this is my first post so I apologize in advance for any formatting issues or community guidelines I might have overlooked.
I have come across the following basic question concerning complex algebraic varieties.
Let $X$ be a complex algebraic variety and $f:X \to\mathbb{C}$ a dominant morphism. Then does there exist a dense open set $U\subset \mathbb{C}$ such that the fibers $f^{-1}(t)$ for all $t\in U$ are all homeomorphic?
I believe the answer involves Whitney stratifications, a topic I am woefully ignorant of. I'm currently reading a bit on this subject, but I would appreciate a direct answer to my question or at least confirmation that I am headed in the right direction.
Here's what I tried:
Pick a stratification of $X$, a filtration by closed subsets $F_{i-1}\subset F_{i}$ such that the complements $S_i=F_{i}\backslash F_{i-1}$ are smooth. First, remove the images of all strata with zero dimensional image under $f$, i.e. if $f\vert_{S_i}$ is not dominant onto $\mathbb{C}$, then remove $f(S_i)$ from $\mathbb{C}$.
For all the remaining strata, remove the critical values of $f\vert_{S_i}$. Now for any $t\in U$, the remaining dense open subset of $\mathbb{C}$ after we've removed all those points, we see that the fibers $f^{-1}(t)\cap S_i$ are all diffeomorphic. Thus, $f^{-1}(t)$ is always a union of the same diffeomorphic pieces.
And that's where I got stuck - I could not find a way to reason that these pieces glue in the 'same way' each time to give homeomorphic fibers. Indeed, it seemed unlikely without any further condition on the stratification, which is why I was reminded of Whitney stratifications.
Thank you!