# explain proof provided

http://pages.uoregon.edu/koch/math432/Universal_Cover.pdf

In the link above, section $4$ $(C \rightarrow X$ is a covering space), I don't understand the last paragraph where the author prove that the restricted map from $<U,\gamma_i>$ to $U$ is an open mapping. Can anyone explain this to me???

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First observe that if $U$ is not path connected, then $\langle U, \gamma \rangle = \langle U_0, \gamma \rangle$, where $U_0$ is the path component containing the point $\gamma(1)$. This allows to only considers opens $\langle U, \gamma\rangle$ where $U$ is path-connected.
Then observe that for $U$ path-connected, the image of $\langle U, \gamma \rangle$ under the projection $\pi \colon C \to X$ is just $U$ now!
Finally, to show that the restriction $\pi' \colon \langle U, \gamma_i\rangle \to U$ is open, take an arbitrary open subset of $\langle U, \gamma_i \rangle$. This is of the form $\bigcup_{\alpha} \langle U_{\alpha}, \tau_{\alpha} \rangle$ for some open subsets $U_{\alpha}$, which we can take path-connected by the above. Now you have $$\pi'\left( \bigcup_{\alpha} \langle U_{\alpha}, \tau_{\alpha} \rangle \right) = \bigcup_{\alpha} \pi' \left( \langle U_{\alpha}, \tau_{\alpha} \rangle \right) = \bigcup_{\alpha} U_{\alpha}$$ which is an open subset of $U$ and thus $\pi'$ is open. Note that you can here permute $\pi'$ and the union $\bigcup$ because $\pi'$ is a bijection!