Suppose X is a normal projective complex variety, (X, $\Delta$) is a klt pair and f : X $\to$ Z is a Mori fiber space given by a contraction of an extremal ray for this pair. Here I mean that the relative dimension is at least 1. Do we know anything about the singularities of Z? Z is normal more or less by construction and we do know that if X was Qfactorial then so is Z. Can we say anything more? Eg: is there a $\Delta'$ on Z with (Z, $\Delta'$) klt? Does Z have rational singularities? What if $\Delta$ = 0, is Z terminal? Canonical?
2 Answers
Under the assumption that $X$ is $\mathbb{Q}$factorial, section 5 of the paper https://arxiv.org/pdf/math/0606666 addressed this issue, which was also proved earlier in Ambro's paper. Basically, if you assume the pair $(X,\Delta)$ is klt, so is the base $(Z,\Delta_Z)$ for some $\Delta_Z$. As the example of Prokhorov shows, this is optimal, namely, the base may not be terminal (canonical) even you assume the $(X,\Delta)$ is. For lc case, I think dlt modification+perturbation reduce the question to the klt case.

$\begingroup$ To Sandor: surely we can forget $\Delta_Z$. But I write $\Delta_Z$ because I want to emphasize in the proof the discriminant divisor will appear, and for many questions we should take it into account. $\endgroup$– CYXUCommented Oct 15, 2010 at 17:32

$\begingroup$ Hi Chenyang! Sure that makes sense, I just meant that for the original question. I actually thought about adding that nevertheless there might be natural divisor that should be added and that I can imagine that it is easier or more natural to prove that a certain pair is klt (etc) than proving it for the ambient variety directly. $\endgroup$ Commented Oct 16, 2010 at 7:15

I think I can answer one of your questions, namely about $Z$ having rational singularities. It does and here is why:
$Z$ is indeed normal by construction.
Since $f$ is a Moricontraction, $K_X$ is $f$nef and $f$big. (In fact, since the relative Picard number is $1$ it is $f$ample). I claim that $R^if_*\mathcal O_X=0$ for all $i>0$. If $f$ were birational, this would be a simple consequence of KawamataViehweg vanishing. In this case we need to work a little more.
First, by Kollár's torsionfreeness theorem (see [Yujiro Kawamata, Katsumi Matsuda, and Kenji Matsuki. Introduction to the minimal model problem. In Algebraic geometry, Sendai, 1985, volume 10 of Adv. Stud. Pure Math., pages 283360. NorthHolland, Amsterdam, 1987.] for the format needed here and see the references there for proofs) these sheaves are torsionfree. (Not in general, just in this case!).
Second, let $U\subseteq Z$ be a nonempty subset over which $f$ is flat. The fibers over $U$ are Fano varieties and hence $H^i(F,\mathcal O_F)=0$ for $i>0$ for a fiber $F$. I probably should have taken a resolution to start with and then one would not have to worry about singularities, but at least for a general $F$ one can say that $(F,\Delta_{F})$ is klt and hence KawamataViehweg vanishing applies, so we get the above vanishing. Anyway, then by Grauert's theorem [Hartshorne, III.12] it follows that $(R^if_*\mathcal O_X)_{U}=0$ for all $i>0$. However, we have already established that these sheaves are torsionfree, so if they are zero on a nonempty open set then they are zero everywhere.
OK, so we get that $R^if_*\mathcal O_X=0$ for all $i>0$. Now we are only almost there because the original definition of rational singularities would require this for a birational morphism and $f$ is decidedly not birational. But this still implies that $Z$ has rational singularities by SJK: A characterization of rational singularities.
Now, regarding whether $Z$ can be klt, canonical, etc. One simplification I can suggest is that you do not need the $\Delta'$ there. If you find a $\Delta'$ such that $(Z,\Delta')$ is klt, dlt, lc, etc, then so is $Z$ since by the $\mathbb Q$factoriality $\Delta'$ will be $\mathbb Q$Cartier. See [KollárMori, 2.35].