0
$\begingroup$

I'm a bit lost in Differential Topology by Victor Guillemin and Alan Pollack. The book states in Chapter 2:

Before proceeding, we must resolve an ambiguity in the definition of $T_x(X)$; will another choice of local parametrization produce the same tangent space? Suppose $\psi : V \rightarrow X$ is another choice [with $\phi$ as the original choice], with $\psi = 0$ as well. Then the map $h = \psi^{-1} \circ \phi$ is a diffeomorphism. Consider $\phi = \psi \circ h$, and take the derivative to get $d\phi_0 = d\psi_0 \circ dh_0$.

When I differentiate I got:

$$\phi = \psi \circ h$$ $$d\phi_0 = d(\psi \circ h)_0$$ $$d\phi_0 = d\psi_{h(0)} \circ dh_0$$

But how do we know that $h(0) = 0$?

$\endgroup$
  • $\begingroup$ It is assumed that $\phi(0) = x = \psi(0)$. This implies $h(0) = \psi^{-1}( \phi( 0 )) = \psi^{-1}(x) = 0$. $\endgroup$ – Dominique R.F. Sep 5 '16 at 6:38
  • $\begingroup$ @DominiqueR.F.: Ok that makes sense. Thanks for the clarification. $\endgroup$ – Dair Sep 5 '16 at 6:43
1
$\begingroup$

It is assumed throughout this argument that the charts are centered at $x$, that is $\phi(0) = x = \psi(0)$. This implies $h(0) = \psi^{-1}( \phi(0 )) = \psi^{-1}(x) = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.