Suppose that $R \to R'$ is a ring map. Let $\mathcal{U}$ be the standard affine open covering of $\mathbf{P}^n_R$, and let $\mathcal{U}'$ be the standard affine open covering of $\mathbf{P}^n_{R'}$. Note that $\mathcal{U}'$ is the pullback of the covering $\mathcal{U}$ under the canonical morphism $\mathbf{P}^n_{R'} \to \mathbf{P}^n_R$. Hence there is a map of {\v C}ech complexes $$ \gamma : \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}', \mathcal{O}_{\mathbf{P}_{R'}}(d)) $$ which is compatible with the map on cohomology by Cohomology, Lemma \ref{cohomology-lemma-functoriality-cech}. It is clear from the computations in the proof of Lemma \ref{lemma-cohomology-projective-space-over-ring} that this map of {\v C}ech complexes is compatible with the identifications of the cohomology groups in question. (Namely the basis elements for the {\v C}ech complex over $R$ simply map to the corresponding basis elements for the {\v C}ech complex over $R'$.) Whence the first statement of the lemma. \medskip\noindent Now fix the ring $R$ and consider two homogeneous polynomials $f, g \in R[T_0, \ldots, T_n]$ both of the same degree $m$. Since cohomology is an additive functor, it is clear that the map induced by multiplication by $f + g$ is the same as the sum of the maps induced by multiplication by $f$ and the map induced by multiplication by $g$. Moreover, since cohomology is a functor, a similar result holds for multiplication by a product $fg$ where $f, g$ are both homogeneous (but not necessarily of the same degree). Hence to verify the second statement of the lemma it suffices to prove this when $f = x \in R$ or when $f = T_i$. In the case of multiplication by an element $x \in R$ the result follows since every cohomology groups or complex in sight has the structure of an $R$-module or complex of $R$-modules. Finally, we consider the case of multiplication by $T_i$ as a $\mathcal{O}_{\mathbf{P}^n_R}$-linear map $$ \mathcal{O}_{\mathbf{P}^n_R}(d) \longrightarrow \mathcal{O}_{\mathbf{P}^n_R}(d + 1) $$ The statement on $H^0$ is clear. For the statement on $H^n$ consider multiplication by $T_i$ as a map on {\v C}ech complexes $$ \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d + 1)) $$ We are going to use the notation introduced in the proof of Lemma \ref{lemma-cohomology-projective-space-over-ring}. We consider the effect of multiplication by $T_i$ in terms of the decompositions $$ \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d)) = \bigoplus\nolimits_{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_i = d} \check{\mathcal{C}}^\bullet(\vec{e}) $$ and $$ \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d + 1)) = \bigoplus\nolimits_{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_i = d + 1} \check{\mathcal{C}}^\bullet(\vec{e}) $$ It is clear that it maps the subcomplex $\check{\mathcal{C}}^\bullet(\vec{e})$ to the subcomplex $\check{\mathcal{C}}^\bullet(\vec{e} + \vec{b}_i)$ where $\vec{b}_i = (0, \ldots, 0, 1, 0, \ldots, 0))$ the $i$th basis vector. In other words, it maps the summand of $H^n$ corresponding to $\vec{e}$ with $e_i < 0$ and $\sum e_i = d$ to the summand of $H^n$ corresponding to $\vec{e} + \vec{b}_i$ (which is zero if $e_i + b_i \geq 0$). It is easy to see that this corresponds exactly to the action of the contragredient of multiplication by $T_i$ as a map $$ (R[T_0, \ldots, T_n])_{-n - 1 - (d + 1)} \longrightarrow (R[T_0, \ldots, T_n])_{-n - 1 - d} $$ This proves the lemma.