**Proof.**
Note that $\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N_ n)$. Choose a presentation

\[ A^{\oplus t} \to A^{\oplus s} \to M \to 0 \]

Applying the right exact functor $\mathop{\mathrm{Hom}}\nolimits _ A(-, N)$ we obtain a complex

\[ 0 \xrightarrow {\alpha } N^{\oplus s} \xrightarrow {\beta } N^{\oplus t} \]

whose cohomology in the middle is $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)$ and such that for $n \geq 0$ the cohomology of

\[ 0 \xrightarrow {\alpha _ n} N_ n^{\oplus s} \xrightarrow {\beta _ n} N_ n^{\oplus t} \]

is $\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$. Let $c \geq 0$ be as in Lemma 15.100.1 for this $A$, $I$, $\alpha $, and $\beta $. By part (3) of the lemma we deduce the Mittag-Leffler property for $(\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n))$. The kernel and cokernel of the maps $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)/I^ n\mathop{\mathrm{Hom}}\nolimits _ A(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$ are killed by $I^ c$ by [art part (6) of the lemma. We find that $\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge $ by part (2) of the lemma. The equality

\[ \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge ) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) \]

follows formally from the fact that $M^\wedge = \mathop{\mathrm{lim}}\nolimits M_ n$ and $M_ n = M^\wedge /I^ nM^\wedge $ and the corresponding facts for $N$, see Algebra, Lemma 10.97.4.

The result for isomorphisms follows from the case of homomorphisms applied to both $(\mathop{\mathrm{Hom}}\nolimits (M_ n, N_ n))$ and $(\mathop{\mathrm{Hom}}\nolimits (N_ n, M_ n))$ and the following fact: for $n > m > 0$, if we have maps $\alpha : M_ n \to N_ n$ and $\beta : N_ n \to M_ n$ which induce an isomorphisms $M_ m \to N_ m$ and $N_ m \to M_ m$, then $\alpha $ and $\beta $ are isomorphisms. Namely, then $\alpha \circ \beta $ is surjective by Nakayama's lemma (Algebra, Lemma 10.20.1) hence $\alpha \circ \beta $ is an isomorphism by Algebra, Lemma 10.16.4.
$\square$

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