If a covering map has a section, is it a $1$-fold cover?

Question

If $q: E\rightarrow X$ is a covering map that has a section (i.e. $f: X\rightarrow E, q\circ f=Id_X$) does that imply that $E$ is a $1$-fold cover?

Answer 1

It follows from your assumptions that $q$ is a 1-sheeted and is a homeomorphism. I'm going to call the map $\pi$ instead of $q$ for the rest of this post.

Assume we have a covering $\pi:X\rightarrow Y$ and $f:Y\rightarrow X$ with $\pi\circ f = Id_X$.

I claim that $f(Y)$ is both open and closed in $X$.

To see it, for any $\hat{p}\in X$, let $p = \pi(\hat{p})$. Choose a neighborhood $U$ around $p$ for which $\pi$ trivializes: $\pi^{-1}(U) = \coprod V_\alpha$ with $\pi|_{V_\alpha}$ a homeomoprhism. and let $V$ be the particular $V_\alpha$ containing $\hat{p}$.

Now, if $\hat{p}\in f(Y)$, then $V\subseteq f(Y)$. This follows from considering the inclusion $i:U\rightarrow Y$. Since both $f|_{U}$ and $\pi^{-1}|_{U}$ are lifts of this inclusion agreeing at $\hat{p}$, they must agree on all of $U$. It follows that $V=\pi^{-1}(U) = f(U)$ as claimed. This shows $f(Y)$ is open.

If, on the other hand $\hat{p}\notin f(Y)$, a very similar argument shows that $V\cap f(Y) = \emptyset$, showing that $f(Y)^c$ is open, that is, that $f(Y)$ is closed.

Putting this together, $f(Y)$ is open and closed. Hence, it is a connected component of $X$. If $X$ itself is connected, this implies $f(Y) = X$ which implies that $\pi$ is a homeomorphism with inverse $f$ so, is in particular, 1 sheeted.

Answer 2

Why an answer to a six year old question? Simply because it is an interesting question and the existing answer applies only under the assumption that $X$ is locally connected.

Without any local niceness assumption we shall prove a more general result. The answer is given as a community wiki because the essential idea is contained in the question Section of a covering projection from a connected space which was closed as a duplicate of the present one.

Differing from the standard definition, we do not require covering maps to be surjective. This interpretation can be found for example in Hatcher's "Algebraic Topology". Note that if $q : E \to X$ is a covering map in this sense, then $X' = q(E)$ is a clopen subset of $X$ and $q' : E \xrightarrow{q} X'$ is a covering map in the standard sense and $q^{-1}(X \setminus X') = \emptyset$.

Let us observe that $1$-fold coverings are nothing else than homeomorphisms. A $1$-fold covering is obviously a bijection. Since all coverings are open maps, we see that $1$-fold coverings are homeomorphisms. The converse is trivial.

By a splitting of a covering map $q : E \to X$ we mean a partition of $E$ into a pair open subsets $(E_1, E_2)$ of $E$ such that the $q \mid_{E_i} : E_i \to X$ are covering maps. Note that if we are given a partition of $E$ into open subsets $E_1, E_2 \subset E$ and covering maps $q_i : E_i \to X$, then the map $q : E \to X$ such that $q \mid_{E_i} = q_i$ is a covering map.

Theorem. Let $q : E \to X$ be a covering map which has a section $f : X \to E$. Then $f(X)$ is clopen in $E$ and $(f(X),E \setminus f(X))$ is a splitting of $q$.

Remark. $q \mid_{f(X)} : f(X) \to X$ is trivially a homeomorphism (its inverse being $f' : X \xrightarrow{f} f(X)$. Thus it is a $1$-fold covering.

Proof. Let $x \in X$. There exists an open neighborhood $U(x)$ of $x$ in $X$ which is evenly covered, i.e. we have $q^{-1}(U(x)) = \bigcup_{\alpha \in A} U_\alpha(x)$ with pairwise disjoint open $U_\alpha(x) \subset E$ such that the restrictions $q^x_\alpha : U_\alpha(x) \to U(x)$ of $q$ are homeomorphisms.

Let $\alpha(x)$ be the unique index such that $f(x) \in U_{\alpha(x)}(x)$. Since $f$ is continuous, there exists an open neighborhood $U'(x) \subset U(x)$ of $x$ such that $f(U'(x)) \subset U_{\alpha(x)}(x)$. Obviously $q^x_{\alpha(x)}(f(U'(x))) = q(f(U'(x))) = U'$, hence $f(U'(x)) = (q^x_{\alpha(y)})^{-1}(U'(x))$.

This shows that $f(U'(x))$ is open in $E$, thus $f(X) = \bigcup_{x \in X} f(U'(x))$ is open in $E$.

As a subset of $U(x)$ also $U'(x)$ is evenly covered, with decomposition $q^{-1}(U'(x)) = \bigcup_{\alpha \in A} U'_\alpha(x)$, where $U'_\alpha(x) = U_\alpha(x) \cap q^{-1}(U'(x)) = (q^x_\alpha)^{-1}(U'(x))$.

If $y \notin f(X)$, we have $$y \in q^{-1}(U'(q(y)) \setminus f(X) = q^{-1}(U'(q(y)) \setminus f(U'(q(y))) = q^{-1}(U'(q(y)) \setminus U'_{\alpha(q(y))}(q(y)) \\= \bigcup_{\alpha \in A \setminus \{\alpha(q(y))\}} U'_\alpha(q(y))$$ which is an open subset of $E$ not intersecting $f(X)$. This shows that $E \setminus f(X)$ is open in $E$.

Moreover, $q' = q \mid_{E \setminus f(X)}$ is a covering map. In fact, the sets $U'(x)$ are evenly covered by $q'$ with sheets $U'_\alpha(x)$, $\alpha \in A \setminus \{\alpha(x)\}$. In fact, take $y = f(x)$ in the above equation. Then $$(q')^{-1}(U'(x)) = (q')^{-1}(U'(q(y)) = q^{-1}(U'(q(y)) \setminus f(X) = \bigcup_{\alpha \in A \setminus \{\alpha(q(y))\}} U'_\alpha(q(y)) \\= \bigcup_{\alpha \in A \setminus \{\alpha(x)\}} U'_\alpha(x) .$$

Corollary 1. Let $q : E \to X$ be a covering map which has a section $f : X \to E$. If $E$ is connected, then $q$ is a homeomorphism.

Proof. $f(X)$ is a nonempty and clopen subset of $E$, thus $f(X) = E$.

Corollary 2. Let $q : E \to X$ be a covering map which has a section $f : X \to E$. If $X$ is connected, then $f(X)$ is a component of $E$.

Proof. $f(X)$ is connected, thus contained in a component $C$ of $E$. Since $f(X)$ is clopen in $E$, the pair $(f(X), C \setminus f(X)$ is a partition of $C$ into open subsets of $C$. Since $f(X) \ne \emptyset$, we see that $C \setminus f(X) = \emptyset$.

Answer 3

Here I assume that both $E$ and $X$ are path connected and locally path connected, and the section $f$ is continuous. Such assumptions are really natural when dealing with the covering map.

Then there is a really quick way to solve the problem.

I will use Munkres Lemma 80.2:

Let $p=id_X:X\rightarrow X, q=f: X\rightarrow E, r=q: E\rightarrow X$. Lemma 80.2 tells us that $f$ is a covering map. Since $f$ is injective, f is a homeomorphism. We have $id_X=q\circ f$. Thus q is a homeomorphism.

Answer 4

A connected covering space $f:E\rightarrow X$ admits no section ( global section) unless $f$ is a homeomorphism.

Edit: looking @Andy's post I'm not so sure of what I said now.