From Introduction to Graph Theory by Douglas B. West:
When they write $V(D)$, do they mean the vertices of the the graph $G$ (which has kernel-perfect orientation $D$), or do they mean the vertices of $D[U]$, which is just $U$? I would this it’s the former, but later on in the proof they say that each vertex in $V(D)-S$ has a successor in $S$, which makes it look like $V(D)$ is indeed $U$. Because the kernel $S$ of $D[U]$ is a kernel for $D[U]$ right, and not for $G$?
I am also confused on what $D’$ is… Is it true that $D’=D[U]$?
EDIT
Alright, I figured it out. By $D'$ they mean the subdigraph induced by $V(D)-S$, and $V(D)=V(G)$. Furthermore, it's a mistake of the book that $f'(x)<f(x)$ for $x\in V(D)-S$; they should have written $x\in U-S$. I also don't see why we would need that fact. All we really need is that $\vert L^*\vert\geq1+d_{D-S}^+(x)$ for $x\in D'$, where $L^*$ is the original list assignment without the colour $a$. Then $L'$ is simply the list assignment with possible additional colours deleted, such that $\vert L'\vert=1+d_{D-S}^+(x)$, which is really all we need for the induction hypothesis.