# Topology of a smoothing of an isolated singularity

Let $$(X,x)$$ be an affine variety with a normal isolated singularity and $$Y$$ be a smoothing of $$X$$ (for this we mean $$Y$$ can be realized as a smooth fibre of a deformation of $$X$$).

Question. Can we say that the topology of $$Y$$ is nontrivial, i.e. $$\pi_i(Y)\neq \{1\}$$ for some $$i\geq 1$$? Here the topology is induced from the analytic structure instead of the Zariski topology.

• In dimension 2, I think this isn't possible: Bhupal, Stipsicz, Szabó, and Wahl classified the singularities that have a rational homology disc smoothing (which is weaker than a contractible one), and none of the singularities they found bound integer homology disc smoothings (which is still weaker than contractible). In higher dimension, the h-cobordism theorem at least tells you that the link of the singularity is a sphere and that the Milnor fibre is a ball. Dec 6, 2022 at 15:30

Edit. This does not answer the question, but it just provides examples of homotopically non-trivial smoothings $$Y$$ in every dimension $$n \geq 1$$. Perhaps someone might find it useful, hence I will not delete it.
Consider the nodal quadric hypersurface $$X=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2=0\} \subset \mathbb{C}^{n+1},$$ admitting the smoothing family $$X_t=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2-t=0\}.$$ Set $$Y=X_1$$, which is a smooth quadric in $$\mathbb{C}^{n+1}$$.
By [1], $$Y$$ is diffeomorphic to the tangent bundle $$TS^{n}$$ which is, in turn, homotopically equivalent to $$S^n$$, see [2]. So we get $$\pi_n(Y)=\pi_n(S^n)=\mathbf{Z}.$$