# bundle isomorphism

Let $M$ be a manifold and let $U_{\alpha}$ and $U_{\beta}$ be coordinate charts with coordinates $x^{\alpha}$ and $x^{\beta}$, respectively.

How to show that $f_{\alpha} : U_{\alpha}\times\mathbb{R}^n\to TM_{|U_{\alpha}}$, $(p,a^{\alpha})\to\sum_{i} a^\alpha_i\frac{\partial}{\partial x^\alpha_i}|_{p}$ is a bundle isomorphism?

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What is your question? – Eric O. Korman Oct 22 '12 at 16:07
What is the relation of $U_\beta$? With what are you having trouble? – levap Oct 23 '12 at 11:01

Checking that $f_{\alpha}$ is a bundle isomorphism is the same as checking two things

• $f_{\alpha}$ is a diffeomorphism ;

• $f_{\alpha}$ commutes with the projection maps $\pi_{\alpha} : \mathrm{U}_{\alpha} \times \mathbb R^n \to \mathrm{U}_{\alpha}$ and $\pi'_{\alpha} : \mathrm{TM}_{|\mathrm U_{\alpha}} \to \mathrm{U}_{\alpha}$ i.e you have to check that $f_{\alpha} \circ \pi_{\alpha} = \pi'_{\alpha} \circ f_{\alpha}$.

By definition of the tangent bundle $$\mathrm{TM}_{|\mathrm U_\alpha} = \bigcup_{x \in \mathrm U_{\alpha}} \{x\} \times \mathbb R^n$$

We deduce that $f_{\alpha}$ is a diffeomorphism. The only thing left to check is the compatibility with the projection maps. But this comes immediately.

Hence $f_{\alpha}$ is a bundle isomorphism.

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