $D$ to them results in zero.

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

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.
You are correct, the Leibniz rule is what's being applied here.

#### joypav

##### Active member
Hope you're enjoying Tu. As you noticed, I mentioned to another user that it's a book I like quite a bit.
I am enjoying it. I've been reading it, along with Munkres "Analysis on Manifolds" as a secondary source, but I find Tu much more easy to follow and understand.

Thank you for the help!