The Stacks project

Proof. The case $d > 1$. Since we have a compatible system of maps $p^*(I^ q) \to \mathcal{O}_ X(q)$ for $q \geq 0$ there are canonical maps $p^*(I^ q/I^{q + \nu }) \to \mathcal{O}_{Y_\nu }(q)$ for $\nu \geq 0$. Using this and pulling back $\varphi $ we obtain a map

such that the composition $M \to H^0(X, p^*M) \to H^0(X, \mathcal{O}_{Y_ n}(n))$ is the given homomorphism $\varphi $ combined with the map $I^ n/I^{2n} \to H^0(X, \mathcal{O}_{Y_ n}(n))$. Since $\mathcal{O}_{Y_ n}(n)$ is invertible on $Y_ n$ the linear map $\chi $ determines a section

with notation as in Remark 51.22.5. The discussion in Remark 51.22.5 shows the cokernel and kernel of $can : p^*(M^\vee ) \to (p^*M)^\vee $ are scheme theoretically supported on $Y_ c$. By Lemma 51.22.3 the map $(p^*M)^\vee (n) \to (p^*M)^\vee (n - 2c)$ factors through $p^*(M^\vee )(n - 2c)$; small detail omitted. Hence the image of $\sigma $ in $\Gamma (X, (p^*M)^\vee (n - 2c))$ comes from an element

By Lemma 51.22.4 part (5), the fact that $M^\vee $ is an $(A, n, c)$-module by More on Algebra, Lemma 15.70.7, and the fact that $n \geq q_0 + (1 + d)c$ so $n - 2c \geq q_0 + (d - 1)c$ we see that the image of $\sigma '$ in $H^0(X, p^*M^\vee (n - (1 + d)c))$ is the image of an element $\tau $ in $I^{n - (1 + d)c}M^\vee $. Write $\tau = \sum a_ i \tau _ i$ with $\tau _ i \in I^{n - (2 + d)c}M^\vee $; this makes sense as $n - (2 + d)c \geq 0$. Then $\tau _ i$ determines a homomorphism of modules $\psi _ i : M \to I^{n - (2 + d)c}/I^{2n - (2 + d)c}$ using the evaluation map $M \otimes M^\vee \to A/I^ n$.

Let us prove that this works¹. Pick $z \in M$ and let us show that $\varphi (z)$ and $\sum a_ i \psi _ i(z)$ have the same image in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. First, the element $z$ determines a map $p^*z : \mathcal{O}_{Y_ n} \to p^*M$ whose composition with $\chi $ is equal to the map $\mathcal{O}_{Y_ n} \to \mathcal{O}_{Y_ n}(n)$ corresponding to $\varphi (z)$ via the map $I^ n/I^{2n} \to \Gamma (\mathcal{O}_{Y_ n}(n))$. Next $z$ and $p^*z$ determine evaluation maps $e_ z : M^\vee \to A/I^ n$ and $e_{p^*z} : (p^*M)^\vee \to \mathcal{O}_{Y_ n}$. Since $\chi (p^*z)$ is the section corresponding to $\varphi (z)$ we see that $e_{p^*z}(\sigma )$ is the section corresponding to $\varphi (z)$. Here and below we abuse notation: for a map $a : \mathcal{F} \to \mathcal{G}$ of modules on $X$ we also denote $a : \mathcal{F}(t) \to \mathcal{F}(t)$ the corresponding map of twisted modules. The diagram

commutes by functoriality of the construction $can$. Hence $(p^*e_ z)(\sigma ')$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - 2c))$ is the section corresponding to the image of $\varphi (z)$ in $I^{n - 2c}/I^{2n - 2c}$. The next step is that $\sigma '$ maps to the image of $\sum a_ i \tau _ i$ in $H^0(X, p^*M^\vee (n - (1 + d)c))$. This implies that $(p^*e_ z)(\sum a_ i \tau _ i) = \sum a_ i p^*e_ z(\tau _ i)$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c))$ is the section corresponding to the image of $\varphi (z)$ in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. Recall that $\psi _ i$ is defined from $\tau _ i$ using an evaluation map. Hence if we denote

the map we get from $\psi _ i$, then we see by the same reasoning as above that the section corresponding to $\psi _ i(z)$ is $\chi _ i(p^*z) = e_{p^*z}(\chi _ i) = p^*e_ z(\tau _ i)$. Hence we conclude that the image of $\varphi (z)$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c))$ is equal to the image of $\sum a_ i\psi _ i(z)$. Since $n - (1 + d)c \geq q_0$ we have $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c)) = I^{n - (1 + d)c}/I^{2n - (1 + d)c}$ by Lemma 51.22.1 and we conclude the desired compatibility is true.

The case $d = 1$. Here we argue as above that we get

and then we use Lemma 51.22.4 part (4) to see that $\sigma '$ is the image of some element $\tau \in I^{n - 2c}M^\vee $. The rest of the argument is the same.

The case $d = 0$. Argument is exactly the same as in the case $d = 1$. $\square$