Restricting a non-constant map to an ample divisor Let $X$ And $Y$ be smooth projective irreducible varieties over the complex numbers. Let $f:X \to Y$ be a non-constant morphism.
Assume that the dimension of $X$ is at least two.

Question. Let $D\subset X$ be an ample divisor. Is the restriction $f|_D : D\to X$ still non-constant?

This is false when $f$ is only assumed to be a rational map (and $D$ is not contained in the locus of indeterminacy of $f$).
Also, if $f$ is finite, then the answer is positive. Finally, if $f$ has at least two positive-dimensional fibres, the answer is also positive.
 A: Let me expand my comment into an answer. I weaken my previous claim a bit to say that if $f \colon X \rightarrow Y$ is any nonconstant morphism and $D$ is an effective ample divisor on $X$, then $f(D)$ cannot be a point.
Suppose $X$ has dimension at least 2, $D$ is an effective ample divisor on $X$, and $f \colon X \rightarrow Y$ is a nonconstant morphism that contracts $D$ to a point. Replacing $Y$ with $f(X)$ if necessary, we can assume $f$ is surjective.
If $Y$ has dimension at least 2, choose a curve $C_Y \subset Y$ which does not pass through the point $f(D)$. Then by taking appropriate hyperplane sections of $f^{-1}(C_Y)$ one can find a curve $C \subset X$ which maps onto $C_Y$. Since $f(D)$ is disjoint from $C_Y$, the divisor $D$ is disjoint from $C$. But this contradicts ampleness of $D$.
(The same argument shows that $f(D)$ cannot have codimension $\geq 2$ in $Y$.)
If $Y$ has dimension 1, then all fibres of $f \colon X \rightarrow Y$ are divisors, and $D$ is contained in a single fibre of $f$. Choose $C$ to be any curve contained in a different fibre. Then $D$ is disjoint from $C$, again a contradiction.
A: This is a partial answer, too long to be a comment.
Let me give a family of examples where $D$ can be found (every effective divisor $D$ works, actually).
Take any smooth, projective variety $X$ admitting a finite endomorphim $f \colon X \to X$ of degree $\geq 2$, and let $f_i \colon X \to X$ be the composition of $f$ with itself $i$ times.
Since $f$ is finite, the same holds for $f_i$ and so $f_i$ does not contract any subvariety of positive dimension. In particular, the restriction of $f_i$ to every (ample) divisor is non-constant for all $i \geq 1$.
Examples:

*

*$X$= Abelian variety, $f \colon X \to X$ is the multiplication by $2$;

*$X=\mathbb{P}^n$, $f \colon X \to X$ is given by $[x_0: \ldots :x_n] \mapsto [x_0^2: \ldots: x_n^2]$.

