**Proof.**
Part (1) is a special case of part (2). The second equality in (2) follows from Lemma 10.9.5. Choose a presentation

\[ \bigoplus \nolimits _{j = 1, \ldots , m} R \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} R \to M \to 0. \]

By Lemma 10.10.1 this gives an exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \bigoplus \nolimits _{i = 1, \ldots , n} N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} N. \]

Inverting $S$ and using Proposition 10.9.12 we get an exact sequence

\[ 0 \to S^{-1}\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}N \]

and the result follows since $S^{-1}M$ sits in an exact sequence

\[ \bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}R \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}R \to S^{-1}M \to 0 \]

which induces (by Lemma 10.10.1) the exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _{S^{-1}R}(S^{-1}M, S^{-1}N) \to \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}N \]

which is the same as the one above.
$\square$

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