The map $N \to N'$ in (3) comes from Constructions, Equation (\ref{constructions-equation-multiply-more-generally}) by taking global sections. \medskip\noindent By Constructions, Equation (\ref{constructions-equation-global-sections-more-generally}) there is a map of graded $A$-modules $M \to \bigoplus_{n \in \mathbf{Z}} H^0(X, \widetilde{M(n)})$. If the generators of $M$ sit in degrees $\geq k$, then the image is contained in the submodule $N' \subset \bigoplus_{n \in \mathbf{Z}} H^0(X, \widetilde{M(n)})$ and we get the map in (4). \medskip\noindent By Algebra, Lemmas \ref{algebra-lemma-graded-Noetherian} and \ref{algebra-lemma-S-plus-generated} the ring $A_0$ is Noetherian and $A$ is generated over $A_0$ by finitely many elements $f_1, \ldots, f_r$ homogeneous of positive degree. Let $d = \text{lcm}(\deg(f_i))$. Then we see that (6) holds for example by Constructions, Lemma \ref{constructions-lemma-where-invertible}. \medskip\noindent Because $M$ is a finite $A$-module we see that $\widetilde{M}$ is a finite type $\mathcal{O}_X$-module, i.e., a coherent $\mathcal{O}_X$-module. Thus part (2) follows from Lemma \ref{lemma-coherent-on-proj-general}. \medskip\noindent We will deduce (1) from (2) using a trick. For $q \in \{0, \ldots, d - 1\}$ write $$ {}^qN = \bigoplus\nolimits_{n + q \geq k} H^0(X, \widetilde{M(q)}(n)) $$ By part (2) these are finite $A$-modules. The Noetherian ring $A$ is finite over $A^{(d)} = \bigoplus_{n \geq 0} A_{dn}$, because it is generated by $f_i$ over $A^{(d)}$ and $f_i^d \in A^{(d)}$. Hence ${}^qN$ is a finite $A^{(d)}$-module. Moreover, $A^{(d)}$ is Noetherian (follows from Algebra, Lemma \ref{algebra-lemma-dehomogenize-finite-type}). It follows that the $A^{(d)}$-submodule ${}^qN^{(d)} = \bigoplus_{n \in \mathbf{Z}} {}^qN_{dn}$ is a finite module over $A^{(d)}$. Using the isomorphisms $\widetilde{M(dn + q)} = \widetilde{M(q)}(dn)$ we can write $$ N' = \bigoplus\nolimits_{q \in \{0, \ldots, d - 1\}} \bigoplus\nolimits_{dn + q \geq k} H^0(X, \widetilde{M(q)}(dn)) = \bigoplus\nolimits_{q \in \{0, \ldots, d - 1\}} {}^qN^{(d)} $$ Thus $N'$ is finite over $A^{(d)}$ and a fortiori finite over $A$. Thus (1) is true. \medskip\noindent Let $K$ be a finite $A$-module such that $\widetilde{K} = 0$. We claim that $K_n = 0$ for $d|n$ and $n \gg 0$. Arguing as above we see that $K^{(d)}$ is a finite $A^{(d)}$-module. Let $x_1, \ldots, x_m \in K$ be homogeneous generators of $K^{(d)}$ over $A^{(d)}$, say sitting in degrees $d_1, \ldots, d_m$ with $d | d_j$. For each $i$ and $j$ there exists an $n_{ij} \geq 0$ such that $f_i^{n_{ij}} x_j = 0$ in $K_{d_j + n_{ij}}$: if not then $x_j/f_i^{d_i/\deg(f_i)} \in K_{(f_i)}$ would not be zero, i.e., $\widetilde{K}$ would not be zero. Here we use that $\deg(f_i) | d | d_j$ for all $i, j$. We conclude that $K_n$ is zero for $n$ with $d | n$ and $n > \max_j (d_j + \sum_i n_{ij} \deg(f_i))$ as every element of $K_n$ is a sum of terms where each term is a monomials in the $f_i$ times one of the $x_j$ of total degree $n$. \medskip\noindent To finish the proof, we have to show that $M \to N'$ is an isomorphism in all sufficiently large degrees. The map $N \to N'$ induces an isomorphism $\widetilde{N} \to \widetilde{N'}$ because on the affine opens $D_+(f_i) = D_+(f_i^d)$ the corresponding modules are isomorphic: $N_{(f_i)} \cong N_{(f_i^d)} \cong N'_{(f_i^d)} \cong N'_{(f_i)}$ by property (6). By Properties, Lemma \ref{properties-lemma-proj-quasi-coherent} we have a canonical isomorphism $\widetilde{N} \to \widetilde{M}$. The composition $\widetilde{N} \to \widetilde{M} \to \widetilde{N'}$ is the isomorphism above (proof omitted; hint: look on standard affine opens to check this). Thus the map $M \to N'$ induces an isomorphism $\widetilde{M} \to \widetilde{N'}$. Let $K = \Ker(M \to N')$ and $Q = \Coker(M \to N')$. Recall that the functor $M \mapsto \widetilde{M}$ is exact, see Constructions, Lemma \ref{constructions-lemma-proj-sheaves}. Hence we see that $\widetilde{K} = 0$ and $\widetilde{Q} = 0$. By the result of the previous paragraph we see that $K_n = 0$ and $Q_n = 0$ for $d | n$ and $n \gg 0$. At this point we finally see the advantage of using $N'$ over $N$: the functor $M \leadsto N'$ is compatible with shifts (immediate from the construction). Thus, repeating the whole argument with $M$ replaced by $M(q)$ we find that $K_n = 0$ and $Q_n = 0$ for $n \equiv q \bmod d$ and $n \gg 0$. Since there are only finitely many congruence classes modulo $n$ the proof is finished.