Dtyjlui

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

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 $pi:Xrightarrow Y$ and $f:Yrightarrow X$ with $picirc f = Id_X$.

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 $Vsubseteq f(Y)$. This follows from considering the inclusion $i:Urightarrow Y$. 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 $Vcap f(Y) = emptyset$, 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.

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:

Let $q : E to X$ be a covering map with connected domain $E$. If $p$ has a section $f : X to E$, then $p$ is a $1$-fold covering (which is the same as a homeomorphism).

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.

Let us first 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.

To prove that $q$ is a homeomorphism, it suffices to show that $f(X) = E$. Then $f circ q circ f = f circ id_X = f = id_E circ f$ which implies $f circ q = id_E$ because $f$ is surjective.

$f(X) = E$ will be proved by showing that $f(X)$ is open and closed in $E$.

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

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

As a subset of $U$ also $U'$ is evenly covered, with decomposition $q^{-1}(U') = bigcup_{alpha in A} U'_alpha$, where $U'_alpha = U_alpha cap q^{-1}(U') = q_alpha^{-1}(U')$.

By construction $q^{-1}(U') cap f(X) = f(U') = U'_{alpha(y)}$.

If $y in f(X)$, we have $y in q^{-1}(U') cap f(X) = U'_{alpha(y)}$ which is an open subset of $E$ contained in $f(X)$. This shows that $f(X)$ is open.

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

A connected covering space $f:Erightarrow X$ 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.