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Tangent space is independent of the coordiante function.

Let $X$ be a smooth n-manifold and $\phi: U\to X$ be a local parametrization at $x$ ($\phi(0)=x$). Define tangent space as $T_x(X)=d\phi_0(\mathbb R^n)$

We claim tangent space is independent of the local parametrization so we take another local parametrization $\psi: V\to X$ and we shrink $U,V$ such that $\phi(U)=\psi(V)$ namely we take a diffeomorphism $h=\psi^{-1}\circ \phi:U\to V$

Such that $\phi=\psi\circ h$ and differentiating it yields:$$d\phi_0=d\psi_0\circ dh_0$$

Question: Why last equation would imply image of $d\psi$ consists image of $d\phi$?

Jale'de jaled
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  • Plus what a confusing book, never gives proper definitions, never gives enough and proper explanations... – Jale'de jaled Nov 15 '20 at 16:15
  • I think it is a wonderful book pedagogically speaking, but not a proper encyclopedic text. This might be helpful, too: https://math.stackexchange.com/questions/970219/prove-that-h-x-x-does-not-depend-on-the-choice-of-local-parametrization/2794375#2794375 – J. Moeller Nov 15 '20 at 23:36
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    I do not know what book you're talking about, but if you're trying to learn this material, it is assumed that you know and understand basic linear algebra and multivariable analysis (including the inverse function theorem). It is an elementary exercise to show that when $f=g\circ h$, then the image of $f$ is a subset of the image of $g$. You should not blame the book here; you should learn the background material first. – Ted Shifrin Dec 02 '20 at 20:10

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To spell it out, if $x\in\operatorname{im}(d\phi_0)$, then $x=d\phi_0(y)$ for some $y$, and so $x=(d\psi_0\circ dh_0)(y)=(d\psi_0)(dh_0(y))$ is in $\operatorname{im}(d\psi_0)$. That is, $\operatorname{im}(d\phi_0)\subset\operatorname{im}(d\psi_0)$.

Also, you may want to check out J. Milnor's Topology from the differentiable viewpoint for what I find to be a clear exposition.

ho boon suan
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