Proof.
Let $\alpha =(\alpha _ n) : (\mathcal{F}_ n) \to (\mathcal{G}_ n)$ be a morphism of $\textit{Coh}(X, \mathcal{I})$. The cokernel of $\alpha $ is the inverse system $(\mathop{\mathrm{Coker}}(\alpha _ n))$ (details omitted). To describe the kernel let
\[ \mathcal{K}'_{l, m} = \mathop{\mathrm{Im}}(\mathop{\mathrm{Ker}}(\alpha _ l) \to \mathcal{F}_ m) \]
for $l \geq m$. We claim:
the inverse system $(\mathcal{K}'_{l, m})_{l \geq m}$ is eventually constant, say with value $\mathcal{K}'_ m$,
the system $(\mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m)_{m \geq n}$ is eventually constant, say with value $\mathcal{K}_ n$,
the system $(\mathcal{K}_ n)$ forms an object of $\textit{Coh}(X, \mathcal{I})$, and
this object is the kernel of $\alpha $.
To see (a), (b), and (c) we may work affine locally, say $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{I}$ corresponds to the ideal $I \subset A$. By Lemma 30.23.1 $\alpha $ corresponds to a map $f : M \to N$ of finite $A^\wedge $-modules. Denote $K = \mathop{\mathrm{Ker}}(f)$. Note that $A^\wedge $ is a Noetherian ring (Algebra, Lemma 10.97.6). Choose an integer $c \geq 0$ such that $K \cap I^ n M \subset I^{n - c}K$ for $n \geq c$ (Algebra, Lemma 10.51.2) and which satisfies Algebra, Lemma 10.51.3 for the map $f$ and the ideal $I^\wedge = IA^\wedge $. Then $\mathcal{K}'_{l, m}$ corresponds to the $A$-module
\[ K'_{l, m} = \frac{a^{-1}(I^ lN) + I^ mM}{I^ mM} = \frac{K + I^{l - c}f^{-1}(I^ cN) + I^ mM}{I^ mM} = \frac{K + I^ mM}{I^ mM} \]
where the last equality holds if $l \geq m + c$. So $\mathcal{K}'_ m$ corresponds to the $A$-module $K/K \cap I^ mM$ and $\mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m$ corresponds to
\[ \frac{K}{K \cap I^ mM + I^ nK} = \frac{K}{I^ nK} \]
for $m \geq n + c$ by our choice of $c$ above. Hence $\mathcal{K}_ n$ corresponds to $K/I^ nK$.
We prove (d). It is clear from the description on affines above that the composition $(\mathcal{K}_ n) \to (\mathcal{F}_ n) \to (\mathcal{G}_ n)$ is zero. Let $\beta : (\mathcal{H}_ n) \to (\mathcal{F}_ n)$ be a morphism such that $\alpha \circ \beta = 0$. Then $\mathcal{H}_ l \to \mathcal{F}_ l$ maps into $\mathop{\mathrm{Ker}}(\alpha _ l)$. Since $\mathcal{H}_ m = \mathcal{H}_ l/\mathcal{I}^ m\mathcal{H}_ l$ for $l \geq m$ we obtain a system of maps $\mathcal{H}_ m \to \mathcal{K}'_{l, m}$. Thus a map $\mathcal{H}_ m \to \mathcal{K}_ m'$. Since $\mathcal{H}_ n = \mathcal{H}_ m/\mathcal{I}^ n\mathcal{H}_ m$ we obtain a system of maps $\mathcal{H}_ n \to \mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m$ and hence a map $\mathcal{H}_ n \to \mathcal{K}_ n$ as desired.
To finish the proof of (1) we still have to show that $\mathop{\mathrm{Coim}}= \mathop{\mathrm{Im}}$ in $\textit{Coh}(X, \mathcal{I})$. We have seen above that taking kernels and cokernels commutes, over affines, with the description of $\textit{Coh}(X, \mathcal{I})$ as a category of modules. Since $\mathop{\mathrm{Im}}= \mathop{\mathrm{Coim}}$ holds in the category of modules this gives $\mathop{\mathrm{Coim}}= \mathop{\mathrm{Im}}$ in $\textit{Coh}(X, \mathcal{I})$. Parts (2) and (3) of the lemma are immediate from our construction of kernels and cokernels.
$\square$
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