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