Cover of a simply connected and lpc space is trivial

I have I question about a reduction step in the proof of the statement that a cover of simply connected and locally path-connected space is trivial (source: "Fundamental Groups and Galois Groups" by Szamuely, Tamás; page 40):

There is said that it's enough to show that for a connected cover $$p: Y \to X$$ the lemma is injective.

My question is why is it enough to show this statement to conclude already lemma 2.4.4?

Remark: I know that there are also ways to show 2.4.4 directly but the point of my interests is to understand why it suffice to show the second statement or in other words why the second statement already imply lemma 2.4.4?

Let $$p : Y \to X$$ be a covering, where $$X$$ is simply connected and lpc. Then $$Y$$ is lpc because $$p$$ is a local homeomorphism. Hence all path components $$Y_\alpha$$ of $$Y$$ are open subsets of $$Y$$.

Moreover, the restrictions $$p_\alpha : Y_\alpha \to X$$ are coverings.

1) $$p_\alpha$$ is surjective.

Let $$x \in X$$. Choose any $$y \in Y_\alpha$$ (which is possible because $$Y_\alpha \ne \emptyset$$). Let $$u : [0,1] \to X$$ be a path from $$p(y)$$ to $$x$$. Then $$u$$ can be lifted to a path $$v : [0,1] \to Y$$ such that $$v(0) = y$$. Clearly $$p(v(1)) = x$$. But we must have $$v([0,1]) \subset Y_\alpha$$ since $$Y_\alpha$$ is the path component of $$Y$$ which contains $$y = v(0)$$.

2) $$p_\alpha$$ is a covering.

Let $$x \in X$$ and $$U$$ be an open path connected neighborhood of $$x$$ which is evenly covered. Then $$p^{-1}(U) = \bigcup_\beta U_\beta$$ with pairwise disjoint open $$U_\beta \subset Y$$ which are mapped by $$p$$ homeomorphically onto $$U$$. Each $$U_\beta$$ is path connected, thus contained in a unique $$Y_{f(\beta)}$$. Hence $$U_\beta \cap Y_\alpha = U_\beta$$ if $$f(\beta) = \alpha$$ and $$U_\beta \cap Y_\alpha = \emptyset$$ if $$f(\beta) \ne \alpha$$. This shows that $$p_\alpha^{-1}(U) = p^{-1}(U) \cap Y_\alpha = \left(\bigcup_\beta U_\beta \right) \cap Y_\alpha = \bigcup_\beta (U_\beta \cap Y_\alpha) = \bigcup_{\beta \text{ with } f(\beta) = \alpha} U_\beta .$$ Therefore $$U$$ is evenly covered by $$p_\alpha$$. Note that the $$U_\beta$$ with $$f(\beta) = \alpha$$ are pairwise disjoint open subsets of $$Y_\alpha$$ which are mapped by $$p_\alpha = p \mid_{Y_\alpha}$$ homeomorphically onto $$U$$.

If we can prove what is claimed to be enough, we know the all connected coverings of $$X$$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $$p_\alpha$$ and deduce that $$X$$ itself is evenly covered.

• Hi, thank you for you answer. One point is unclear for me: When you show that $p_\alpha : Y_\alpha \to X$ are coverings how do you see that $p_\alpha^{-1}(U) = \bigcup_{\beta \text{ with } f(\beta) = \alpha} U_\beta$ isn't empty? Therefore why there exist a $\beta$ with $f(\beta) = \alpha$ and $\#\{\beta \vert f(\beta) = \alpha\}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$? – KarlPeter Jan 10 at 1:49
• You are right, my proof contains a gap. I forgot to show that the $p_\alpha$ are surjective. I shall edit my answer. – Paul Frost Jan 10 at 12:24
• Note that $p_\alpha^{-1}(U)$ is non-empty because $p_\alpha$ is surjective. – Paul Frost Jan 10 at 13:00