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Let $X$ be a complex algebraic curve, assumed to be connected, smooth and complete. Let $f: X \rightarrow X$ be a surjective morphism. Define a backward complete set for $f$ as a subset $S$ of $X$ such that $f^{-1}(S) \subset S$ (I am not sure if it is the standard terminology).

If $f$ has infinitely many finite backward complete sets, is $f$ necessarily an automorphism?

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    $\begingroup$ If $X$ has genus $\ge2$, then its only endomorphisms are automorphisms. So you're really asking about curves of genus $0$ and $1$. For elliptic curves, endomorphisms are homomorphisms (isogenies) composed with translations, so it's easy to answer your question. For $\mathbb P^1$, this is pretty standard stuff. $\endgroup$
    – Joe Silverman
    Commented Sep 13, 2019 at 11:00
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    $\begingroup$ Thanks. So for elliptic curves, the answer is yes, am I right? (if $f$ is an isogeny of degree $d$ followed by a translation, $f^n$ is an isogeny of degree $d^n$ followed by a translation, and $(f^n)^{-1}(P)$ has cardinality $d^n$ for any point $P$. Thus even the existence of a single finite backward complete set implies that $d=1$ and $f$ an automorphism.) But what's the answer for ${\mathbb P}^1$? $\endgroup$
    – Joël
    Commented Sep 13, 2019 at 11:17

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For a map $f:\mathbb P^1\to\mathbb P^1$ of degree $d\ge2$, there are three cases: (1) There are no finite backward invariant sets. (2) There is one such set consisting of a single point.Moving that point to infinity, we have $f(x)\in\mathbb{C}[x]$ is a polynomial of degree $d$. (3) There are two such points. Moving them to $0$ and $\infty$, the map $f$ has the form $f(x)=cx^{\pm d}$.

This is a standard first result in complex dynamics. The proof is quite easy using the Riemann-Hurwitz genus formula, which for $f:\mathbb P^1\to\mathbb P^1$ says that $$ 2d-2 = \sum_{P\in\mathbb P^1} (e_P(f)-1). $$ If $\{P_1,\ldots,P_n\}$ is a backward invariant set, then each $f^{-1}(P_i)$ must consist of a single point, so the points in the set are totally ramified, i.e., $e_{P_i}(f)=d$. Hence $$ 2d-2 = \sum_{P\in\mathbb P^1} (e_P(f)-1) \ge \sum_{i=1}^n (e_{P_i}(f)-1) = n(d-1).$$ This proves that $n\le2$, and a more careful analysis of the $n=1$ and $n=2$ cases yields the results mentioned earlier.

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    $\begingroup$ Thanks. By the way, do you know ab basic reference (a textbook maybe) which covers carefully this kind of "first standard results in complex dynamics"? For instance, the same question you answered but for higher dimensional varieties, or (what I am really looking for) with the endomorphism $f$ replaced by a self-correspondence. What I have found by looking randomly on the web goes very fast, after the first definitions, to more advanced stuff such as Julia and Fatou set, or otherwise arithmetic questions. $\endgroup$
    – Joël
    Commented Sep 14, 2019 at 11:48
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    $\begingroup$ @Joël There are lots of references for $\mathbb P^1$, but for $\mathbb P^n$ and other higher dimensional varieties, not so much. You might try: Fornaess, John Erik and Sibony, Nessim. Complex dynamics in higher dimension. I, Ast\'erisque 222 (1994), 201-231. Complex dynamics in higher dimension. II, Modern Methods in Complex Snalysis (Princeton, NJ, 1992), Ann. of Math. Stud. 137, Princeton Univ. Press, 1995, 135-182. $\endgroup$
    – Joe Silverman
    Commented Sep 14, 2019 at 12:18

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