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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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closed as off-topic by abx, user44191, Pace Nielsen, Sean Lawton, Friedrich Knop Apr 17 at 6:13

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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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