Let$X$ and$Y$ be topological spaces and let$\tau$ be a topology on the set-theoretic product$X \times Y$ such that:
- The first projection$p \colon X \times Y \to X$ is continuous and open with respect to$\tau$.
- For each$x \in X$, the embedding$i_x \colon Y \to X \times Y$,$y \mapsto (x, y)$ induces a homeomorphism of$Y$ onto the fiber$p^{-1}(x) = \{ x \} \times Y \subset X \times Y$, equipped with the subspace topology induced from$\tau$.
Question: Is$\tau$ necessarily equal to the product topology? If not, does the situation change if we assume that$X$ and$Y$ are, for example, locally compact and Hausdorff?
Here is my intuition behind the question: The second property tells us that the fibers$p^{-1}(x)$ carry the correct topology, i.e. the same as the one induced by the product topology.Dustin Clausen argues in the comments tothis question on MathOverflow that openness of a continuous map means the fibers are varying continuously over the base space.So the first property should hopefully mean that we can spread out the "correctness of the fibers" horizontally.
2 Answers2
Your assumptions are consistent with a fibre bundle with base$X$ and fibre$Y$, but non-trivial fibre bundles are not homeomorphic to the product$X\times Y$. For example, the mobius band is a$Y=\mathbb{R}$ bundle over$X=S^1$ not homeomorphic to$S^1\times \mathbb{R}$.
Let$X=[0,1]$ and$Y=\{0,1\}$. Equip$Z=X\times Y$ with the topology induced by sets of the form$(U\times Y)\setminus F$ where$U$ is open in$X$ and$F$ is a finite set.
Then$Z\to X$ is continuous and open, and fibers$\{x\}\times Y$ are discrete just like$Y$, but$Z$ is not Hausdorff: for all$x\in X$, the elements$(x,0)$ and$(x,1)$ don't have disjoint neighborhoods.
You mustlog in to answer this question.
Explore related questions
See similar questions with these tags.