2
$\begingroup$

Let $p:Y\to X$ be a covering map of topological spaces where $X$ is assumed to be locally path connected (and hence the same is true of $Y$) but neither $X$ nor $Y$ is assumed to be connected. In this context, what is the correct definition of the regularity of $p$?

My thoughts on this are as follows. Because $X$ is locally path connected, it can be proved that for each path component $X_0$ of $X$, and each connected component $Y_0$ of $Y$ meeting $p^{-1}(X_0)$, the induced map $Y_0\to X_0$ is a covering map (where now we do have connectedness of source and target). For such a covering (which I'll call connected), there is the standard notion of regularity: for any $x_0\in X_0$ and any $y_0\in Y_0$ lying over $x_0$ (via $p$), the image of $\pi_1(Y_0,y_0)$ in $\pi_1(X_0,x_0)$ is normal (that this is independent of $x_0$ and the $y_0$ lying over it uses the connectedness of $Y_0$). It is equivalent to insist that the group $\mathrm{Aut}(Y_0/X_0)$ of deck transformations act transitively on every fiber of $Y_0\to X_0$ (and again this uses connectedness). With this in mind, I think that the most reasonable definition of regularity without global connectedness hypotheses is: for each $x\in X$ and each $y\in p^{-1}(x)$, the image of $p_*:\pi_1(Y,y)\to \pi_1(X,x)$ is normal in the target. Using that the inclusion of a connected component into $Y$ (or $X$) induces an isomorphism on $\pi_1$'s (using the same base point for source and target), one can show that this definition is equivalent to asking that for all $X_0$ and $Y_0$ as above, $Y_0\to X_0$ is regular.

The thing I find dissatisfying about this definition, however, is that it's unclear what it says about the action of the global group of deck transformations $\mathrm{Aut}(Y/X)$ on the fibers of $p$. Each $\mathrm{Aut}(Y_0/X_0)$ will act transitively on the fibers of $Y_0\to X_0$, but given a collection of deck transformations of the various components of $Y$, although we will get an automorphism of $Y$, we won't be able to get between two points of a fiber $p^{-1}(x)$ which lie in different components this way.

Alternatively we could call $p:Y\to X$ regular if $\mathrm{Aut}(Y/X)$ acts transitively on all fibers. This has the satisfying feature that the induced continuous map $Y/\mathrm{Aut}(Y/X)\to X$ is a homeomorphism...but again, it seems to have the disadvantage that it doesn't obviously (to me) imply regularity of the various connected covers $Y_0\to X_0$ arising from $p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.