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Suppose that $V$ is a complex analytic manifold of dimension 3 with mild singularities, say it is an orbifold (i.e. has only quotient singularities). Let $C$ be a complex irreducible curve in $V$. Suppose that $V'$ is a blow up of $V$ along this curve that contracts only one divisor in $V'$. By this I mean that there is a holomorphic map $V'\to V$ that is an isomorphism on the preimage of $V\setminus C$ and the preimage of each point of $C$ is a curve in $V'$.

Question. How to prove that $b_2(V')-b_2(V)=1$ using preferably a purely topological reasoning?

Comments. I would be grateful for an idea of the proof or for a reference. Note that in the case when $C$ is smooth and does not intersect singularities of $V$, the proof is easy. I am confident that the statement is correct, and interested in a simple proof of it. If might be that the statement holds even if $V$ has more complicated singularities (though according to the answer of Remke below this is not always the case), and definitely it is not important that the dimension of $V$ is 3. I would prefer to get a proof of the statement rather to get a counterexample by relaxing the condition on singularities.

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I am not sure whether $b_2(V')-b_2(V)=1$ always holds. Anyway, in the book of Peters and Steenbrink you can find "the Mayer-Vietoris sequence of the discriminant square". If $E$ is the exceptional divisor then you have an exact sequence of mixed Hodge structures $$ H^1(V')\oplus H^1(C)\to H^1(E) \to H^2(V)\to H^2(V')\oplus H^2(C)\to H^2(E) \to H^3(V) \to \dots$$

Now if $C$ is smooth and does not intersect the singular locus of $V$ then $E$ is a $\mathbb{P}^1$-bundle over $C$. Hence the map $H^1(C)\to H^1(E)$ is surjective. One can easily show that the map $H^2(V)\to H^2(E)$ is not the zero map; that $H^2(C)\to H^2(E)$ is injective; the image of the first map is not contained in the image of the second and that $h^2(C)=1, h^2(E)=2$ holds, so the above sequence reduces to $$ 0 \to H^2(V)\to H^2(V')\oplus \mathbb{C}\to \mathbb{C}^2\to 0.$$ Now, if $C$ passes through the singularities of $V$ little of the above remains true, e.g., $E$ might be a conic bundle over $C$ with reducible fibers over the points where $C$ intersects the singular locus of $V$. In this case $h^2(E)>2$ holds. The additional classes in $H^2(E)$ typically contribute to the kernel of $H^3(V')\to H^3(V)$ and reduce the dimension of the weight 2 part of $H^3$, but I expect that in some cases they might force $h^2(V')-h^2(V)>1$.

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Remke, thanks. Though using mixed Hodge structures is not allowed in my question :) Since I don't consider algebraic manifold, just a complex analytic. I'll change accordingly my question. – aglearner Sep 9 '11 at 21:13
The above mentioned long exact sequence also exist in cases where $V$ and $V'$ are nice enough topological spaces. However in that case it is only an exact sequence of vector spaces. The main reason to restrict to singular algebraic varieties is that even in this case one might construct examples such that $b_2(V')-b_2(V)>1$ holds. – Remke Kloosterman Sep 9 '11 at 21:30

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