MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 4 added 125 characters in body

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 both for an idea of the proof and 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 casecase), and definitely it is not important that the dimension of $V$ is 3. But anyway I am interested mainly in the case when would prefer to get a proof of the statement holdsrather to get a counterexample by relaxing the condition on singularities.

show/hide this revision's text 3 added 223 characters in body

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 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 both for an idea of the proof and for a reference. Note that in the case when $C$ is smooth and does not intersect singularities of $V$, the proof is easy. I think, am confident that it should be sufficient to impose only the condition, that $V$ statement is normal (correct, and not 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 an orbifold)not always the case. But anyway I am interested mainly in the case when the statement holds.
show/hide this revision's text 2 added 24 characters in body

Suppose that $V$ is a complex $3$-fold analytic manifold of dimension 3 with mild singularities, say it is an orbifold (i.e. has only quotient singularities). Let $C$ be a complex 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?

I would be grateful both for an idea and for a reference. Note that in the case when $C$ is smooth and does not intersect singularities of $V$, the proof is easy. I think, that it should be sufficient to impose only the condition, that $V$ is normal (and not that $V$ is an orbifold).
show/hide this revision's text 1