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This is a question that I originally posted on Math Stack Exchange. After a couple of days I have not received any comments or answers, and after thinking about it more I realize that this question is probably better suited for this community. I have forwarded the original post to this post.

Let $X$ be a projective $\mathbb{Q}$-factorial variety ("variety" is irreducible and reduced over a field of characteristic zero; not necessarily algebraically closed, but we can add this if necessary), let $\pi:X\to Y$ be a map with irreducible fibres, where $Y$ is a projective variety, and let $F$ be the reduction of a fibre of $Y$ (so $F$ is a projective variety). Let $K_X$ denote the canonical divisor of $X$.

My question is this:

If $C$ is a curve contained in $F$, then do we have $K_F\cdot C\leq K_X\cdot C$?

Maybe there are more conditions that need to be placed on $\pi$. I've tried to keep the question somewhat general to see if there is a general answer, but my motivation comes from the case where $\pi$ is an elementary MMP step and $X$ is a spherical variety.

There is a combinatorial proof for the case of toric varieties and MMP; see Lemma 7.1 of "Approximating rational points on toric varieties" by McKinnon and Satriano. But this proof is very specific to the toric case because it uses the fan combinatorics. I would hope that there is a general proof of this that works for more general varieties.

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    $\begingroup$ Welcome new contributor. Heuristically this seems likely. If $F$ is a divisor, then contractibility of $F$ implies that the intersection number of any irreducible curve against the $\mathbb{Q}$-Cartier divisor class of $F$ is nonpositive; now apply adjunction. If $F$ is not a divisor, consider the blowing up $\widetilde{X}$ of $X$ along $F$ with its exceptional divisor $\widetilde{F}$. Unfortunately, the singularities of $F$ and $X$ intercede. In the toric case, there is always a small resolution of $X$ that is the coarse moduli space of a smooth DM stack (but not so in general). $\endgroup$
    – Jason Starr
    Commented Jun 29, 2023 at 20:31
  • $\begingroup$ Thank you for the comment, @JasonStarr. If $X$ is smooth and $F$ is a divisor, then using the adjunction formula makes sense to me. But even when $X$ is smooth and $F$ not a divisor, I'm not sure how the blowing up process will affect things. Here is what I think so far: if $g:X'\to X$ is the blow up, then we get the exceptional divisor $F'$ and $C'$ with $g_*C'=C$. Now $K_{X'}=g^*K_X+bF'$ for some $b>0$. So $$C\cdot K_X = C'\cdot g^*K_X = C'\cdot K_{X'}-bC'\cdot F' \geq C'\cdot K_{F'}-bC'\cdot F'$$ and $-bC'\cdot F'$ should be negative. I want to say that this is $\geq C\cdot K_F$. $\endgroup$
    – Dave
    Commented Jun 30, 2023 at 18:17

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