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Consider a holomorphic map $h: X \to E$ between compact, connected, complex analytic manifolds Let $p: \tilde{E}\to E$ be the universal cover, and denote by $\tilde{h}: \tilde{X}\to\tilde{E}$ the pull-back of $h$ via $p$.

Then why do $h$ and $\tilde{h}$ have the same fibres? Since $\tilde{X}$ is the covering space of $X$, shouldn't the fiber of $h$ be the quotient of the fiber of $\tilde{h}$?

The statement is from here p.4, last paragraph.

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    $\begingroup$ If you want to delete your question for whatever reason, either temporarily or permanently, you can use the delete button (you can undelete it later). Please, do not vandalize your own posts. $\endgroup$ Apr 15, 2019 at 7:46
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    $\begingroup$ @EmilJeřábek I certainly agree with your comment - vandalizing questions is a bad thing. I will just point out that the OP needs to register first. As the asker in this case is an unregistered user, they cannot delete the question. Meta: Unregistered users should be able to delete their own answers. Related discussion on this site: Should MathOverflow require registration to ask a question? $\endgroup$ Apr 15, 2019 at 7:55
  • $\begingroup$ Sorry, I wasn’t aware of that. $\endgroup$ Apr 15, 2019 at 8:18

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Fibers are always preserved by pullbacks, because fibers are themselves pullbacks and compositions of pullbacks are pullbacks.

More precisely, if

$\require{AMScd} \begin{CD} A @>>> B\\ @VVV @VVV\\ C @>>> D \end{CD}$

is a pullback diagram, and

$\require{AMScd} \begin{CD} K @>>> A\\ @VVV @VVV\\ F @>>> C \end{CD}$

too, then

$\require{AMScd} \begin{CD} K @>>> B\\ @VVV @VVV\\ F @>>> D \end{CD}$

is one as well (this is one case of the so-called Pullback Lemma, very general category theory result). Apply this to $F=\{*\}$ then $K$ is the fiber of $A\to C$ if and only if the second square is a pullback. Therefore, if it is the fiber of $A\to C$, it is also the fiber of $B\to D$ by the third square.

Here take $A=\tilde{X}, B=X, D=E,C=\tilde{E}$, $F$ any point whose fiber you wish to consider and $K$ said fiber over $\tilde{h}$, then $K$ is also the fiber of $p($that point $)$ over $h$.

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