Question:
Let $(E,M ,\mathcal P, F)$ a locally trivial bundle. Define an equivalence relation $\sim$ on $E$ that identifies points in the same fiber of $\mathcal P$, i.e., $p_1 \sim p_2$ iff $\mathcal P(p_1) = \mathcal P(p_2)$. Show that $M$ is homeomorphic to $E/\sim$.
Attempt: Results I have used so far
Lemma 1: Let $(E,M ,\mathcal P, F)$ be a locally trivial bundle. Then $\mathcal P : E \to M$ is a continuous, open surjection.
Lemma 2: Let $\mathcal Q : X \to Y$ be a quotient map, $Z$ e topological space and $f : X \to Z$ a continuous map constant on each fiber of $\mathcal Q$. Then there exists a unique continuous map $\bar f : Y \to Z$ such that $\bar f \circ \mathcal Q = f$.
Consider $\mathcal Q : E \to E/\sim $ the quotient map with respect to $\sim$ given.Since $$[p_1] = [p_2] \iff p_1 \sim p_2 \iff \mathcal P (p_1) = \mathcal P(p_2)$$ then $\mathcal P$ is constant on each fiber of $\mathcal Q$. By Lemma 1 $\mathcal P$ is continuous then it follows from Lemma 2 that there exists a unique continuous map $\bar {\mathcal P}$ such that $\bar {\mathcal P} \circ \mathcal Q = \mathcal P$. As $\mathcal P$ (again Lemma 1) is surjective then $\bar {\mathcal P}$ is a bijection.
It is left to prove that $\bar {\mathcal P}^{-1}$ is continuous.
1) It seems I need to show that $\bar {\mathcal P}$ is an open mapping.
Any ideas?
I believe this proves the result. I will write it here for future reference.
Let $U \subseteq E/\sim$ be any open set in $E/\sim$. Since $\mathcal Q$ is continuous, $\mathcal Q^{-1} (U)$ is open. Now by Lemma 1 we have that $\mathcal P$ is an open maping then $\bar{\mathcal P}(U) =\mathcal P(\mathcal Q^{-1}(U))$ is open, showing that $\bar {\mathcal P}$ is an open maping. As $\bar{\mathcal P}$ is continuous and surjective we may conclude that $\bar {\mathcal P}^{-1}$ is continuous. Thus showing that $M$ is homeomorphic to $E/\sim$.