Is a closed subset of a trivial bundle with all fibers homeomorphic a fiber bundle itself? Let $X$ and $Y$ topological spaces and $\pi:X\times Y\to X$ the projection. Assume that $P\subset X\times Y$ is a closed subset, $\iota:P\to X\times Y$ the embedding and assume that $p=\pi\iota:P\to X$ is surjective, such that all fibers $p^{-1}(x)$ are homeomorphic.
Is $p$ a fiber bundle?
More specifically, if this is the case, is the local trivialization of $p$ induced by restricting the homeomorphisms $\pi^{-1}(U)\cong U\times Y$ (where $U\subset X$ open) which we have for the trivial bundle $\pi$, to $p^{-1}(U)$?
Edit: This seems not to be true in general. In the application I have in mind, all fibers are $\mathbb C$-vector spaces of the same finite dimension, $X$ and $Y$ are (possibly non-irreducible) affine varieties and $P$ is closed in the Zariski topology.
 A: I don't think this is true.  Take $X = Y = \mathbb{R}$ and $$P = \{(x,y) \in \mathbb{R}^2 : \lvert y \rvert \leq \lvert x \rvert\} \cup \{(0,y) \in \mathbb{R}^2 : 0 \leq y \leq 1\}.$$  Then all the fibers look like closed intervals, and $P$ is not a fiber bundle over $X = \mathbb{R}$: there is no local trivialization at $0$. 
Response to edit: I still don't think it's true.  Take $X = Y = \mathbb{A}^1_\mathbb{C}$, say $X = \operatorname{Spec} \mathbb{C}[x]$ and $Y = \operatorname{Spec} \mathbb{C}[y]$, so $X \times Y \cong \mathbb{A}^2_\mathbb{C} = \operatorname{Spec} \mathbb{C}[x,y]$ and $P = V(xy-1) \cup V(x,y)$.  Then all the fibers are points (= o-dimensional $\mathbb{C}$-vector spaces), but the $\mathbb{C}[x,y]/(x(xy-1), y(xy-1))$ is not flat over $\mathbb{C}[x]$ because $x$ is a zero divisor in $\mathbb{C}[x,y]/(x(xy-1), y(xy-1))$.  So, assuming whatever you mean by a "fiber bundle" in algebraic geometry should require flatness, $P$ is not a fiber bundle.  
By the way, if you want positive-dimensional fibers, you can just take $P \times \mathbb{A}^n_\mathbb{C} \to X$ instead.
