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I try to proof the following statement: Let $(W_p, F_p, t_p)$ and $(W_p', F_p', t_p')$ be two local trivialization around the same point $p \in M$, then $F_p$ and $F_p'$ are diffeomorphic.

I showed that $(W_p \cap W_p') \times F_p$ and $(W_p \cap W_p') \times F_p'$ are diffeomorphic by using the fact that $t_p' \circ t_p^{-1}$ is a diffeomorphism. But how can I conclude from here that $F_p$ and $F_p'$ are diffeomorphic? I assume that I should do something with the projection onto the second factor $\text{pr}_2$ and the map $\text{pr}_2 \circ t_p' \circ t_p^{-1}: (W_p \cap W_p') \times F_p \to F_p'$ and show that this a diffeomorphism. But I don't know what the inverse of $\text{pr}_2$ would be. Can someone help or can tell me if my ideas are correct or not?

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Consider $i_p:F_p\rightarrow W_p\cap W'_p\times F_p$ defined by $i_p(x)=(p,x)$ and $i'_p:F_p\rightarrow W_p\cap W'_p\times F'_p$ defined by $i'_p(x)=(p,x)$. Write $f=pr_2\circ (t'_p\circ t_p^{-1})\circ i_p$ and $g=pr_2\circ (t_p\circ {t'_p}^{-1})\circ i'_p$.

$f(g(x))=f(pr_2(t_p\circ {t'_p}^{-1})(i'_p(x))=pr_2(t'_p\circ t_p^{-1})(i_p(pr_2(t_p\circ {t'_p}^{-1})(i'_p(x)))$.

Since $i_p\circ pr_2:p\times F_p\rightarrow p\times F_p$ is the identity, you deduce that

$f(g(x))=pr_2(t'_p\circ t_p^{-1})(t_p\circ {t'_p}^{-1})(i'_p(x)))=pr_2(i'_p(x))=x$.

Similarly,show that $g\circ f=Id$.

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  • $\begingroup$ In the fourth line do you mean $pr_2 \circ i_p$? The other way around it doesn't make sense. By your argument I don't have to show that the map you constructed is a diffeomorphism, right? I am not sure about this since I don't know about the inverse of $pr_2$. $\endgroup$ – JDoe Jul 14 '16 at 14:44
  • $\begingroup$ $i_p\circ pr_2:p\times F_p\rightarrow p\times F_p$ is the identity $\endgroup$ – Tsemo Aristide Jul 14 '16 at 14:47
  • $\begingroup$ Okay now it is clear. Do I have to restrict the map to $p$ or can I chose an arbitrary point $q \in W_p \cap W_p'$? And can you say something about my concerns about $pr_2$? $\endgroup$ – JDoe Jul 14 '16 at 14:57

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