Regarding fibrations between smooth manifolds

[aalma]

Definitions:
1. A map $p : X → Y$ of smooth manifolds is called a trivial fibration with fiber $Z$ which is also a smooth manifold, if there is a diffeomorphism $θ : X → Y ×Z$ such that $p$ is the composition of $θ$ with the natural projection $pr_1:Y × Z → Y$.
2. A map $p: X →Y$ is a locally trivial fibration with fiber $Z$ if for all $y \in Y$ there exists an open neighborhood $U⊂Y$ of $y$ such that $p: p^{−1}(U) → U$ is a trivial fibration with fiber $Z$.
3. A smooth map $p : X → Y$ is called a quotient map if the following conditions are fulfilled:
1. $U ⊂Y$ is open iff $p^{−1}(U)$ is open in X.
2. $f : U →R$ is smooth iff $f ◦p$ is a smooth function on $p^{−1}(U)$.
Now, I need to show that any locally trivial fibration is a quotient map.

Let $p:X\to Y$ a locally trivial fibration.
Well I need to verify the two conditions for a quotient map

1. $U ⊂ Y$ is open iff $p^{-1}(U)$ is open in $X$.
2. $f : U → R$ is smooth iff $f ◦ p$ is a smooth function on $p^{-1}(U)$.

For the first condition, suppose that $U$ is an open subset of $Y$. We want to show that $p^{-1}(U)$ is open in $X$. By the definition of a locally trivial fibration, for every $y \in U$, there exists an open neighborhood $V_y$ of $y$ such that $p^{-1}(V_y)$ is diffeomorphic to $V_y × Z$ via a diffeomorphism $θ_y$. Since $U$ is an open subset of $Y$, we can cover $U$ by the open neighborhoods $V_y$. That is, $U = ∪_y V_y$.
Now consider $p^{-1}(U)$. For any point $x \in p^{-1}(U)$, we have $p(x) ∈ U$, so there exists a $V_y$ containing $p(x)$. Since $p^{-1}(V_y)$ is diffeomorphic to $V_y × Z$ via $θ_y$, we can write $x = θ_y(y', z)$ for some $y' \in V_y$ and $z \in Z$. Moreover, since $V_y$ is an open neighborhood of $y$, we can assume that $y'$ lies in a smaller open set $U_y$ contained in $V_y$. Therefore, we have $x = θ_y(y', z) ∈ U_y × Z$, which is contained in $p^{-1}(V_y)$. Thus, we have shown that every point $x \in p^{-1}(U)$ is contained in an open set of the form $U_y × Z$, which is diffeomorphic to an open set in $Y × Z$. Therefore, $p^{-1}(U)$ is an open subset of $X$.
Does this seem okay?
Next, let's consider the second condition. Suppose that $f : U → R$ is a smooth function on $U$. We want to show that $f ◦ p$ is a smooth function on $p^{-1}(U)$. For any point $x \in p^{-1}(U)$, we have $p(x) ∈ U$, so there exists a $V_y$ containing $p(x)$. Since $p^{-1}(V_y)$ is diffeomorphic to $V_y × Z$ via $θ_y$, we can write $x = θ_y(y', z)$ for some $y' \in V_y$ and $z \in Z$. Moreover, since $f$ is smooth on $U$, we can write $f(y') = g_y(y')$ for some smooth function $g_y$ on $V_y$. Then we have:
$(f ◦ p)(x) = f(p(x)) = f(y') = g_y(y') = (g_y ◦ p')(y', z)$
where $p'$ is the projection $Y × Z → Y$. We note that $(g_y ◦ p')$ is a smooth function on $V_y × Z$, which is diffeomorphic to $p^{-1}(V_y)$ via $θ_y$. Therefore, $(g_y ◦ p')(y', z)$ is a smooth function on $p^{-1}(V_y)$, which contains $x$. Since this is true for any $V_y$ containing p(x), we conclude that $f ◦ p$ is a smooth function on $p^{-1}(U)$.
What do you think?Thanks in advance for any tips..

 

[mathwonk]

i apologize if this is unhelpful, but i hope it may be. First step is to realize it suffices to prove this for a trivial fibration p:YxZ-->Y, in which case it follows almost immediately from the definition of the product topology and the fact that p restricts to a diffeomorphism from Yx{q}-->Y, for any point q in Z.