Let $\pi:\mathcal X\to B$ be a deformation of a compact complex manifold $X=\pi^{-1}(0)$, then for any $t\in B$, the first Chern class $c\_1(X)=c\_1(X\_t)$?


I know the Chern class of a manifold depends on the complex structure, for example, the same diffeomorphism type of $\mathbb C^1$ with a different complex structure may have a different Chern class, but also the same differential manifold with different complex sturctures may have the same Chern class, for example, the deformations of a Calabi-Yau manifold have the same Chern class $c\_1=0$ while the complex structures are different. So my question is: is the Chern class a deformation invariant?



Deformation invariance of Chern classes
This is actually true for *all* Chern classes, but you must first say how you identify $H^\*(X,\mathbb{Z})$ and $H^\*(X\_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball (say). Then the restriction maps $H^\*(\mathscr{X},\mathbb{Z})\rightarrow H^\*(X,\mathbb{Z})$ and $H^\*(\mathscr{X},\mathbb{Z})\rightarrow H^\*(X\_t,\mathbb{Z})$ are isomorphisms, so you can consider everything in $H^\*(\mathscr{X},\mathbb{Z})$. But now from the exact sequence $0\rightarrow \pi ^\*T\_B\rightarrow T\_{\mathscr{X}}\rightarrow T\_{\mathscr{X}/B}\rightarrow 0$ and the triviality of $T\_B$ you get that $c\_i(X)$ and $c\_i(X\_t)$ are the restrictions of the same class, namely $c\_i(T\_{\mathscr{X}})$.


