# Morphism of covering

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?

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 $$\tilde u(1) \in A \cap X_b \subset P \cap X_b$$.
• 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. Feb 2, 2021 at 10:45