Principal G-bundle induces a homeomorphism between orbit space and base

Question

I am reading Chapter 14 of the algebraic topology book by Tammo tom Dieck. However, it is not going smooth since the very beginning. I will give the definitions first:
Def (Principal G-bundle). Let $G$ be a topological group and $E$ a topological space equipped with a right $G$ action ($r:E\times G\to E, r(x,g):=xg$). A principal G-bundle is a continuous map $p:E\to B$ such that \

1. for any $g\in G$ and $x\in E$, we have $p(xg)=p(x)$
2. for each $b\in B$, there is an open neighborhood $U\subseteq B$ containing $b$ and a $G$-homeomorphism $\varphi_U:p^{-1}(U)\to U\times G$.

Then he claimed that there is a homeomorphism between $E/G$ and $B$.
My question. I defined $h:E/G\to B$ by sending the orbit $\bar{x}$ of $x\in E$ to $p(x)$. This is well-defined by axiom 1 of the principal $G$-bundle. I did prove that it is continuous. However, I am having difficulty proving that it is open, and bijective.

1. Surjectivity: because for each $b\in B$, one finds an open neighborhood $U\subseteq B$ and homeomorphism $\varphi_U:p^{-1}(U)\to U\times G$. Hence for $(b,g)\in U\times G$, there is some $x\in p^{-1}(U)$ being mapped to $(b,g)$. But this does not conclude that $p(x)=b$.
2. Injectivity: Given $h(\bar{x_1})=h(\bar{x_2})$, to show that $\bar{x_1}=\bar{x_2}$, is to show that these two orbits intersect, i.e. there exists some $g\in G$ such that $x_1 g=x_2$. By the condition, we have $p(x_1)=p(x_2)$. So I tried to use the local trivialization, i.e. given $b=p(x_1)=p(x_2)$, there is some $U$ and a homeomorphism $\varphi_U:p^{-1}(U)\to U\times G$. Now $x_1,x_2$ are both in $p^{-1}(U)$, but I have no idea how to show that they are the same.
3. Openness: Given $\bar{U}\subseteq E/G$ open, we have $q^{-1}(\bar{U})\subseteq E$ is open by the quotient topology where $q$ is the quotient $E\to E/G$. I think next it suffices to show that $p:E\to B$ is open. But I have no clue.
4. Continuous: Given $U\subseteq B$ open, we wish to have $h^{-1}(U)\subseteq E/G$ open as well. By quotient topology, this is equivalent to proving that $q^{-1}(h^{-1}(U))\subseteq E$ is open. Since $p=h\circ q$, we see that $q^{-1}(h^{-1}(U))=p^{-1}(U)$, which is open by the continuity of $p$.

I apologize for the triviality of the question, as in many places the proof of this part is skipped. Any help is appreciated! Thanks in advance.

Answer 1

Let us first show that the fibers of $p$ are precisely the $G$-orbits.

Suppose that $x,y \in E$ are in the same $G$-orbit, that is, $xg = y$ for some $g \in G$. Then $p(x) = p(xg) = p(y)$, implying $x, y$ are in the same fiber.

Conversely, suppose that $x, y$ belong to the fiber $E_b = p^{-1}(\{b\})$. Then there exists an open set $U$ containing $b$ such that there exists a $G$-homeomorphism $\varphi_U \colon p^{-1}(U) \to U \times G$ that agrees with the projection onto the first factor. Restricting yields a $G$-homeomorphism between the fiber $E_b = p^{-1}(\{b\})$ and $G$. This means that $G$ acts transitively on $E_b$; in particular, there exists $g \in G$ such that $xg = y$.

Finally, note that $p \colon E \to B$ is a surjective quotient map, since we may cover $E$ with open sets on which $p$ is a projection map. Thus $p$ descends to a homeomorphism from the orbit space $E/G$ to the base space $B$.