# Why is $E \sqcup F$ not path connected?

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.

Thanks

• By definition $E \sqcup F$ has at least two components, namely $E$ and $F$. – Ben Steffan Mar 25 at 22:34
• 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 – William Mar 25 at 23:28
• Yes you are right, thanks! @William – 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.