Proving that the map from a topological space to a fiber bundle is open. There are two pieces to a question, the first:
Let G be a group acting on a space $E$. Show that the quotient map $E \rightarrow E/G$ is open.
The second:
Let p : E → X be a fiber bundle. Show that p is an open map.  
My thoughts are the following.  First we see what a "fiber" is: If for E and X as topological spaces we let $p:E \rightarrow X$ be a mapping, $U \subseteq X$ and $U$ is a neighborhood of $x$, then $p^{-1}(x) \in E$ where $p$ is called the fiber over $x$.  
The fiber bundle is one such that:
$\forall y \in E \space\space \exists U \subset X \text{ where }\phi_{U}:p^{-1}(U) \text{ is homeomorphic to the product space } F \times U \text{ where } F \text{ is the fiber bundle }$
Now I address the second piece first as I have a better idea of why this would be true, however could use some clarification.  On this page:  https://en.wikipedia.org/wiki/Fiber_bundle
It says: "Every fiber bundle is an open map, since projections of products are open maps."  While I don't see why the projections of products are open maps (unless they are just referring to topological spaces as top. spaces are both open and closed), I am wondering if p is an open map as by the definition of a fiber bundle we have that since the product space $F \times U$ is open as $U$ is an open neighborhood of $x$ then since the pre-image $p^{-1}(x) \in E$ is homeomorphic to $F \times U$ then $p$ has to be an open map.
As for the first part, since $E$ is a Topological space and hence it is open, then if I mod out some closed set it would still be open.  Is there more to it than this?
Thanks,
Brian
 A: The first part of your question is confusing only because by definition quotient maps are open... 
Because $p$ is a surjection, $p^{-1}(U)$ for $U\subset B$ (open) as defined cover $E$ and is open ($p$ is continuous). From the property of the fiber bundle, $p|_U$ is open because $p=\pi \circ \phi_U$, where $\phi_U$ is a homeomorphism, which makes it open due to the composition of open maps is open (and if you don't get why $\pi$ is open, try proving that a projection map is always open. It is essentially by the definition of the product topology.)
So now take an arbitrary open set $O$ in $E$. We want to show $p(O)$ is open. For $x\in p(O)$, there exists a $U_x$ such that $p^{-1}(U_x)$ is an open set. From the above discussion we know the restriction of $p$ to any $p^{-1}(U_x)$ is open. Because $p^{-1}(U_x)\cap O$ is open, $ p|_{p^{-1}(U_x)}(O)$ therefore is an open set. To finish observe $$p(O)= \bigcup_{x\in p(O)} p|_{p^{-1}(U_x)}(O).$$ This implies $p(O)$ is open.
A: If the group action is continuous, your first question is answered in Lee's Introduction to Smooth Manifolds. It's Lemma 9.15 in the first edition.
If by "group" you mean "smooth Lie group," you can apply the lemma directly.
If by "group" you mean "countable group," observe that any countable group is naturally a smooth Lie group of dimension zero. (I'm restricting to countable groups because Lee defines a manifold to be second-countable. I don't know if this restriction is necessary.)
