For an abstract differential manifold $M$ of dimension $m$, for every point $p \in M$, we can define a tangent vector to M at p as a function: $$v:C_p \to \mathbb{R}^m, $$ where $C_p$ is the set of all charts of $M$ around $p$, with the property that for every two charts $\varphi_1, \varphi_2 \in C_p,$ we have that $$v(\varphi_2) = (dc)_{\varphi_1(p)}(v(\varphi_1)), $$ where $c = \varphi_2 \circ \varphi_1^{-1}$ is the transition function (that is smooth) and $(dc)_{\varphi_1(p)}$ is the classical differential of the function $c$ in the point $\varphi_1(p)$.

Next, we can define that tangent space of M at p as $$T_p(M) = \{v \ \mid \ v \text{ is a tangent vector to } M \text{ at } p \}, $$ which has a natural vector space structure over $\mathbb{R}$: point-wise addition and multiplication of functions, i.e. $$(v+w)(\varphi):= v(\varphi) + w(\varphi) \text{ and } (c \cdot v)(\varphi):= c \cdot (v(\varphi)),$$ for every $\varphi \in C_p$ and every $v,w \in T_p(M)$.

So far, this definition seems a bit awkward and difficult to work with, but nevertheless, now I want to define the differential of a smooth function:

For a smooth function $f:M \to N$ between two differential manifolds, we define the differential of $f$ at $p \in M$ as the unique function $$(df)_p : T_p(M) \to T_{f(p)}(N) $$ such that for every $\varphi \in C_p$ and for every $\chi \in C_{f(p)}$, we have that $$((df)_p(v)) (\chi) = (d(\chi \circ f \circ \varphi^{-1}))_{\varphi(p)}(v(\varphi)), \forall v \in T_p(M), $$ where $(d(\chi \circ f \circ \varphi^{-1}))_{\varphi(p)}$ is the classical differential of a function from $\mathbb{R}^m$ to $\mathbb{R}^n$.

My question is, how are we sure that this $(df)_p$ exists (how can we find such a function such that that special property is satisfied) and that it is well defined (that we define this $(df)_p$ independently of the charts $\varphi$ and $\chi$)?

Moreover, how can we prove that this function is actually linear (after we have defined it). I am asking this because, in order to prove that it is linear, I would choose a chart $\varphi \in C_p$ and $v,w \in T_p(M)$, and then, for every $\chi \in C_{f(p)}$, we have that $$((df)_p(v+w))(\chi) = (d(\chi \circ f \circ \varphi^{-1}))_{\varphi(p)}((v+w)(\varphi)), $$ and then linearity follows because the classical differential is a linear function. However, is this proof complete? (I am asking this because I have chosen only a particular chart $\varphi$ and proved the linearity, although for every such chart the argument is the same).


There are two separate things to show: (i) you want to show that your definition of $((df)_p(v))(\chi)$ is independent of the choice of chart $\varphi\in C_p$ and (ii) you want to show that $(df)_p(v)$ is a tangent vector in the sense that you've defined it. For (i), suppose $\varphi_1, \varphi_2 \in C_p$. Then \begin{align*} (d(\chi\circ f\circ \varphi_2^{-1}))_{\varphi_2(p)}(v(\varphi_2)) &= (d(\chi\circ f\circ \varphi_2^{-1}))_{\varphi_2(p)}((d(\varphi_2\circ\varphi_1^{-1}))_{\varphi_1(p)}(v(\varphi_1)))\\ &= (d(\chi\circ f\circ\varphi_1^{-1}))_{\varphi_1(p)}(v(\varphi_1)) \end{align*} using your definition of a vector in the first equality, and properties of the classical differential in the second equality.

For (ii), suppose $\chi_1, \chi_2 \in C_{f(p)}$. Then picking any $\varphi\in C_p$, \begin{align*} ((df)_p(v))(\chi_2) &= (d(\chi_2\circ f\circ \varphi^{-1}))_{\varphi(p)}(v(\varphi)) \\ &= (d(\chi_2\circ\chi_1^{-1}))_{\chi_1(p)}((d(\chi_1\circ f\circ \varphi^{-1}))_{\varphi(p)}(v(\varphi))) \\ &= (d(\chi_2\circ\chi_1^{-1}))_{\chi_1(p)} (((df)_p(v))(\chi_1)) \end{align*} Here, the first equality is the definition of the differential, the second uses properties of the classical differential, and the third is again the definition of the differential. Hence $(df)_p(v)$ satisfies your criterion for being a tangent vector to $N$ at $f(p)$.

Now that you've proved $(df)_p(v)$ is defined independent of $\varphi\in C_p$ and $\chi\in C_{f(p)}$, you only need verify linearity in one choice of charts, as you said.


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