Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal. Consider the category $\mathcal{A}$ of $I$-adically complete $R$-modules, see Algebra, Definition 10.96.2. Let $\varphi : M \to N$ be a morphism of $\mathcal{A}$. The cokernel of $\varphi $ in $\mathcal{A}$ is the $I$-adic completion $(\mathop{\mathrm{Coker}}(\varphi ))^\wedge $ of the usual cokernel (as $I$ is finitely generated this completion is complete, see Algebra, Lemma 10.96.3). Let $K = \mathop{\mathrm{Ker}}(\varphi )$; since $\varphi $ is continuous this is a closed submodule of $M$. By Lemma 110.10.1 below we see that $K$ is $I$-adically complete and hence is the kernel of $\varphi $ in $\mathcal{A}$.
Lemma 110.10.1. Let $I \subset R$ be an ideal of a ring. Let $M$ be an $I$-adically complete $R$-module and let $N \subset M$ be a submodule. The following are equivalent
$N$ is closed in $M$,
$N = \bigcap _{n \geq 1} (N + I^ n M)$,
$M/N$ is $I$-adically complete
If $I$ is finitely generated, these conditions imply that $N$ is $I$-adically complete.
Proof. The equivalence of (1) and (2) follows from Formal Spaces, Lemma 87.4.2. The equivalence of (2) and (3) is Algebra, Lemma 10.96.10. Assume $I$ is finitely generated and (1) holds. Let $N^\wedge $ be the $I$-adic completion of $N$. Since $M$ is $I$-adically complete we have a factorization
of the inclusion map. Since $N^\wedge \to M$ is continuous and $N$ is dense in $N^\wedge $ and $N$ is closed in $M$, we see that $N^\wedge \to M$ factors through $N$. Thus $N^\wedge = N \oplus C$ for some $A$-module $C$. Thus $N$ is a direct summand of the $I$-adically complete module $N^\wedge $, see Algebra, Lemma 10.96.3 (this is where we use that $I$ is finitely generated), and hence $N$ is $I$-adically complete. $\square$
We will give an example that shows that $\mathop{\mathrm{Im}}\not= \mathop{\mathrm{Coim}}$ in the category $\mathcal{A}$ in general. We take $R = \mathbf{Z}_ p = \mathop{\mathrm{lim}}\nolimits _ n \mathbf{Z}/p^ n\mathbf{Z}$ to be the ring of $p$-adic integers and we take $I = (p)$. Consider the map
where the left hand side is the $p$-adic completion of the direct sum. Hence an element of the left hand side is a vector $(x_1, x_2, x_3, \ldots )$ with $x_ i \in \mathbf{Z}_ p$ with $p$-adic valuation $v_ p(x_ i) \to \infty $ as $i \to \infty $. This maps to $(x_1, px_2, p^2x_3, \ldots )$. Hence we see that $(1, p, p^2, \ldots )$ is in the closure of the image but not in the image. By our description of kernels and cokernels above it is clear that $\mathop{\mathrm{Im}}\not= \mathop{\mathrm{Coim}}$ for this map.
Lemma 110.10.2. Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal. The category of $I$-adically complete $R$-modules has kernels and cokernels but is not abelian in general even when $R$ is Noetherian.
Proof. See above. $\square$
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