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I have a question about some hypothesis in a classical theorem about coverings.

The theorem is the following:

Let $B$ be a connected base space, and $X$ and $Y$ be coverings of $B$ and $g, f: X\to Y$ two $B$-morphisms. If there exists $b\in B$ such that $f_{|X_b} = g_{|X_b}$ then $f=g$.

This is easy to prove is $X$ is connected. If you don't assume $X$ connected, then it is also easy to prove if you assume $B$ locally connected, as in that case every connected component of $X$ is open and closed, and therefore maps surjectively onto $B$ and thus, every connected component of $X$ contains a point of the fiber $X_b$.

This is of course a standard assumption.

But the reference I'm looking at (Douady, Algèbres et théorie galoisiennes) does not make any assumption (other than $B$ connected) and simply states that "every connected component of $X$ contains a point in the fiber $X_b$".

And I fail to see how to prove that whithout the local connectedness assumption, am I missing something?

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This is only a partial answer: It is true if $B$ is path connected.

Let $P$ be a path component of $X$. Note that in general the components of $X$ do not coincide with its path components; but if we prove that $P \cap X_b \ne \emptyset$ for all $b$, then also $C \cap X_b \ne \emptyset$ for all components $C$ and all $b$. Pick $x_0 \in P$ and let $b_0 = p(x_0)$, where $p : X \to B$ denotes our covering projection. Then $x_0 \in C \cap X_{b_0}$. Now let $b \in B$. Choose a path $u$ in $B$ such that $u(0) = b_0$ and $u(1) = b$. Lift $u$ to a path $\tilde u$ in $X$ such that $\tilde u(0) = x_0$. Since $A =\tilde u([0,1])$ is path connected, we get $A \subset P$. But $\tilde u(1) \in A \cap X_b \subset P \cap X_b$.

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  • $\begingroup$ Yes, this works too. I'm still wondering if the theorem is false without any additionnal hypothesis on $B$, be it path connected, or locally connected. In any case I feel like the author is going a bit too fast with his claim, but I'm still wondering whether it's true if you only assume $B$ connected, I've failed to find a counter example. $\endgroup$
    – Nissokam
    Feb 2, 2021 at 10:45
  • $\begingroup$ @Nissokam I doubt that it is true without additional assumptions. But it will not be easy to find counterexamples. $\endgroup$
    – Paul Frost
    Feb 2, 2021 at 10:58

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