Inequality about the euler characteristic of surfaces Let $S$ a complex algebraic projective surfaces. Suppose that there exist a surjective morphism $p:S \rightarrow B$ such that $p$ has connected fibers and $B$ is a smooth curve. Call $F_{\eta}$ the generic fiber of $p$. 
Show that this inequality holds: $\chi_{top}(S) \ge \chi_{top}(B)\chi_{top}(F_{\eta})$.  
My idea is to use the following formula:
$$\chi_{top}(S)=\chi_{top}(B)\chi_{top}(F_{\eta})+\sum\limits_{s \in \Sigma} (\chi_{top}(F_s)-\chi_{top}(F_{\eta})),$$ 
where $\Sigma$ is the set of points of $B$ such that the fiber $F_{s}$ is singular. Now is it true that $\chi_{top}(F_s)\ge \chi_{top}(F_{\eta})$?  
 A: You are right: $\chi_{top}(F_s) \ge \chi_{top}(F_\eta)$ for each non-smooth fibre $F_s$ and the general smooth fibre$F_{\eta}$. A proof may proceed along the following steps:
The map $P: S \longrightarrow B$ is flat. Hence the  holomorphic Euler characteristic $\chi (\mathscr O_{F_b})$ of its fibres is locally constant as a function of $b \in B$ (Grauert's semicontinuity theorem), hence constant. Because for $F_b$ a smooth fibre $\chi_{top}(F_b) = 2 * \chi (\mathscr O_{F_b})$ also the topological Euler numbers of the smooth fibres are constant. 
For any embedded, not necessarily smooth curve $C \subset S$ the adjunction formula for the arithmetic genus 
$$p_{arith}(C) = 1 - \chi (\mathscr O_C) = 1 + \frac {1}{2} \ deg (\kappa_S \otimes \mathscr O_S(C)|C)$$ 
implies 
$$1 - \chi (\mathscr O_C) \ge 1 - \chi (\mathscr O_{C_{red}}).$$
On the other hand, the topological Euler number for $C$ and its reduction $C_{red}$ are the same. 
We claim $ \chi_{top}(F_b) \ge 2 * \chi (\mathscr O_{F_b})$ for any $b \in B$. For the proof we may now restrict to $F_b$ a reduced, connected curve $C$. Computing the Betti numbers of $C$: Because $C$ is connected we have $b_0(C) = 1$. From the exponential sequence and its long cohomology sequence we obtain the injection $H^1(C, \mathbb Z) \longrightarrow H^1(C, \mathscr O_C)$. One shows $b_1(C) \le 2 * h^1(C, \mathscr O_C)$. Furthermore, one shows that the normalization $\nu: C^{norm} \longrightarrow C$ induces an isomorphism $H^2(C, \mathbb Z) \longrightarrow H^2(C^{norm}, \mathbb Z) \cong \mathbb Z^{\oplus N}$ with $N$ the number of irreducible components of $C$. Hence $b_2(C) = N$. We get 
$$\chi_{top}(C) = b_0(C) - b_1(C) + b_2(C) \ge 1 - 2*h^1(C,\mathscr O_C) + N = 2*(1 - h^1(C,\mathscr O_C))$$
or 
$$\chi_{top}(C) \ge 2* \chi(\mathscr O_C).$$
For a non-smooth fibre $F_s$ we get
$$\chi_{top}(F_s) = \chi_{top}((F_s)_{red}) \ge 2 * \chi(\mathscr O_{(F_s)_{red}}) \ge 2 * \chi(\mathscr O_{F_s}) = 2 * \chi(\mathscr O_{F_\eta}) = \chi_{top}(F_\eta), q.e.d.$$
(See Barth, W.; Hulek, K.; Peters, Ch.; van de Van, A.: Compact Complex Surfaces. Chap III, Prop. 11.4 and Chap. II, Prop 2.1)
