My question involves topological spaces $X$, $Y$ and $Z$, two coverings $p : Y \rightarrow X$ and $q: Z \rightarrow X$ of $X$ and a morphism $f: Y \rightarrow Z$ of coverings, i.e. a map which fulfills $p = q \circ f$.

My question is: Is $f$ itself a covering map $Y \rightarrow Z$ ?

First, I thought it should be, because given an element $z \in Z$ we have an open neighborhood $U \subseteq X$ with $q(z) \in U$ such that $p^{-1}(U) = \bigcup_{i \in I}U_i$ and $q^{-1}(U) = \bigcup_{j \in J} V_j$ are disjoint unions of open subsets with $p|_{U_i} : U_i \rightarrow U$ and $q|_{V_j} : V_j \rightarrow U$ homeomorphims. Because of $p = q \circ f$ I get $$ \bigcup_{i \in I}U_i = p^{-1}(U) = f^{-1}(q^{-1}(U)) = \bigcup_{j \in J} f^{-1}(V_j).$$ Now there is exactly one $j \in J$ with $z \in V_j$ and if I let $I':= \{i \in I \:|\: f(U_i) \subseteq V_j\}$ for all $i \in I'$ I have $f|_{U_i} = (q|_{V_j})^{-1} \circ p|_{U_i}$ (right?) which then would be a homeomorphism. But now I still have to show something like $f^{-1}(V_j) = \bigcup_{i \in I'} U_i$ to get a disjoint union of opens which are mapped homeomorphically onto $V_j$ and this is where I started to doubt my first assessment. I am not able to show why it is impossible for some $i \in I$ that $f(U_i) \cap V_j \neq \emptyset$ without $f(U_i) \subseteq V_j$. I believe if I assumed $X$ to be locally connected then I could choose the $U_i$ to be connected and I should be done.

So, in fact there are two things I would like to know, namely if my argument up to the point I stated it is valid and if one can go on and show the "proposition" without assuming local connectedness.

EDIT: As was pointed out, the answer to my question is negative if $f$ is not surjective. Is there a chance of saving it by assuming surjectivity and continuity of $f$?

  • $\begingroup$ I don't understand why $f$ has to be onto. For example, take $X = \{x\}$, and take $Z = \{x_1, x_2\}$. Take $Y = X$, take $p = id$, and take $q(x_1), q(x_2) = x$. Define $f(x) = x_1$. Why is this not a counterexample? $\endgroup$ – user2093 Feb 15 '12 at 18:56
  • $\begingroup$ You are absolutely right, this is a fine counter-example. When I finished writing the question, I had the same idea, but due to some issues with my internet connection I must have accidently asked the question. I edited the question. $\endgroup$ – Matthias Klupsch Feb 15 '12 at 19:08
  • $\begingroup$ But then you could ask whether you can identify sufficient conditions on $f$ and $q$ that would make $f$ a covering map. Obviously there is no question when $f$ is a homeomorphism. $\endgroup$ – user2093 Feb 15 '12 at 19:12
  • $\begingroup$ :) Sure... you posted that as I was editing my comment. $\endgroup$ – user2093 Feb 15 '12 at 19:17
  • $\begingroup$ So it seems that $f$ being continuous and surjective is enough. Take $z\in Z$. Say $U_z$ is a sufficiently small open neighborhood around $z$ in $Z$, mapping homeomorphically onto an open neighborhood $q(U_z)$ in $X$. The preimage of $U_z$ under $f$ in $Y$ coincides with that of $q(U_z)$ under $p$. Take any full open connected $V$ in $p^{-1}(q(U_z))$. Then $f(V) = U_z$, and $f|_V = q^{-1}\circ p|_V$, which is a homeomorphism. $\endgroup$ – user2093 Feb 15 '12 at 19:22

If $p:Y\to X$ and $q:Z\to X$ are coverings of a locally connected topological space $X$, then any morphism $f:Y\to Z $ of those coverings is a covering.
We may assume $X$ connected and $Y=X\times D, Z=X\times E$ both trivial ($D,E$ are discrete spaces).
The morphism $f$ is then of the form $f: X\times D \to X\times E: (x,d) \mapsto (x, \phi (x,d))$.
The key point is that $\phi (x,d)$ actually doesn't depend on $x$ because $X$ is connected and $\phi (x,d)\in E$, a discrete space.
Hence we may write $\phi (x,d)=\Phi (d)$ and we see that our map $f$ is $$f: Y=X\times D \to Z= X\times E :(x,d)\mapsto (x,\Phi (d))$$
It is a covering because the inverse image of the open subset $X\times \lbrace e \rbrace \subset Z$ is $X \times \Phi^{-1} (e))$, a disjoint sum of copies of the open set, and because these open sets cover $Z$.

On surjectivity: a point of view
A covering space $f:Y\to Z $ is a continuous map such that $Z$ can be covered by open subsets $U$ over which $f$ is a trivial covering space $U\times D$ (where $D$ is a discrete topological space).
This allows for the possibility that $D$ is empty, so that a covering space need not be surjective.
Of course definitions are arbitrary and if you want you can demand that a covering be surjective.
I consider that a clumsy condition which will constantly force you to add to perfectly natural proofs a surjectivity check.

  • $\begingroup$ Thanks for your answer. Concerning the surjectivity issue you have a good point. In the texts I read about the topic the authors always assumed surjectivity, but indeed it seems to become more natural without it. Thanks for that insight, too. $\endgroup$ – Matthias Klupsch Feb 16 '12 at 7:52
  • $\begingroup$ I agree that the definition of covering map should not include surjective. Hatcher's book agrees with that. $\endgroup$ – Ronnie Brown Apr 23 '12 at 11:06
  • $\begingroup$ Ah, I'm very happy to have your support, @Ronnie. $\endgroup$ – Georges Elencwajg Apr 23 '12 at 12:46
  • $\begingroup$ @GeorgesElencwajg: can you kindly clarify why $\Phi$ is surjective? I can't see why $\Phi^{-1}(e)$ is nonempty. $\endgroup$ – yojusmath Oct 16 '13 at 21:21
  • $\begingroup$ I never said that $\Phi$ was surjective. On the contrary, my "point of view" in the last section defends the thesis that covering spaces need not be surjective. $\endgroup$ – Georges Elencwajg Oct 16 '13 at 21:33

A recent paper

Brazas, Jeremy, "Semicoverings: a generalization of covering space theory", Homology, Homotopy and Applications, 14 (2012) 33--63, (electronic)

gives a generalisation of covering maps, so that your $f: Y \to Z$ now becomes a semi-covering!


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