If $p:\tilde{X}\to X$ is a 2-fold cover, $E=\tilde{X}\times \Bbb R/\sim$, then $\tilde{X}\to E$,$x\to (x,1)$ is embedding Let $p:\tilde{X}\to X$ be a two fold covering map, and consider the quotient space $E=\tilde{X}\times \Bbb R/\sim$, where $(x,t)\sim (x',-t)$ if $p(x)=p(x')$ and $x\neq x'$. Let $q:\tilde{X}\times \Bbb R\to E$ denote the quotient map, and let $f:\tilde{X}\to E$ be the map defined by $f(x)=q(x,1)$. I want to show that $f$ is a topological embedding. It is clearly a continuous injection, so it suffices to show that $f$ is a closed map, but it doesn't work so well. Is there another approach?
 A: Here is an overly-sophisticated take on the situation which may (perhaps only at some undetermined point in the future) help you to understand what's going on geometrically. Let $O(n)$ denote group of $n$-by-$n$ orthogonal matrices.

*

*You can think of $\widetilde X$ as a principal $O(1)$-bundle over $X$, the point being that $O(1)$ is just the two element group $\{ \pm 1\}$.

*Given a principal $O(n)$-bundle $P$ over $X$, one can construct an associated $n$-dimensional Riemannian vector bundle $E$ by taking the quotient of $P \times \mathbb{R}^n$ by the $O(n)$ action defined by $u(p,v) = (pu^{-1},uv)$.

*One then has a natural isomorphism $P \to F(E)$, where $F(E)$ is the orthonormal frame bundle of $E$.

*In the specific case $n=1$, the orthonormal frame bundle is actually equal to the the unit sphere bundle $S(E) \subseteq E$.

I'm at least half-serious when I say that this is a helpful way to understand what's going on.
A: You correctly state that it suffices to show that $f$ is a closed map. So let $C \subset \tilde X$ be closed. We want to show that $f(C)$ is closed in $E$. This means that $q^{-1}(f(C)) = q^{-1}(q(C \times\{1\}))$ is closed in $\tilde X \times \mathbb R$.
Since $p$ is a two-fold covering map, there is a unique deck transformation $d : \tilde X \to \tilde X$ such that $d(x) \ne x$ for all $x \in \tilde X$ (it flips the two points in each fiber $p^{-1}(y)$). See my answer to Does there exist a double cover with trivial deck transformation group?
The equivalence relation $\sim$ is therefore given by
$$(x',t') \sim (x,t) \Leftrightarrow (x',t') = (x,t) \text{ or } (x',t') = (d(x),-t) .$$
For $x \in C$ we have $ q^{-1}(q(x,1)) = \{ (x,1), (d(x),-1) \}$, thus
$$q^{-1}(q(C \times\{1\})) = C \times \{1\} \cup d(C) \times  \{-1\}$$
which is certainly closed.
