**Proof.**
Proof of (1). If $M$ is generated by $x_1, \ldots , x_ r$ then $x_1, \ldots , x_ r$ define global sections of $\underline{M}$ which generate it, hence $\underline{M}$ is of finite type. Conversely, assume $\underline{M}$ is of finite type. Let $U \in \mathcal{C}$ be an object which is not sheaf theoretically empty (Sites, Definition 7.42.1). Such an object exists as we assumed $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is not the empty topos. Then there exists a covering $\{ U_ i \to U\} $ and finitely many sections $s_{ij} \in \underline{M}(U_ i)$ generating $\underline{M}|_{U_ i}$. After refining the covering we may assume that $s_{ij}$ come from elements $x_{ij}$ of $M$. Then $x_{ij}$ define global sections of $\underline{M}$ whose restriction to $U$ generate $\underline{M}$.

Assume there exist elements $x_1, \ldots , x_ r$ of $M$ which define global sections of $\underline{M}$ generating $\underline{M}$ as a sheaf of $\underline{\Lambda }$-modules. We will show that $x_1, \ldots , x_ r$ generate $M$ as a $\Lambda $-module. Let $x \in M$. We can find a covering $\{ U_ i \to U\} _{i \in I}$ and $f_{i, j} \in \underline{\Lambda }(U_ i)$ such that $x|_{U_ i} = \sum f_{i, j} x_ j|_{U_ i}$. After refining the covering we may assume $f_{i, j} \in \Lambda $. Since $U$ is not sheaf theoretically empty we see that $I \not= \emptyset $. Thus we can pick $i \in I$ and we see that $x = \sum f_{i, j}x_ j$ in $M$ as desired.

Proof of (2). Assume $\underline{M}$ is a $\underline{\Lambda }$-module of finite presentation. By (1) we see that $M$ is of finite type. Choose generators $x_1, \ldots , x_ r$ of $M$ as a $\Lambda $-module. This determines a short exact sequence $0 \to K \to \Lambda ^{\oplus r} \to M \to 0$ which turns into a short exact sequence

\[ 0 \to \underline{K} \to \underline{\Lambda }^{\oplus r} \to \underline{M} \to 0 \]

by Lemma 18.41.1. By Lemma 18.24.1 we see that $\underline{K}$ is of finite type. Hence $K$ is a finite $\Lambda $-module by (1). Thus $M$ is a $\Lambda $-module of finite presentation.
$\square$

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