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.



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



Restricting a non-constant map to an ample divisor
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.


