# Same fiber of induced covering map [closed]

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.

## closed as off-topic by abx, user44191, Pace Nielsen, Sean Lawton, Friedrich KnopApr 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 $$h$$.