Evaluation maps for moduli of stable maps

Let $\overline{M}_{0,n}(\mathbb{P}^N,d)$ be the moduli space of stable maps of degree $d$ from curves of genus zero with $n$-marked points to $\mathbb{P}^N$.

Consider the product of the evaluation maps:

$$ev:=ev_1\times ...\times ev_n:\overline{M}_{0,n}(\mathbb{P}^N,d)\rightarrow (\mathbb{P}^N)^{n},\quad [C,f,(x_1,...,x_n)]\mapsto (f(x_1),...,f(x_n)).$$

Is is true that $ev$ is proper and $ev_{*}\mathcal{O}_{\overline{M}_{0,n}(\mathbb{P}^N,d)} = \mathcal{O}_{(\mathbb{P}^N)^{n}}$?

• For properness: If one has maps $f: X \to Y$ and $g: Y \to Z$ with $g \circ f$ proper and $g$ separated, then $f$ is proper. If $g \circ f$ is even projective, then $f$ is projective. (This is Hartshorne's "property $\mathcal{P}$".) – TKe Mar 23 '15 at 21:48

For $N>1$ and large $n$ the map $ev$ will not be surjective for dimension reasons: $$\text{dim} \overline{M}_{0,n}(\mathbb{P}^N,d) = Nd + N + d + n-3$$ whereas $$\text{dim} (\mathbb{P}^N)^{n} = N n.$$ Thus the pushforward of the structure sheaf cannot be the structure sheaf on the target.
On the other hand for $n=1$ the map $ev=ev_1$ is surjective and flat (see for example the book "An invitation to quantum cohomology" by Kock and Vainsecher, Lemma 2.5.1). So all one would need to show in this case is that the fibres of $ev$ are geometrically connected (by Exercise 28.1.H. in Vakil's Foundations of Algebraic Geometry).
Idea of a probably much too complicated proof: Given two closed points $\mu,\nu$ in the fibre of a point $p \in \mathbb{P}^N$ they can be connected by a rational curve $g: C \to \overline{M}_{0,n}(\mathbb{P}^N,d)$ because this variety is rational. Via $ev_1$ this induces a curve $f: C \to \mathbb{P}^N$. Choose fixed $N+1$ points $q_i \in \mathbb{P}^N$ such that for all but finitely many points $c\in C$ the points $f(c),q_1, \ldots, q_{N+1}$ are in general linear position. Let $\tilde C \subset C$ be the set of points satisfying this, then $\tilde C$ is still connected. There exists a unique $A(c) \in \text{PGL}(N)$ satisfying $A(f(c))=p, A(q_i)=q_i$. By postcomposing the family induced by $\tilde C \to \overline{M}_{0,n}(\mathbb{P}^N,d)$ with the map $A(c)$ we obtain a family connecting $\mu,\nu$ contained in the fibre of $ev$ over $p$.
Edit: I think that for $n\leq N+2$ and $ev$ surjective, the above arguments can be expanded to conclude the desired property of $ev$ when restricting to the preimage of the set $U \subset (\mathbb{P}^N)^n$ of tuples of points $(p_1, \ldots, p_n)$ in general linear position. The action of $\text{PGL}(N)$ on $\mathbb{P}^N$ induces a transitive action on $U$ and also an action on $\overline{M}_{0,n}(\mathbb{P}^N,d)$ (by postcomposition of the stable map) such that $ev$ is equivariant. Then by generic flatness it should hold that $ev$ is flat over $U$ (see the arguments in the book of Kock and Vainsecher). It is also still proper as the restriction (on the target) of a proper map.
For the connectedness of its fibres one copies the argument above, where now one considers the maps $p_1=ev_1 \circ g, \ldots, p_n=ev_n \circ g : C \to \mathbb{P}^N$ and chooses general $q_1, \ldots, q_{N+2-n}$ fixed. For a general point in $C$ the points $p_1, \ldots, p_n, q_1, \ldots, q_{N+2-n}$ will be in general linear position and the argument goes as before.
• Thanks al lot. From your argument it seems that if $ev:=ev_1\times ...\times ev_n$ is surjective then the push-forward of the structure sheaf is the structure sheaf. Am I rigth? – JdiNirv Mar 24 '15 at 15:44
• So I think that surjectivity is not enough: if you look at the case $n=2, d=1, N=2$ then $ev$ is surjective as through any two points $p,q$ in $\mathbb{P}^2$ there is a line. If $(p,q)$ is not in the diagonal, there is exactly one such line, so the fibre of $ev$ is a point, for $p=q$ however the fibre is positive-dimensional (elements are in correspondence with lines through $p=q$). Hence $ev$ is not flat as fibre dimension jumps. However, I think that over a dense open subset of the target, $ev$ has the desired property. I will try to edit my answer accordingly. – JoS Mar 24 '15 at 17:21