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Let $f:\tilde{M}\to M$ be a $d$-fold covering map and $N\subset M$ be a codimension 2 submanifold. What is the number of path-connected components of $f^{-1}(N)$ (assuming $N$ is connected)?

usr1988
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  • My question boils down to the following: By the definition of a covering map, there is a neighborhood $U\subset M$ of $p\in N\subset M$ such that $f^{-1}(U)$ has $d-$path connected components. If it is possible to shrink $N$ so that $N\subset U$, then obviously $f^{-1}(N)$ has $d-$path connected components. But I'm not sure this is indeed possible. – usr1988 Jul 26 '18 at 10:05
  • What do you mean by "shrinking" $N$? – Paul Frost Jul 26 '18 at 10:54

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It seems you want to show that $N' = f^{-1}(N)$ has $d$ path components. But $N'$ may have only one. Example:

$M = \tilde{M} = S^1 \times S^2, f(z,\xi) = (z^d, \xi)$ (i.e. $f$ is the product of the standard $d$-fold covering of $S^1$ and the covering $id : S^2 \to S^2$).

Then $N = S^1 \times \{ a \}$ is a connected codimension $2$ submanifold such that $f^{-1}(N) = N$.

Another example is $M = \mathbb{RP}^3$ = $3$-dimensional real projective space with universal cover $p : S^3 \to \mathbb{RP}^3$ which has two sheets. The circle $S = S^1 \times \{(0,0) \} \subset S^3 \subset \mathbb{R}^4$ is mapped by $p$ to $p(S^1) = S /\sim$, where $\xi \sim \xi'$ if $\xi = \pm \xi'$. Therefore $p(S) \approx \mathbb{RP}^1 \approx S^1$. $p(S)$ is a a connected codimension $2$ submanifold of $\mathbb{RP}^3$ such that $p^{-1}(p(S)) = S$.

Paul Frost
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  • Thanks a lot, this is a very clear counter example. – usr1988 Jul 26 '18 at 12:08
  • By the way, what if M is not a product manifold? Do you think $N'$ does not have $d-$path connected components? – usr1988 Jul 26 '18 at 12:19
  • Although I don't have a concrete example, I am convinced we have the same situation. – Paul Frost Jul 26 '18 at 12:25
  • I added an example. – Paul Frost Jul 26 '18 at 12:49
  • In your examples, each path-connected component of $f^{-1}(N)$ is diffeomorphic to $N$, right? I think that this is not true in general. For instance, consider a covering $f:\mathbb{R}\times S^2\to S^1\times S^2$ and take $N=S^1\times {pt}\subset S^1\times S^2$. Then $f^{-1}(N)=\mathbb{R}\times {pt}$, which is not diffeomorphic to $N$. Do you know in which case each path-connected component of $f^{-1}(N)$ is diffeomorphic to $N$? – usr1988 Aug 03 '18 at 08:16
  • No, I do not know any criterion for that (except the trivial one that $N$ is contained in an evenly covered open set). Let us restrict to connected $N$. Then the restriction $p_C : C \to N$ to any component of $f^{-1}(N)$ is a covering. Note that even if $C$ is diffeomorphic to $N$, $p_C$ need not be a diffeomorphism. – Paul Frost Aug 03 '18 at 09:44