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I have a couple of questions about arguments in the proof of Lemma 2.6 (see absol page 199, rel p 9) from Shigeyuki Kondo's paper Enriques surfaces with finite automorphism groups:

The setup: Let $Y$ be a complex Enriques surface $Y$, so in following we work in complex setting, ie $Y$ is aspecially a compact (in modern terms proper) complex smooth surface.

LEMMA (2.6). Let $\pi: Y \to \Bbb P^1$ be a special elliptic pencil of an Enriques surface $Y$. (Ie, the general fiber of $\pi$ is smooth elliptic and special means that it has a $2$-section $R$ which is nodal curve (see Def 2.4 in the linked paper). Let $X$ be the $K3$ surface coming as double cover of $Y$ and $\sigma: X \to X$ its covering involution. Denoted by $\widetilde{\pi}: X \to \Bbb P^1$ the base change of $\pi: Y \to \Bbb P^1$. (BTW: This map is called "base change", but bc along what? Isn't this just the composition $\pi \circ p$? Or it there tacitly an involution of $\Bbb P^1$ involved?) Then there is an elliptic pencil $J(\pi): J(Y) \to \Bbb P^1$ which satisfying certain conditions (i),(ii),(iii),(iv) given in the linked paper.

I struggle with several steps in the presented proof. Could somebody look through if one or other point can be resolved:

(A): What is the argument that the inverse image $p^{-1}(R)$ of the $2$-section $R$ of $Y$ - which exists as $Y$ special - is given by a disjoint union of two nodal curves $R_+,R_-$ (here: nodal curve = smooth rational curve, ie $\Bbb P^1$; older terminology, see p 196).
Intuitivelly: Why the double cover $p:X \to Y$ resolves "doubleness" of the $2$-section $2$? In other words $R_+,R_-$ become honest sections of $X \to \Bbb P^1$.

(B): Assuming (A) has been resolved, why there exist automorphism $t: X \to X$ defined as $t(x):=r^+(\widetilde{\pi}(x)-r^-(\widetilde{\pi}(x)$ where $r^+, r^-: \Bbb P^1 \to X$ corresp to honnest sections $R_+,R_-$ from previous part. Note, that $"-"$ should come from fibrewise group structure.
Intuitivelly, this should correspond to a flip of $R_+$ and $R_-$.

I mean, there are strictly speaking two potential problems with this construction: Firstly, since $\pi: Y \to \Bbb P^1$ is elliptic pencil, only the general, but not every fibre of $\pi$ is elliptic curve - so has group structure - but does also the general fibre of $\widetilde{\pi}= \pi \circ p: X \to \Bbb P^1$ inherit group structure, recall $p:X \to Y$ is double cover?
Furthermore, assume we can manage the first problem, what about the fibres, which are not elliptic curves - so carry no group structure? There the minus $"-"$ in def. of $t$ seemingly don't make sense. Or, can one invoke there some extension result?

(C): Later one manages to construct an involution $\tau: X \to X$ and set $S =X/ \langle \tau \rangle $. Is it clear that $S$ admits elliptic pencil induced from $\widetilde{\pi}$? The thing is, $X$ is branched $2$-cover of $S$. General fibre $X_b= \widetilde{\pi}^{-1}(b)$ of $\widetilde{\pi}$ is an elliptic curve , is the quotient $S_b =X_b/ \langle \tau \rangle $ of such general fibre still elliptic curve?

(D): Next one introduces $J(\pi):J(Y) \to \Bbb P^1$ as minimal model of the pencil $S \to \Bbb P^1$ from prev. Q (C). Is it clear why the anticanonical linear system $\vert -K_{J(Y)} \vert $ must contain a fibre of $J(\pi)$? From what does it follow?

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    $\begingroup$ The "branched double cover $p:X\rightarrow Y$" is not branched. $\endgroup$
    – abx
    Commented Nov 4 at 16:46
  • $\begingroup$ @abx: Right, thanks, so its a honnest etale map $\endgroup$
    – user267839
    Commented Nov 4 at 16:54
  • $\begingroup$ Think it's standard, but let me add discussion in I, 18 in Barth's Compact Complex Surfaces adressing this point that the existence of such etale double cover is indicated by existence of $2$-torsion in integral homology. To translate it to algebro geometric context one can find $2$-torsion in Picard group of $Y$, from this there is a machinery to construct cyclic covers, see also Chap 17 $\endgroup$
    – user267839
    Commented Nov 4 at 17:16
  • $\begingroup$ So, this settles (A). For (B), note that a birational automorphism of a K3 surface is actually biregular. $\endgroup$
    – abx
    Commented Nov 4 at 18:13
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    $\begingroup$ $J(Y)$ is $\mathbb{P}^2$ blown up at the 9 base points of a pencil of cubic curves, so $\lvert -K_{J(Y)}\rvert$ consists of the fibers of this pencil. No offense, but I think you should start by reading some basic introduction to surfaces. $\endgroup$
    – abx
    Commented Nov 4 at 20:07

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