Glueing of algebraic surface along two intersecting curves My question arises as part of understanding an analogue of the normalization of singular curves. Assume that $C$ is such a curve and that $p\in C$ is a singular point of $C$ (and the only one for simplicity). Then the normalization $\widetilde C\to C$ glues two points of $\widetilde C$, or, to rephrase this a little, it glues $\widetilde C$ along two points. (This is what happens to a nodal curve. If $C$ is cuspidal, then a point together with a tangent direction is mapped to a point but let us stick to the nodal case, again for simplicity).
Now, let us consider the following: Let $S$ be a surface having a curve singularity along a smooth curve $C$, say a node. Then the normalization of $C$ is a map from a smooth surface $\widetilde S$ to $S$ that maps two curves $C_1,C_2\cong C, C_1\cap C_2=\emptyset$ in $S$ to the curve $C$. Again, we may rephrase this and say that it glues $\widetilde S$ along the curves $C_1,C_2$. The condition $C_1\cap C_2=\emptyset$ comes from the fact that the curve $C$ in $S$ is smooth.
As just indicated, if the curves $C_1,C_2$ on $\widetilde S$ intersect, then I expect the curve $C$ which is the singular locus of $S$, to be singular as well. Also the condition $C_1,C_2\cong C$ should become $C_1,C_2\cong \widetilde C$ in this case so that we still have smooth curves in $\widetilde S$. ($\widetilde C$ denotes the normalization of $C$.)
My question now is the following:
0) Following the suggestion by @Get Off The Internet, given a smooth surface $S'$ and a smooth curve $C'$ together with a $2-1$ map from $C'$ to another curve $C$, which may have singularities, does there exist a surface S having singularities along C, whose normalization is the original surface $S'$?
1) Can we say anything about the singularity of $S$ at the image of the intersection point $C_1\cap C_2\in \widetilde S$? If not, how about the singularity of $C$ in this point?
2) The normalization $\widetilde S\to S$ provides a desingularization of $S$ and it is minimal due to the universal property of the normalization. If we embedded $S\hookrightarrow V$ where $V$ is a smooth variety, then we can blow-up $V$ several times and end up with $\widetilde S\hookrightarrow \widetilde V$ where $\widetilde V$ denotes the blown-up $V$. It is clear that over $C_1,C_2$ there lies a $\mathbb P^1$-bundle that is contracted. Just think of the case of a nodal curve $C$. However, this is much less clear over the point of intersection of $C_1,C_2$. I expect that over this point the $\mathbb P^1$-bundles just intersect and nothing bad happens but I can't see any formal reason for that. Can anyone help me with that?
3) Is there a way to make our lifes simpler, i.e. to make the curves $C_1,C_2$ disjoint again? Probably by blowing up $\widetilde S$ in the point of intersection of this curves?
Any help will be greatly appreciated. Comments on the problem are also very welcome.
Here is a kind of minimal example that I have in mind: Let $C$ be a smooth curve and consider the product $\widetilde S = C\times C$ together with the projection $S\to C$. Whether on the first or the second factor doesn't matter but let us say its on the first factor. Then we have two sections of $S$. The first one is the section $s_p: q\mapsto (q,p)$ (for $p\in C$) and the other one is $d: q\mapsto (q,q)$. Let us denote their images by $S_p,D$, respectively. Obviously, $S_p,D\cong C$ and they intersect in the point $(p,p)$ and nowhere else. If we glue them, we get a surface $S$ that we may still consider as sitting over $C$. The fibres of $S\to C$ are then nodal curves degenerating to a cuspidal one (=the fibre over $p$). However, I'm really uncertain about singularities of $S$ and how its singular locus looks like. My impression is that I have nodes along $C\setminus\{p\}$ and that the singular locus is a cuspidal curve whose normalization is $C$. But I am unable to give a formal argument for this and therefore I am not even sure whether this is correct. Also whether this reflects the general situation is not clear to me. I just thought this example might help.
This does also lead me to the formulation of part 3) of my question. Blowing up $\widetilde S$ in $(p,p)$ separates the two sections $S_p$ and $D$. But then I added a curve in $\widetilde S$ which comes from the exceptional divisor of the blow-up and looks more complicated to me. At least my confusion increases.
Even partial answers or some intuition what might be happening will be useful to me.
 A: There are some misconceptions embedded in your questions, so I think it would be best to clarify them, and then reconsider your questions. 
First of all, if you start with an integral curve, you are correct in saying that its normalization is nonsingular. But the inverse image of the singular points may be several points. If there is just one singular point, and that one is a node, then there are just two points above it, and the original curve is uniquely determined by those two points. If there are more points upstairs, then there always exists a singular curve with that normalization, but it is not unique. It is specified by giving a suitable subring of the semilocal ring of the set of points above.
For surfaces, the normalization of a singular surface may not be nonsingular. In general it is just normal, and may have a finite number of isolated (normal) singular points.
A typical projection  of a nonsingular surface will have a double curve with a finite number of pinch points and triple points. If $S$ is the surface, its normalization will have a curve $C'$ lying over $C$, together with a $2$-$1$ map from $C'$ to $C$. The inverse question is not obvious to me. You might take that to be the first of your questions, namely, given a smooth surface $S'$ and a smooth curve $C'$ on $S'$, together with a $2$-$1$ map from $C'$ to another curve $C$, which may have singularities, does there exist a surface $S$ having singularities along $C$, whose normalization is the original surface $S'$? Even if $C'$ is two disjoint copies of $\mathbb{P}^1$, and $C$ is just one $\mathbb{P}^1$, I do not see the answer immediately. 
