relative canonical divisor VS relative differential sheaf

Let $X$ be a normal variety, $f\colon X\rightarrow C$ a flat surjective morphism onto a smooth curve $C$ with connected fibers. After replacing $C$ by a finite covering, we may assume that $f$ has reduced fibers. Assume that there exists a Cartier divisor $D$ on $X$ with a nonzero morphism $\Omega_{X/C}^r\rightarrow \mathcal{O}(D)$, where $r$ is the relative dimension of $f$ and $\Omega^r_{X/C}$ is the r-th exterior power of the relative differential sheaf $\Omega_{X/C}$. My question is that if we can find an effective Weil $\Delta$ on $X$ such that $K_{X/C}+\Delta=D$. Of course, it is true if $f$ is smooth, and it also holds if $f$ is semi-stable.

I think this may be induced by the semi-stable reduction theorem, but I don't know how to lift our morphism of sheaves to some blow-up of $X$. Any comments or useful references are welcome.

This is true and actually has nothing to do with the morphism. It's a simple fact about divisors and their associated reflexive sheaves.

So, $\omega_{X/C}$, the reflexive sheaf of rank $1$ associated to the Weil divisor $K_{X/C}$ is the reflexive hull of $\Omega_{X/C}^r$. In particular, there exists a natural morphism $$\Omega_{X/C}^r \to \omega_{X/C},$$ which is neither necessarily injective nor surjective, but can be decomposed as a surjective morphism followed by an injective one. $$\Omega_{X/C}^r \twoheadrightarrow (\Omega_{X/C}^r)/({\rm torsion}) \hookrightarrow \omega_{X/C},$$

It is easy to see that if you have a morphism $$\Omega_{X/C}^r \to \mathscr O(D),$$ to a torsion-free sheaf, then it factors through $(\Omega_{X/C}^r)/({\rm torsion})$. On the other hand, since $\omega_{X/C}$ is the reflexive hull of $\Omega_{X/C}^r$ and hence also of $(\Omega_{X/C}^r)/({\rm torsion})$, any reflexive sheaf (e.g., a line bundle) that contains $(\Omega_{X/C}^r)/({\rm torsion})$, also contains $\omega_{X/C}$, which translated to divisors means that $K_{X/C}\leq D$.

This means that what you would like is actually true. In fact, $D$ doesn't even have to be Cartier, it works if it is Weil divisor.

Appendix (in response to the request in the comments):

As I mentioned, the morphism $\Omega_{X/C}^r \to \omega_{X/C}$ is simply the double dual. For any sheaf $\mathscr F$ you have a natural map $\mathscr F\to \mathscr F^{\vee\vee}$. This is it.

Another way to think about it is to consider the embedding of the open set $\imath: U=X\setminus {\rm Sing}\, X\hookrightarrow X$ and then $\mathscr F=\imath_*\imath^*\mathscr F$ so the above map is the natural map given by $1\to \imath_*\imath^*$.

For more on these issues, look at these MO answers: S2 property and canonical sheaf.

• Hello Sándor. Thanks a lot. What I mean are exactly as you said and I have corrected it and wish it's better now. Your argument is also something that comes to my mind, but I am not so sure how the natural morphism $\Omega_{X/C}^r\rightarrow \omega_{X/C}$ can be constructed. I am new on this subject, so this may be a stupid question for experts. Sep 11, 2016 at 11:27
• Hi Chieh, I'll include an explanation for that in the answer. It's easier to write math there.... Sep 11, 2016 at 19:02
• Hi Sándor. Thanks. If I don't misunderstand, you just define $\omega_{X/C}$ to be the double dual of $\Omega_{X/C}^r$. The definition in my mind is a slightly different, which was taken to be $\omega_X\otimes f^*(-\omega_C)$. do these two definitions coincide? Sep 11, 2016 at 19:27
• It's not how I define it. Whatever way you define it, this follows. Your $X$ is normal and hence smooth in codim $1$. All of these are clearly equal on the smooth part and then since the sheaves in question are reflexive, they agree everywhere. By the way, what you wrote is an awful abuse of notation!!! Please don't mix sheaves with divisors. On one hand it raises the hair on my back, but more importantly it can lead to mistakes. Sep 11, 2016 at 20:10
• Thanks so much for your patience and really nice response. So sorry for my notation. The sheaf should be $\omega_X\otimes f^*{\omega_C^*}\simeq \mathcal O(K_X-f^*K_C)$. Anyway, thanks a lot. Sep 11, 2016 at 20:37