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Let $X$ be a smooth complex algebraic variety. Consider the following properties (RP) and (JC) of $X$:

(RP) Rolle Property. Let $f:X\to X$ be a morphism such that the differential $d_x f: T_x X \to T_x X$ is an isomorphism for all (closed) points $x\in X$. Then $f$ is (set theoretically) injective (on closed points).

Notice that (RP) is reminiscent of Rolle's theorem from elementary calculus: if $f(a)=f(b)$ then there's a point $c$ at which $f'(c)=0$.

(JC) Jacobian Conjecture. Let $f:X\to X$ be a morphism such that the differential $d_x f: T_x X \to T_x X$ is an isomorphism for all (closed) points $x\in X$. Then $f$ is an isomorphism.

Clearly, in general, property (JC) is stronger than (RP).

In the case of $X=\mathbb{A}^n$ the above properties are equivalent (thm 1.4) and are notoriously an open problem. For some varieties, they're obviously false: $f:\mathbb{C}^*\to \mathbb{C}^*$, $z\mapsto z^2$, has invertible differential at every point but $f$ is not injective (Another example, with $X$ projective, is a nontrivial isogeny of an elliptic curve).

Notice that if $X$ admits a nontrivial étale cover by itself then (RP) and (JC) are false.

Q1. For which varieties are the properties (RP) and/or (JC) above known to be true? (Or, what are some interesting counterexamples?)

Q2. Is there a variety that satisfies (RP) but not (JC)?

Remark: for $X$ projective, whenever $f:X\to X$ is set theoretically bijective, and the differential condition is satisfied, then $f$ is already an isomorphism.

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    $\begingroup$ Do you include Abelian varieties among the "interesting counterexamples"? $\endgroup$ Commented Jul 8, 2021 at 9:42
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    $\begingroup$ At any rate, the study of smooth projective varieties admitting a non-trivial étale endomorphism is an interesting problem in itself. See for instance projecteuclid.org/journals/osaka-journal-of-mathematics/… for the 3-fold case. $\endgroup$ Commented Jul 8, 2021 at 9:51
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    $\begingroup$ I wouldn't count Abelian varieties as interesting counterexamples in this context, because they always admit the $x\mapsto -x$ isogeny, so they trivially do not satisfy (RP) or (JC). $\endgroup$
    – Qfwfq
    Commented Jul 8, 2021 at 10:03
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    $\begingroup$ By Zariski's main theorem, an injective morphism of smooth complex varieties is an open embedding. From this it is easy to deduce that an injective endomorfism of a smooth complex variety is an isomorphism. Hence RP and JC are equivalent. $\endgroup$
    – Angelo
    Commented Jul 8, 2021 at 11:31
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    $\begingroup$ @Qfwfq: sorry, I was thinking only about the projective case $\endgroup$ Commented Jul 8, 2021 at 13:59

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