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It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

Addendum: The reason that $X'$ exists is that if $h\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $f$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

Addendum: The reason that $X'$ exists is that if $h\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $h$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

Addendum: The reason that $X'$ exists is that if $h:\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $f$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

Addendum: The reason that $X'$ exists is that if $h\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $f$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.

Addendum: The reason that $X'$ exists is that if $h:\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $f$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.