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.