Contraction morphism for surface I am reading Mori and Kollár's 'Birational geometry of algebraic varieties'. I am confused when reading the proof of Thm 1.28, which is the classification of contraction morphism for surfaces. 
Here are my questions:


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*In the second part of the proof, when we are dealing with the case when $C^2=0$, the proof tried to show that $|mC|$ has dimension of global section being larger than 2 when $m$ is large. But I thought when you have $C^2=0$, it says you can move this curve around a little bit and get a (linearly equivalent) curve $C'$ which does not meet $C$, and such a $C'$ should tell you that $dim|C|\geq 1$. If so, it seems the process in the proof is redundant?


(I would greatly appreciate if someone can give me an example of a irreducible curve $C^2=0$ on a surface $X$ with $h^0(X,\mathcal{O}_X(C))=1$.)


*Also in the second part, we established that the linear system $|mC|$ has no fix point(nor component), hence gives a map $\pi$ to projective space. Then by setting $cont_R:X\to Z$ to be the Stein factorisation of $\pi$, the proof says the fiber of $cont_R$ is linearly equivalent to $C$. I do not know how the Stein factorisation get rid of the multiple $m$ in the linear system.


Thanks for the help!.
 A: *

*"I thought when you have $C^2=0$, it says you can move this curve around a little bit and get a (linearly equivalent) curve $C'$ which does not meet $C$..."


That is not true; the correct statement is rather that if $C^2=0$ and if you can move $C$ to another curve $C'$, then $C$ and $C'$ will not meet.
For an example where your statement fails, let $f:X \rightarrow Y$ be a fibration from a surface to a curve, and suppose that $f$ has a multiple fibre supported on a smooth curve $C$. (This really can happen, e.g. for elliptic surfaces.) Then $C^2=0$, but $C$ does not move in a linear system.


*"I do not know how the Stein factorisation get rid of the multiple m in the linear system."


On one hand, you seem to be confused when you speak of "get[ting] rid of the multiple $m$". Here $C$ is the thing that is getting contracted, and $mC$ is what is "doing the contracting", so to speak. The fact that they are proportional is really just a coincidence arising from the fact that we're considering the case $C^2=0$ here. In general, if $C$ is the fibre of a contraction and $D$ is a curve in the linear system defining the contraction, then the only relation between them is that $C \cdot D=0$. In particular their classes can be linearly independent. Note also that the rest of the proof doesn't actually use the fact that the contraction is defined by $|mC|$!
On the other hand, it does seem to be true that something a little funny is happening here. The statement that $\sum_i a_i [C_i] = [C]$ does not seem to be completely justified at this point in the proof. The reason is essentially the same as the example I gave above: $C$ could be the support of a multiple fibre, in which case we would have  $\sum a_i [C_i] = k[C]$ for some $k >1$. However, if you read the next line  of the proof it doesn't matter, because all they need is that each $[C_i]$ belongs to the ray $R$. (And indeed the hypothesis that $R \cdot K_X <0$ shows that the situation I just described can't actually occur.)
I hope that helps!
