Topology of varieties in a flat family

Question

I am learning about flat deformations of schemes at the moment. I understand that while it is not immediately clear from the definition why this should be a good definition it turns out that these have many properties that one wants a family of schemes to have. To get the correct picture, I want to ask if my intuition regarding topology is correct: Consider a flat and projective morphism $f: X \to T$ where $T$ is variety over $\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$ and assume that $T(\mathbb{K})$ is connected (in the classical topology). Then my intuition says that things can only change when passing through singular varieties in the family. Here are my concrete questions:

1. Let $X_t$ be smooth for all $t\in T$. Is it true that all $X_t(\mathbb{K})$ are homeomorphic to each other?

2. Let $t_0\in T(\mathbb{K})$ and $Y=X_{t_0}$. Assume that $X_t$ is smooth for every $t\neq t_0$. Let $Y_{sm}$ be the nonsingular locus of $Y$ and assume that the space $Y_{sm}(\mathbb{K})$ is connected. Is it then true that every $X_t(\mathbb{K})$ is connected? (This makes maybe only sense for $\mathbb{K}=\mathbb{R}$.)

Answer 1

It seems to me that your questions are quite difficult in the generality you asked them. Certainly this is the case for me, a person without a proper understanding of the flatness condition. Here is a post from MathOverflow I find useful for me.

Let $X_t$ be smooth for all $t\in T$. Is it true that all $X_t(\mathbb{K})$ are homeomorphic to each other?

Assume $T$ is irreducible. Using resolution of singularities and base change, I think we can assume $T$ to be smooth. In this situation, since your family is projective, and all the fibers are smooth, I think that this implies that $f$ is a proper submersion, and hence the answer is yes.

Let $t_0\in T(\mathbb{K})$ and $Y=X_{t_0}$. Assume that $X_t$ is smooth for every $t\neq t_0$. Let $Y_{sm}$ be the nonsingular locus of $Y$ and assume that the space $Y_{sm}(\mathbb{K})$ is connected. Is it then true that every $X_t(\mathbb{K})$ is connected? (This makes maybe only sense for $\mathbb{K}=\mathbb{R}$.)

My answer considers only the case $K = \mathbb{C}$, complex numbers, and dimension of fibers of $f$ strictly positive with $Y_{sm}$ nonempty. Since $Y_{sm}$ is connected, it is irreducible, and hence $Y$ itself is irreducible. Assume that the generic fiber $X_t$ has $k > 1$ connected components. If we apply the Stein factorization to the map $f$, we get a finite morphism $g: T' \to T$, of degree $k$, and such that $t_0' = g^{-1}(t_0)$ is just one point. But then this point must have multiplicity $k$, which implies that the whole fiber $Y$ is multiple. This contradicts $Y_{sm}$ nonempty.

Maybe one can suppose $\dim T = 1$ in this argument, and then everything is more intuitive.

Thanks very much! So regarding the second question: This only works over the complex numbers because Stein factorization only works over complex numbers? Would $\dim T =1$ help for the case of $K=\mathbb{R}$?

In the real case, the answer can be negative, as the family of cubic plane curves giving the universal deformation of the nodal cubic shows.

That I don't understand: If $Y$ is the nodal cubic, then $Y_{sm}(\mathbb{R})$ is not connected. So it doesn't satisfy the assumptions.

Yes, you're right. I just mixed up the real and complex cases. So I can say nothing in the real case.