Let $B$ be an arc-connected and locally arc-connected space. Suppose that $p:E\to B$ and $q: F\to B$ are covering spaces. Let $r:E\sqcup F\to B$ be the function such that $r(x)=p(x)$ for all $x\in E$ and $r(x)=q(x)$ for all $x\in F$. Show that $r:E\sqcup F\to B$ is a covering space.

"Take $x\in B$, then there is an open $U$ of $B$ such that $x\in U$ and $p^{-1}(U)=\sqcup_{\alpha\in A}V_{\alpha}$, where $V_{\alpha}\in E$ for all $\alpha\in A$, in addition, there is also an open $W$ of $B$ such that $x\in W$ and $q^{-1}(W)=\sqcup_{\beta\in B}S_{\beta}$

So, $p^{-1}(U)=\sqcup_\alpha V_\alpha$ with $V_\alpha\subseteq E$ and such that $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism.
Restricting the codomains to $U\cap W$, we still obtain homeomorphisms $V_\alpha\cap p^{-1}(W)\to U\cap W$,
and we will explicitly have $$r^{-1}(U\cap W)=\bigsqcup_\alpha (V_\alpha\cap p^{-1}(W))\ \sqcup\ \bigsqcup_\beta (S_\beta\cap q^{-1}(U))\,.$$ "

That is from another question. But I am interested in how to show that $E \sqcup F$ is not path connected.


  • 3
    $\begingroup$ By definition $E \sqcup F$ has at least two components, namely $E$ and $F$. $\endgroup$ – Ben Steffan Mar 25 at 22:34
  • $\begingroup$ Your title should be a concise summary of the question you're asking, not the first few of sentences of set-up (people can always get more detailed information by reading your question). A title like "Why is $E\sqcup F$ not path connected?" would be a more appropriate. Here are some other helpful tips for asking a good question: math.stackexchange.com/help/how-to-ask $\endgroup$ – William Mar 25 at 23:28
  • $\begingroup$ Yes you are right, thanks! @William $\endgroup$ – User96 Mar 25 at 23:35

Using the fact that the interval $I$ is connected you can show that any path-connected space is also connected, as follows:

Suppose $X$ is path connected, but suppose it is not connected so we can write $X$ as a disjoint union $E\sqcup F$ where $E$ and $F$ are non-empty open subsets. Let $\gamma\colon I \to X$ be a path such that $\gamma(0)\in E$ and $\gamma(1)\in F$. Then $\gamma^{-1}(E)$ and $\gamma^{-1}(F)$ are disjoint, non-empty open subsets of $I$, whose union is $I$: in other words $I$ is not connected, which is a contradiction. Therefore if $X$ is path-connected it must also be connected.

Now for your problem take the contrapositive: if $E\sqcup F$ is not connected then it is not path-connected either.

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