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In Huybrechts' book Complex geometry: An introduction p.269, Proposition 6.2.2, the author gives a proof of the following theorem

(Ehresmann) Let $\pi:\mathcal X\to B$ be a proper family of differential manifolds. If $B$ is connected, then all fibers are diffeomorphic.

His proof can be summarised as

Step 1: Connect any two points of $B$ by a smooth arc, we may assume that the base $B$ is an interval $(-\varepsilon,1+\varepsilon)$. Then we only need to show that the fibers $X_0\cong X_1$.
Step 2: By the submersion condition, locally in $\mathcal X$, the map $\pi$ looks like the projection $\mathbb R^{m+1}\to \mathbb R$, so we can lift the vector field $\frac{\partial}{\partial t}$ on $B$ to the local chart $\mathbb R^{m+1}$.
Step 3: By compactness, there exist finitely many open sets $U_i\subset \mathcal X$ covering the fiber $X_0$ such that $\frac{\partial}{\partial t}$ can be lift to a vector field $v_i$ on $U_i$. Using a partition of unity, one constructs in this way a vector field $v$ on $\cup_iU_i$ that projects to $\frac{\partial}{\partial t}$ on some neighborhood of $0\in(-\varepsilon,1+\varepsilon)$.
Step 4: Since the family is proper, there exists a point $\tau>0$ such that $X_{\tau}$ is contained in $\cup_iU_i$. Using the compactness of $[0,1]$, it suffices to show that $X_0$ and $X_{\tau}$ are diffeomorphic. A diffeomorphism $X_0\to X_{\tau}$ is provided by the flow associated to the vector field $v$.

My question lies in Step 4: for a proper family, why does it suffice to show that $X_0$ and $X_{\tau}$ are diffeomorphic?

As we know, Kähler property is stable under small differential deformations, then we can also find a $\tau>0$ such that $X_{\tau}$ is Kähler, in this way any deformation of $X$ is Kähler, which is obviously a mistake by a result of Hironaka.

Then why does this method apply to Step 4 of the proof?

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    $\begingroup$ It is not enough to show that they are diffeomorphic. He is showing that we can provide a fiber preserving diffeomorphism between $X_0 \times I$ and an open set of $X$ containing $X_0$, $I$ an interval. So $X$ is locally trivial, a fiber bundle over the interval. $\endgroup$
    – Ben McKay
    Commented May 9, 2022 at 9:54
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    $\begingroup$ Since all fibers are diffeomorphic, all fibers are diffeomorphic to a Kaehler manifold just when one fiber is a Kaehler manifold. But they are only diffeomorphic as real manifolds, not as complex manifolds. The complex structure can vary, and does in many elementary examples, already on complex surfaces. $\endgroup$
    – Ben McKay
    Commented May 9, 2022 at 9:56
  • $\begingroup$ @BenMcKay, for the diffeomorphism $F:X_0\times I\to \mathcal X;(x,t)\mapsto F(x,t)$, let $F_t(x):=F(x,t)$, then $F_t:X_0\to X_t$ provides a diffeomorphism between $X_0$ and $X_t$, which implies all the fibers over $B$ are diffeomorphic, isn't it? $\endgroup$
    – Tom
    Commented May 9, 2022 at 11:04
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    $\begingroup$ Yes, that is right. But $F_t$ need not be holomorphic, since the vector field is produced using a partition of unity, so only $C^{\infty}$. $\endgroup$
    – Ben McKay
    Commented May 9, 2022 at 11:53
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    $\begingroup$ To prove the result stated, that all fibers are $C^{\infty}$ diffeomorphic, his proof is fine. He splits up his interval into finitely many subintervals, and proves diffeomorphisms one step at a time through each of these subintervals. What I should have written is that the stronger result, which is what is usually called Ehressmann's theorem, states that the map is a $C^{\infty}$ fiber bundle, for which we need to ensure the local triviality. $\endgroup$
    – Ben McKay
    Commented May 9, 2022 at 15:29

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