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Let $\phi : Y \rightarrow X$ be a $n$-sheet covering of the 8 figure. Prove that there exist $\psi : Z \rightarrow X$ an at most $n!$- sheet regular covering such that the diagram commutes (where $\mu$ is a covering). $\require{AMScd}$ \begin{CD} Z @>\mu>> Y\\ @V \psi V V @VV \phi V\\ X @>>Id> X \end{CD}

(Sorry, but I don't know how to draw diagonal arrows with this code. It should have an arrow $\psi$ from $Z$ to $X$)

I have been thinking of doing this by induction. Assuming the statement true for $(n-1)$-sheeted coverings. We enumerate the $n$ vertices of $Y$.

I want to apply the statement to $Y_1',\ldots, Y_n'$, where $Y_j'$ is $Y$ without the vertex $j$. Here I glue the corresponding arrows arriving and leaving (from the viewpoint of the vertex $j$, $a^{-1}$ continues as $a$ and the same with $b$, where $a$ and $b$ are the arrows that project to each loop in figure 8. If there is any loop in $j$, I simply ignore it).

Each time I will obtain an at most (n-1)!-sheeted regular covering $Z_j$ satisfying the conditions. But I don't see how to glue them to obtain the at most $n\times (n-1)!$-sheeted regular covering.

This idea comes to me just by the numbers involved. Maybe I should proceed in a different way. Any advice? Thanks

GingFreecss17
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    Using the correspondence between (path-connected) covering projections over (sufficiently nice) spaces $X$ and conjugacy classes of subgroups of $ \pi_1(X)$, what you are trying to show reduces to proving that if $H$ is a subgroup of finite index $n$ of a group $G$ , then $H$ contains a normal subgroup of index at most $n!$. This can be proved by considering the kernel of the homomorphism $G \to S_n$ you get from the left (or right) action of $G$ on the cosets of $H$. – Rob Arthan Oct 07 '21 at 22:20
  • @RobArthan Thank you, sir. Your comment has been extremely useful. – GingFreecss17 Oct 08 '21 at 05:19

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