Suppose $S$ is an algebraic surface (possibly projective) over an algebraically closed field $k$. Suppose $D\_i$ are irreducible smooth curves (rational, if you want) with negative self-intersection and simple normal crossings such that there is a morphism $f:S\rightarrow S'$ which collapses them to a point.


Can we form an effective divisor $D=\sum d\_i D\_i$ such that $D\cdot D\_i<0$ for all $i$? Note that if I had said $\leq$ rather than $\lt$ this is the content of the Proposition in page 83 of [Reid's Park City lectures](http://www.maths.warwick.ac.uk/~miles/surf/ParkC/ch4.ps). I ask for strict inequality.


My belief is that this is possible, but I don't manage to give a proof nor find a reference or even to find a counter-example... This would be very useful for many arguments in birational geometry of surfaces.



Numerically negative exceptional divisor on a surface.
The intersection matrix $(D\_i\cdot D\_j)$ is negative definite and hence invertible. So the system of equations: $\sum\_i d\_i D\_i\cdot D\_j=\alpha\_j$ is solvable for any set of $\alpha\_j$'s, in particular for negative ones. It is an easy consequence of the negative definite property that if $\alpha\_j<0$ for all $j$, then $d\_i>0$ for all $i$.


A really slick way to do this (in any dimension) if $S$ is quasi-projective, $f$ is projective, and $S'$ is $\mathbb Q$-factorial the following:
Let $H$ be an effective (reduced if you want) ample divisor on $S$ and consider $f\_\*H$ (as a subscheme). Since $S'$ is $\mathbb Q$-factorial, this is a $\mathbb Q$-Cartier divisor, so $mf\_\*H$ is Cartier for some $m$. Therefore $f^\*(mf\_\*H)-mH$ is effective, supported on the exceptional divisor and (strictly) negative on every curve contracted by $f$.


