Let $f:X\to B$ be a fibration with singular fibers $F_1,\cdots,F_s$. We define $\Omega_{X/B}$ by the sequence $$0\to f^*\Omega_B\to \Omega_X\to \Omega_{X/B}\to 0.$$ I know that $f^*\Omega_B\simeq\mathscr O_X(f^*K_B+D(f))$, where $D(f):=\sum^s_{i=1}(F_i-F_{i,red})$, and $$c_1(\Omega_{X/B})=K_X-f^*K_B-D(f)=K_{X/B}-D(f).$$
Q How to get $\Omega^{**}_{X/B}\simeq K_{X/B}(-D(f))\hookrightarrow K_{X/B}$.