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$\DeclareMathOperator\Hom{Hom}$I have some questions on a proof of Proposition II.3.10 & notations from Rational Curves on Algebraic Varieties by Janos Kollar (page 117).

We work in setting exposed in 3.1 Definition (p. 113):

Let $C$ be a proper curve without embedded points (for us a curve is an integral scheme of dimension $1$, proper over $k$, all of whose local rings are regular), $X$ a smooth variety and $f: C \to X$ a morphism. Let $B \subset C$ be a closed subscheme with ideal sheaf $I_B$ and $g = f \rvert _B$.

The Proposition uses a couple of notations (same notations as above):

Let $F: C \times \Hom(C,X,g) \to X$ be the universal morphism. For later applications we also consider the induced morphism $F^{(2)}: C \times C \times \Hom(C,X,g) \to X \times X$ (not relevant for us).

Let $p, q \in C$ be closed points and $f : C \to X$ a morphism such that $f \rvert _B = g$.

If $p, q \not \in B$, then by (1.2.16) and (1.9)

$$T_{C \times \Hom(C,X,g)} \otimes k(p, [f])=T_C \otimes k(p) + H^0(C, f^*T_X \otimes I_B).$$

Few words on notations: $T_?$ is the tangent sheaf (the dual to Kähler $\Omega_?$), $\Hom(C,X g)= \{f \in \Hom(C,X) \vert f \rvert _B=g \}$ and $[f] \in \Hom(C,X g)$.

Let $df(s): T_C \otimes k(s) \to T_X \otimes k(f(s))$ be the differential of $f$ at $s \in C$,

$$\phi(p,f): H^0(C, f^*T_X \otimes I_B) \to f^*T_X \otimes k(p)$$

the evaluation map.

The following is a reformulation of (1.2.19):

3.4 Proposition (Notation as above). Then

$$dF(p,[f])=df(p)+ \phi(p,f).$$

Now we are ready for Prop. 3.10:

3.10 Proposition (Notation as in (3.3)). Assume that $C \cong \mathbb{P}^1$ and $\lvert B \rvert \le 2$ and write $f*T_X \otimes I_B = \sum \mathcal{O}(a_i)$. Then

$$\#\{i \vert a_i \ge 0 \} = \operatorname{rank} dF(p, [f]) \ \forall p \in \mathbb{P}^1 - B.$$

Proof. Since $\vert B \vert \le 2$, $H^0(C, T_{\mathbb{P}^1} \otimes I_B) \to T_{\mathbb{P}^1} \otimes k(p)$ is surjective (???). Therefore

$$\operatorname{rank} \phi(p,f) =\operatorname{rank} dF(p, [f]).$$

Questions:

Q₁: Why does $\lvert B \rvert \le 2$ imply $H^0(C, T_{\mathbb{P}^1} \otimes I_B) \to T_{\mathbb{P}^1} \otimes k(p)$ is surjective? Firstly what is $\lvert B \rvert$? The cardinality of $B$ as set or the dimension of dimensional linear system induced by $B$ as subscheme?

I tend to say that it's the latter one. Nevertheless, why does it imply surjectivity of $H^0(C, T_{\mathbb{P}^1} \otimes I_B) \to T_{\mathbb{P}^1} \otimes k(p)$?

Q₂: What is exactly the rank $\operatorname{rank} \phi(p,f)$ of $\phi(p,f): H^0(C, f^*T_X \otimes I_B) \to f^*T_X \otimes k(p)$? Rank as what? Linear maps? Over which field?

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    $\begingroup$ Q1. $B$ is a closed subscheme, so its ideal sheaf will be $I_B=\mathcal{O}_{\mathbb{P}^1}(-p-q)$ (where $p$ can be equal to $q$), so $|B|$ is meant to denote the degree of the closed subscheme, i.e., the numbers of points accounting for multiplicities. Since $T_{\mathbb{P}^1}\cong\mathcal{O}_{\mathbb{P}^1}(2)$, this line bundle will continue to be globally generated (implying the surjectivity you want) after we twist by up to 2 points, hence the assumption. Q2. Both maps are linear maps of vector spaces over an algebraically closed field, so rank means the rank of the image. $\endgroup$
    – Frank
    Feb 5, 2020 at 7:48
  • $\begingroup$ @Frank:About the last point I'm not sure. You say "Both maps are linear maps of vector spaces over an algebraically closed field, so rank means the rank of the image". Did Kollar in his book somewhere assumed that base field $k$ is algebraically closed? Is it really neccessary to make this assumption? $H^0(C, T_{\mathbb{P}^1} \otimes I_B)$ has always structure of a vc over base field and the $k(p)$-vc $ T_{\mathbb{P}^1} \otimes k(p)$ obtains also $k$-space structure by transitivity. Where do we explicitly need alg. closeness? $\endgroup$
    – user267839
    Feb 5, 2020 at 20:14
  • $\begingroup$ Yes the assumption is made on p113 for the entire section. Probably a lot of statements hold more generally but I'd have to think a bit about it $\endgroup$
    – Frank
    Feb 5, 2020 at 21:16

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