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Going through the proof of the proposition in the title.

Let $M_1^n, M_2^m$ be differentiable manifolds and let $\varphi : M_1 \to M_2$ be a differentiable mapping. For every $p \in M_1$ and for each $v \in T_p M_1$, choose a differentiable curve $\alpha : (-\epsilon,\epsilon) \to M_1$ with $\alpha(0) = p$, $\alpha'(0) = v$. Take $\beta = \varphi \circ \alpha$. The mapping $d \varphi_p : T_p M_1 \to T_{\varphi(p)} M_2$ given by $d \varphi_p (v) = \beta'(0)$ is a linear mapping that does not depend on the choice of $\alpha$.

There's a very specific bit I can't figure, which I'd like you'd help me to elaborate.

Proof: Let $x : U \to M_1$ and $y : V \to M_2$ be a parameterizations at $p$ and $\varphi(p)$, respectively. Expressing $\varphi$ in these parameterization, we can write $$ y^{-1} \circ \varphi \circ x (q) = (y_1(x_1,\ldots,x_n),\ldots,y_m(x_1,\ldots x_n)) $$ where $$ \begin{array}{l} q = (x_1,\ldots,x_n) \in U \\ (y_1,\ldots,y_m) \in V \end{array}. $$ On the other hand, expressing $\alpha$ in the parameterization $x$, we obtain $$ x^{-1}\circ \alpha(t) = (x_1(t),\ldots,x_n(t)). $$ Therefore, $$ y^{-1} \circ \beta(t) = (y_1(x_1(t),\ldots,x_n(t)),\ldots,y_m(x_1(t),\ldots x_n(t))) $$

Here now the very bit I don't understand

It follows the expression for $\beta'(0)$ with respect to the basis $ \left\{ \left( \frac{\partial}{\partial y_i}\right) _0\right\}$ of $T_{\varphi(p)}M_2$, associated to the parameterization $y$, is given by $$ \beta'(0) = \left(\sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0), \ldots, \sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0) \right). $$

The rest of the proof is clear

How exactly is the basis $ \left\{ \left( \frac{\partial}{\partial y_i}\right)_0 \right\}$ used to derive an expression for $\beta'(0)$?

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Note that $\beta:(-\epsilon,\epsilon)\to M_2$. So $\beta(0)\in M_2$ and since $M_2$ is of dimension $m$ so $\beta(t) = (y_1(x_1(t),\ldots,x_n(t)),\ldots,y_m(x_1(t),\ldots x_n(t)))....(1)$

Also $\{(\frac{\partial}{\partial y_i})|_t:1\le i\le m\}$ is a basis of the tangent space $T_{\beta(t)}M_2$ at $\varphi(p)=\beta(t)$ for any $t\in(-\epsilon,\epsilon)$.

Now differentiating $(1)$ at $t=0$ we obtain,

$$\beta'(0) = \left(\sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0), \ldots, \sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0) \right).$$ With respect to the basis means, $$\beta'(0) = \sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0)\frac{\partial}{\partial y_1}+ \ldots+\sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0)\frac{\partial}{\partial y_m} .$$

Hope this works.

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  • $\begingroup$ Need some time to digest your answer. $\endgroup$ – user8469759 Mar 21 at 12:06

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