Let $B = \{x\_1,\dots,x\_{d-2},y\_1,\dots,y\_k\}$ be a subscheme of $d-2+k$ distinct points of $\mathbb{P}^1$, and $g:B\rightarrow \mathbb{P}^2$ be a morphism mapping $x\_1,\dots,x\_{d-2}$ to a fixed point $p\_0$ and $y\_1,\dots,y\_k$ to general points $q\_1,\dots,q\_k\in\mathbb{P}^2$.


Assume that there exists a morphism $f:\mathbb{P}\rightarrow\mathbb{P}^2$ of degree $d$ restricting to $g$ on $B$, so that the space $Hom\_d(\mathbb{P}^1,\mathbb{P}^2,f\_{|B} = g)$, of morphisms restricting to $g$ on $B$, is non empty.


Do we have that $h^1(\mathbb{P}^1,f^{\*}T\_{\mathbb{P}^2}\otimes\mathcal{I}\_B) = 0$ for $d\gg 0$? Would this imply that $Hom\_d(\mathbb{P}^1,\mathbb{P}^2,f\_{|B} = g)$ has dimension $h^0(\mathbb{P}^1,f^{\*}T\_{\mathbb{P}^2}\otimes\mathcal{I}\_B) = d-2k+6$ and that it is smooth at $f$?


Thank you very much.



Tangent space to spaces of maps
I think this is not true, at least if $k\geq 6$. The Euler exact sequence pulled back to $\mathbb{P}^1$ is
$$0\rightarrow \mathscr{O}\_{\mathbb{P}^1}\rightarrow \mathscr{O}\_{\mathbb{P}^1}(d)^3\rightarrow f^\*T\_{\mathbb{P}^1}\rightarrow 0\,.$$Thus $H^1(f^\*T\_{\mathbb{P}^1}\otimes \mathscr{I}\_B)=H^1(f^\*T\_{\mathbb{P}^1}(-d-k+2))$ is zero iff $\alpha :H^1(\mathscr{O}\_{\mathbb{P}^1}(-d-k+2)\rightarrow H^1(\mathscr{O}\_{\mathbb{P}^1}(-k+2)^3$ is surjective, or equivalently, by Serre duality, $\alpha^{\vee} :H^0(\mathscr{O}\_{\mathbb{P}^1}(k-4))^3\rightarrow H^0(\mathscr{O}\_{\mathbb{P}^1}(k-4+d) )$ is injective. Now $\alpha ^{\vee}$ is given by $\alpha ^{\vee}(A,B,C)=AF+BG+CH$, where $f(x)=(F(x),G(x),H(x))$.


Choose the coordinates on $\mathbb{P}^2$ so that $p\_0=(0,0,1)$; then $F$ and $G$ vanish at each $x\_i$, so $F(x)=Q(x)\prod (x-x\_i)$ and $G(x)=R(x)\prod (x-x\_i)$, with $\deg Q=\deg R=2$. But this implies $RF-QG=0$, so as soon as $k\geq 6$ $\alpha ^{\vee}$ is not injective, whatever $d$ is.


