Let $X$ be a complex manifold, and $f:X \to D$ a proper map with connected fibers to the unit disc $D \subset \mathbb C$. Assume that $f$ is submersive (algebro-geometric lingo: smooth) over $D^* = D \setminus \{0\}$.
Under which conditions (possibly after shrinking $D$) can I say that the central fiber $X_0 = f^{-1}(0)$ is a deformation retract of $X$?
If $X_0$ is smooth, this follows easily by Ehresmann's theorem.
Yesterday I thought I found a reference in Voisin's Hodge Theory and Complex Algebraic Geometry II, that this works if $f$ has only isolated critical points. But today I can't find that anymore. Is it true anyway? Is it true in more general settings?
Motivation. This is one of the starting assumptions when reading about the formalism of vanishing cycles in Le formalisme des cycles évanescents, SGA 7-2 by Deligne.
This question was also posted on mathoverflow.net, where one finds references (though I didn't really find a clean statement in the references).
In the comments of that question, there is also the following argument using gradient descent: Choose a Riemannian metric $g$ on $X$, and consider the vector field $$\nabla |f|^2 = \hat g^{-1}(d|f|^2),$$ where $\hat g: TX \to T^*X, v \mapsto g(v,-)$ is the identification given by $g$. For any $p \in X$ the vector $f_* \nabla|f|^2_p$ is a multiple of $$\begin{pmatrix}x(f(p)) \\ y(f(p))\end{pmatrix}$$ where $x,y$ are the real coordinates on $D$, since that is the direction in which $|f|^2$ increases the most. By the escape lemma, the flow associated to $- \nabla|f|^2$ is a total flow $\Phi: \mathbb R \times X \to X$, because all integral curves approach $X_0$, hence they cannot escape. Choosing a reparametrization $\varphi: [0,1) \to \mathbb R_{\geq0}$ to obtain a retract $$\Psi:[0,1] \times X \to X, \quad \Psi(t,p) = \Phi(\varphi(t),p).$$ For $t=1$ we take the limit of the corresponding integral curve, which exists because $f:X \to D$ is proper.