# Is $\mathbb{C}^n$ rigid?

Let $$\pi:X\to S$$ be a smooth family of complex manifolds (in the sense of deformation theory) such that $$\pi^{-1}(0)\cong\mathbb{C}^n$$ and $$S\subset \mathbb{C}$$ is a neighborhood of $$0$$. Is $$\pi$$ trivial? That is, is $$X\cong \mathbb{C}^n\times S$$ possibly after shrinking $$S$$?

I know that a smooth family of compact complex manifolds with $$H^1(M,\mathcal{T}_M)=0$$ is trivial (where $$M=\pi^{-1}(0)$$) but I'm not sure whether this extends to this non-compact situation.

Example. $$\pi: \{(z,w)\in {\mathbb C}^2: |zw|<1\}\to {\mathbb C}$$, $$\pi(z,w)=z$$.
Edit. Similarly, to get a nontrivial deformation of $${\mathbb C}^n$$, consider $$X=\{(z_0, z_1,...,z_n)\in {\mathbb C}^{n+1}: |z_0 z_1|<1\}$$ and let $$\pi$$ be the projection of $$X$$ to $${\mathbb C}$$ which the 1-st coordinate line in $${\mathbb C}^{n+1}$$. Then $$\pi^{-1}(t)$$ (for $$t\ne 0$$) will be biholomorphic to $$B \times {\mathbb C}^{n-1}$$, where $$B\subset {\mathbb C}$$ is an open disk. Hence, these fibers are not biholomorphic to $$\pi^{-1}(0)={\mathbb C}^{n}$$.