Let $ \left\{U_i \right\}$ be an open cover of $X$. On some online sources and some MSE questions, the map $\coprod _iU_i\rightarrow X$ is given as an example for a local homeomorphism which is not a covering map.

However, the definition of a covering map is not always given. If I define a covering map to be a surjective local homeomorphism, it seems to me the above is is a covering map.

Am I wrong? If not, what are some alternative definitions of covering maps which do not include this example?

  • 1
    $\begingroup$ This question doesn't deserve downvotes. It's a simple misunderstanding. $\endgroup$ Commented Jun 2, 2016 at 16:05

1 Answer 1


The definition of "covering map" is not just "surjective local homeomorphism". A map $p:Y\to X$ is a covering map if for each $x\in X$, there is a neighborhood $U$ of $X$ and a partition of $p^{-1}(U)$ into disjoint open sets $U=\bigcup V_i$ such that the restriction of $p$ to each $V_i$ is a homeomorphism $V_i\to U$. (Some authors also require $p$ to be surjective.)

This implies $p$ is a local homeomorphism, but it is stronger, because the condition is local on $X$ not just on $Y$. That is, we don't just have a neighborhood of each point of $Y$ in which $p$ is a homeomorphism, but a neighborhood of each point of $X$ such that $p$ is a homeomorphism on each set of an open partition of the inverse image. In particular, for instance, consider $X=\mathbb{R}$ and $Y=(-\infty,1)\sqcup (0,\infty)$, with $p:Y\to X$ the obvious map. Then note that if you take any interval $U$ around $0$ or $1$ in $X$, the inverse image will have two components, one of which maps homeomorphically to $U$ but the other of which will only map to a subset of $U$. So this $p$ is not a covering map, even though it is a local homeomorphism.

  • $\begingroup$ @EricWofsey, extremely sorry. I must have pressed the downvote button by mistake. I apologize. $\endgroup$
    – Hmm.
    Commented Jun 2, 2016 at 19:24

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