# How do I know that $h(0)$ is $0$?

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 $h(0) = 0$?

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

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