In Chapter III, section $5$ of Hartshorne, we want to calculate the cohomology groups of sheaves $\mathcal{O}_X(n)$ on a projective space $X=\mathbb{P}_A^r$, where $A$ is a noetherian ring.
We are going to prove the following statement: $$H^i(X,\mathcal{O}_X(n))=0, 0<i<r, \forall n\in \mathbb{Z}.$$
First we can claim that $$H^i(X,\mathcal{F})_{x_r}=0,$$ where $\mathcal{F}=\bigoplus_{n\in \mathbb{Z}}\mathcal{O}_X(n).$
We consider the exact sequence of sheaves $$0\rightarrow \mathcal{F}(-1)\rightarrow \mathcal{F}\rightarrow \mathcal{F}_H\rightarrow 0,$$ where $H$ is the hyperplane $x_r=0$, and $\mathcal{F}_H=\bigoplus_{n\in \mathbb{Z}}\mathcal{O}_H(n)$.
Then it gives a long exact sequence of $A$-modules as follows, $$0 \rightarrow H^{r-1}(X,\mathcal{F}(-1)) \xrightarrow{\times x_r} H^{r-1}(X,\mathcal{F})\rightarrow H^{r-1}(X,\mathcal{F}_H)$$ $$\xrightarrow{\delta} H^r(X,\mathcal{F}(-1)) \xrightarrow{\times x_r} H^r(X,\mathcal{F})\rightarrow H^r(X,\mathcal{F}_H) \rightarrow 0,$$ note that $H^{r-2}(X,\mathcal{F}_H)=0$ by the induction hypothesis.
In order to show that the first $\times x_r$ above is an isomorphism, we need to prove $\delta$ is injective.
So now my question is: Why this $\delta$ is defined by $\times x_r^{-1}$ as Hartshorne said?
Although we know that im$(\delta)=$ ker$(H^r(X,\mathcal{F}(-1)) \xrightarrow{\times x_r} H^r(X,\mathcal{F})$
is generated by $x_0^{l_0}x_1^{l_1}\cdots x_{r-1}^{l_{r-1}}x_r^{-1}$ in degree $n=\Sigma_{i=0}^{r-1} l_i-1$, but how to conclude that $\delta$ is exactly the map $\times x_r^{-1}$?