$\def\fra{\mathfrak{a}}
\def\c{\mathrm{c}}$I found the answer to be quite subtle, for the question brings up the issue that $\fra$-adic completions may not be $\fra$-adically nor $\hat{\fra}$-adically complete (see the containments \eqref{contain}, which in general are strict). You mention in your first “thought” two possible topologies on $\hat{A}$; namely, the ones induced by the filtrations $\{i(\fra^n)\}$ and $\{(\fra^e)^n\}$. Although in both topologies the sum in $\hat{A}$ is continuous, only with the latter one can prove continuity of the product in $\hat{A}$ (since $(\fra^e)^n$ is an ideal but $i(\fra^n)$ is not, as you acknowledge). The following two lemmas will explain why. Despite this, the filtration $(\fra^e)^n$ is still not the correct one for $\hat{A}$ (‘correct’ here means ‘satisfying a certain universal property,’ see Lemma 6 below).
Given $G$ an abelian group, a filtration $G\supset G_1\supset G_2\supset\cdots$ of subgroups always defines a topology on $G$ (that depends on the filtration); see Definition 1 and Lemma 2 here plus the example in between. Moreover:
Lemma 1. An abelian group along with the topology induced by a filtration of subgroups is a topological abelian group.
This is true in the non-abelian case as well—even when the subgroups are non-normal—, but the additive notation is the one that interests us, so we state it so.
Proof. On the one hand, note that $-(x+G_n)=-x+G_n$ is open, whence $-:G\to G$ is continuous. Denote $\sigma:G\times G\to G$ to the sum. We have to show that $\sigma^{-1}(x+G_n)$ is open, where $x\in G$. Let $(y,z)\in\sigma^{-1}(x+G_n)$, i.e., $y+z+G_n=x+G_n$. It suffices to show that $(y+G_n)\times(z+G_n)\subset \sigma^{-1}(x+G_n)$. Let $(f,g)\in (y+G_n)\times(z+G_n)$. Then $f+g\in y+G_n+z+G_n=y+z+G_n=x+G_n$. $\square$
In the case that interests us, we have:
Lemma 2. Let $A$ be a ring with a filtration $A\supset\fra_1\supset\fra_2\supset\cdots$ of ideals. Then $A$, along with the topology induced by the filtration $(\fra_n)$, is a topological ring.
Proof. We already know that the sum is continuous (Lemma 1). Denote $\mu:A\times A\to A$ to the multiplication map. Let $x\in A$. We show that $\mu^{-1}(x+\fra_n)$ is open. Let $(y,z)\in\mu^{-1}(x+\fra_n)$, i.e, $yz+\fra_n=x+\fra_n$. It suffices to show that $(y+\fra_n)\times(z+\fra_n)\subset\mu^{-1}(x+\fra_n)$. Let $(f,g)\in (y+\fra_n)\times(z+\fra_n)$. Then $f+g\in (y+\fra_n)(z+\fra_n)=yz+\fra_n=x+\fra_n$. $\square$
In the next result, we settle the multiplication on $\hat{A}$. Given an abelian topological group $G$, denote $G'$ to the set of Cauchy sequences on $G$. Then $G'$ is a group with the obvious sum. Let $G_0\subset G'$ be the subset of Cauchy sequences converging to zero. Then $G_0$ is a subgroup (to prove these two claims, use continuity of $+_G$ to show that for any neighborhood $U\subset G$ of zero there is another neighborhood $V\subset G$ of zero with $V+V\subset U$). We define $\hat{G}=G'/G_0$.
Lemma 3. Let $A$ be a first-countable topological ring.
For each $x\in A$ and each neighborhood $V\subset A$ of zero, there is another neighborhood $U\subset A$ such that $xU\subset V$.
$A'$ becomes a ring with term-wise multiplication.
$A_0$ is an ideal of $A'$ (notations from before).
Therefore, $\hat{A}=A'/A_0$ is a ring whenever $A$ is first-countable.
Proof. Proof of part 1: To look for a contradiction, suppose it's false. Let $U_n$ be a countable neighborhood basis of $A$ at $0$. We can assume $U_n$ is decreasing. For each $n$, there is $s_n\in U_n$ with $xs_n\not\in V$. In particular, $s_n\to 0$. Since multiplication in $A$ is continuous, $xs_n\to 0$. Contradiction.
Proof of part 2: Let $(x_n),(y_n)\in A'$. We have to show that $(x_ny_n)$ is Cauchy. Let $W\subset A$ be a neighborhood of zero. Let $V\subset A$ be a neighborhood of zero with $V+V+V+V\subset W$ (use continuity of the sum $G\times G\times G\times G\to G$). Since the product in $A$ is continuous, there is a neighborhood $N\subset A$ of zero with $NN\subset V$. There is $k$ such that $x_n-x_m, y_n-y_m\in N$ for all $n,m\geq k$. By part 1, there is a neighborhood $U\subset A$ of zero with $x_kU, y_kU\subset V$. Moreover, we can suppose $U\subset N$. There is $\ell$ such that for all $n,m\geq\ell$, we have $x_n-x_m,y_n-y_m\in U$. Finally, for $n,m\geq\max(k,\ell)$, we have
\begin{align*}
x_ny_n-x_my_m
&=x_ny_n-x_my_n+x_my_n-x_my_m\\
&=(x_n-x_m)y_n+x_m(y_n-y_m)\\
&\in U(y_k+N)+(x_k+N)U\\
&\subset Uy_k+UN+x_kU+NU\\
&\subset Uy_k+NN+x_kU+NN\\
&\subset V+V+V+V\\
&\subset W.
\end{align*}
Proof of part 3 (similar to part 2): Let $(x_n)\in A'$, $(s_n)\in A_0$ and $W\subset A$ be a neighborhood of zero. Since sum is continuous in $A$, there is a neighborhood $V\subset A$ of zero such that $V+V\subset W$. By continuity of the product, there is a neighborhood $N\subset A$ of zero with $NN\subset V$. There is $k$ such that $x_n-x_m\in N$ for all $n,m\geq k$. By the first part, there is a neighborhood $U\subset A$ of zero with $x_kU\subset V$. We can suppose $U\subset N$. There is $\ell$ with $s_n\in U$ for all $n\geq\ell$. Hence, for all $n\geq \max(k,\ell)$, we have
$$
x_ns_n\in(x_k+N)U= x_kU+NU\subset x_kU+NN\subset V+V\subset W. \quad\square
$$
Exercise 4. Let $A$ be a first-countable topological ring, and let $I\subset A$ be an ideal. Then:
if $I$ is open, $\hat{I}\subset \hat{A}$ is an ideal.
A sufficient condition for $I$ to be open is that for all sequences $s_n$ in $A$ converging to zero, the set $\{s_n\}\cap I$ is non-empty. (Hint: to show that for $x\in I$ one has $x\in\operatorname{int}_AI$, note that it suffices to show it for $x=0$.)
We finally come to our case of interest: Let $A$ be a ring, $\fra\subset A$ be an ideal, and suppose $A$ has the $\fra$-adic topology. Then there are containments of additive groups
$$
\label{contain}\tag{1}
i(\fra)\subset\fra^n\hat{A}\subset\hat{\fra}^n\subset\widehat{\fra^n}.
$$
(With the OP's notation, $(\fra^e)^n=\fra^n\hat{A}$.) It happens that these containments aren't equalities in general. A particular case where the first containment is strict is $A=R[x]$, where $R$ is any non-zero ring. For cases where the other two contaiments are strict, see Bourbaki, Commutative Algebra, Ch. III, Exercise 2.12, p. 235. Not everything is lost: in the case $\fra$ is finitely generated, one has $\fra^n\hat{A}=\hat{\fra}^n=\widehat{\fra^n}$, for $\fra^n$ will also be finitely generated, so we may apply this result (take $M=A$). Now, the second additive group in \eqref{contain} is an ideal (by definition), whereas the two last ones are ideals by Exercise 4 (alternatively, note that $\widehat{\fra^n}$ is the kernel of the canonical homomorphism $\hat{A}\to A/\fra^n$).
Summing up: if we topologize $\hat{A}$ with respect to the filtrations of ideals given by $\fra^n\hat{A}, \hat{\fra}^n$ or $\widehat{\fra^n}$, by Lemma 2 $\hat{A}$ will be a topological ring. Also, one can verify by hand that the inverse image along $i$ of the ideals $\fra^n\hat{A}, \hat{\fra}^n,\widehat{\fra^n}$ equals $\fra^n$ in all cases. Hence, $i$ is a continuous ring homomorphism of topological rings when $A$ has the $\fra$-adic topology and $\hat{A}$ has the topology induced by any of the three filtrations $\fra^n\hat{A}, \hat{\fra}^n,\widehat{\fra^n}$. Moreover, with the topology induced by any of these, $\hat{A}$ is Hausdorff, since $(0)\subset\bigcap\fra^n\hat{A}\subset\bigcap \hat{\fra}^n\subset\bigcap\widehat{\fra^n}=(0)$, where the last equality is Remark 4.1 from here.
So which filtration is better on $\hat{A}$? The answer is: the last one, because among all three last filtrations in \eqref{contain}, only $(\widehat{\fra^n})$ gives $\hat{A}$ a topology that makes this ring always (i) satisfy a universal property (Lemma 6) and (ii) have a topological ring structure where every Cauchy sequence converges (proven here; see this for examples of non-convergent Cauchy sequences with the topologies coming from the filtrations $(\fra^n\hat{A})$ and $(\hat{\fra}^n)$).
We will prove the universal property in full generality. So from now on $A$ will be an arbitrary first-countable topological ring.
Given a first-countable abelian topological group $G$, for any subset $S\subset G$, denote
$$
\tag{2}\label{subset}
\hat{G}\supset\hat{S}=\{\xi\in\hat{G}\mid\text{for all } (x_n)\in\xi,\; x_n \in S\text{ for all }n\gg 0\}.
$$
The sets $\hat{S}$ can be used to topologize $\hat{G}$ (see Lemmas 2, 3 and 7 here), in a way that turns $\hat{G}$ into a first-countable topological group. In our case, the topology on $\hat{A}$ will be the one induced on the underlying additive group $(\hat{A},+)$ by the underlying first-countable topological group $(A,+)$. This way, we have:
Lemma 5. Let $A$ be a first-countable topological ring. The ring $\hat{A}$ is topological.
This can be seen as a generalization of Lemma 2 for more general ring-structure-compatible topologies.
Proof. The sum is continuous in $\hat{A}$ (see here). We are going to show, using a similar argument, that the product $\mu:\hat{A}\times\hat{A}\to\hat{A}$ is continuous. Let $W\subset A$ be a neighborhood of the identity and pick $([x_n],[y_n])\in\mu^{-1}([z_n]+\hat{W})$, i.e., $[x_ny_n]-[z_n]\in\hat{W}$. Pick a countable neighborhood basis $(U_n)$ of $A$ at zero.
$\underline{\text{Claim}}\colon$ There is $N$ such that for all $n\geq N$,
$$
(x_n+U_N)(y_n+U_N)-z_n+U_N\subset W.
$$
To look for a contradiction, suppose there's no such $N$. Then, for each $N$, there is an $n_N\geq N$ and $a_N,b_N,c_N\in U_N$ with
\begin{align*}
W&\not\ni(x_{n_N}+a_N)(y_{n_N}+b_N)-z_{n_N}+c_N\\
&=x_{nN}y_{nN}+\underbrace{a_Ny_{n_N}+x_{n_N}b_N+a_Nb_N+c_N}_{\text{goes to }0\text{ whenever }N\to+\infty}-z_{n_N},
\end{align*}
since $a_N,b_N,c_N\to 0$ and $(x_{n_N}),(y_{n_N})$ are Cauchy (Exercise 6 here), and $A_0\subset A'$ is an ideal (Lemma 3). This means $\hat{W}\not\ni[x_{n_N}y_{n_N}-z_{n_N}]$. By the mentioned Exercise 6 from the link, $[(x_{n_N}y_{n_N}-z_{n_N})_N]=[(x_ny_n-z_n)_n]$, so contradiction.
Next, we claim that
$$
([x_n]+\hat{U}_N)\times([y_n]+\hat{U}_N)\subset
\mu^{-1}([z_n]+\hat{W}).
$$
To see this, let $[a_n],[b_n]\in\hat{U}_N$ and let $s_n\in A$ be a sequence convergent to zero. In particular, $s_n\in U_N$ for all $n\gg 0$. Thus,
$$
(x_n+a_n)(y_n+b_n)-z_n+s_n\in W,\quad\forall n\gg 0.
$$
Hence, $([x_n]+[a_n])([y_n]+[b_n])-[z_n]\in\hat{W}$, as desired. $\square$
A sequentially complete topological ring is a Hausdorff topological ring where every Cauchy sequence converges.
Lemma 6 (Universal property of the completion of a first-countable topological ring). Let $A$ be a first-countable topological ring. The map $A\to\hat{A}$ is universal among continuous ring homomorphisms from $A$ into sequentially complete topological rings.
Proof. The Hausdorff property for $\hat{A}$ is Remark 4.1 this post; convergence of Cauchy sequences in $\hat{A}$ is proven here. Let $\psi:A\to B$ be a morphism of topological rings, with $B$ is sequentially complete. Apply Lemma 9 from the first link to get a unique continuous and additive factorization $\tilde{\psi}:\hat{A}\to B$. Since $\psi$ is unital, so is $\tilde{\psi}$. It is left to see that $\hat{\psi}$ is multiplicative. This follows from the definition of $\hat{\psi}$, multiplicativity of $\psi$ and continuity of the product in $B$. $\square$
In conclusion: in the particular case that the topology on $A$ is the $\fra$-adic one for an ideal $\fra\subset A$, the ring $\hat{A}$, in order to satisfy the universal property, needs to be topologized with respect to the filtration $\widehat{\fra^n}$. Indeed, one can inspect the proof of the linked Lemma 9 above to see the criticality in the use of the $\widehat{\fra^n}$'s to deduce continuity of $\tilde{\psi}$.
We now address the issue about which one is the correct topology on $\hat{M}$, the completion of an $A$-module.
Lemma 7. Let $A$ be a ring with a filtration $(\fra_n)$ of ideals. An $A$-module $M$, topologized with respect to the filtration $(\fra_nM)$, is a topological $A$-module.
Proof. The sum in $M$ is continuous by Lemma 1. The proof of the continuity of $A\times M\to M$ is very similar to that of Lemma 2. $\square$
Lemma 8. Let $A$ be a topological ring and let $M$ be a topological $A$-module.
- Let $V\subset M$ be a neighborhood of zero, and let $x\in M$, $f\in A$. If $A$ is first-countable, there is a neighborhood $U\subset A$ of zero such that $Ux\subset V$. If $M$ is first-countable, there is are neighborhood $W\subset M$ of zero such that $fW\subset V$.
Suppose $A$ and $M$ are first-countable.
The additive group $M'$ is an $A'$-module with term-wise scalar multiplication.
Both $A'M_0$ and $A_0M'$ are contained in $M_0$ (the former means that $M_0\subset M'$ is an $A'$-submodule).
Part 2 and first claim from part 3 imply that $\hat{M}=M'/M_0$ is an $A'$-module. The second claim from part 3 means $A_0\subset\operatorname{Ann}_{A'}\hat{M}$, so $\hat{M}$ is a module over $\hat{A}=A'/A_0$. The map $\hat{A}\times\hat{M}\to\hat{M}$ is given by $[f_n][x_n]=[f_nx_n]$.
Proof. Very similar to Lemma 3: the proof of part 1 is basically the same. The proofs of parts 2 and 3 are carried out almost entirely similar except that when proof of Lemma 3 “there is a neighborhood $N\subset A$ of zero with $NN\subset V$” in the module case we have to say instead “there are neighborhoods $N\subset A$, $\tilde{N}\subset M$ of zero with $N\tilde{N}\subset M$.” We leave the details to the reader. $\square$
Lemma 9. Let $A$ be a first-countable topological ring, and let $M$ be a sequentially complete topological $A$-module. There is a canonical $\hat{A}$-module structure on $M$ extending the $A$-module structure; namely, $[f_n]x=\lim_nf_nx$, for $[f_n]\in\hat{A}$ and $x\in M$. If $M$ is first-countable, this is the unique topological $\hat{A}$-module structure extending the $A$-module structure.
Proof. Uniqueness is easy: if we denote $\phi:A\to\hat{A}$ to the canonical morphism, then, for $[f_n]\in\hat{A}$ and $x\in M$, we have $[f_n]x=(\lim_n\phi(f_n))x=\lim_n \phi(f_n)x=\lim_n f_nx$. Hence, the topological $\hat{A}$-module structure is uniquely determined by the $A$-module structure. (Note we didn't use first-countability.)
We show existence. First we equip $M$ with an $A'$-module structure. Given $(f_n)\in A'$ and $x\in M$, we claim that $(f_nx)$ is Cauchy in $M$. Let $V\subset M$ be a neighborhood of zero. By Lemma 8, part 1, there is a neighborhood $U\subset A$ of zero with $Ux\subset V$. There is $k$ with $f_n-f_m\in U$ for all $k\geq n,m$, whence $f_nx-f_mx=(f_n-f_m)x\in Ux\subset V$ for all $n,m\geq k$. Thus, we can define $A'\times M\to M$ by $(f_n)x=\lim_nf_nx$. Next, we claim that $A_0\subset\operatorname{Ann}_{A'}M$. Let $s_n\in A$ be convergent to zero, and $x\in M$. Using the same $U$ and $V$ as before, we have $s_nx\in Ux\subset V$ for all $n\gg 0$, whence $0=\lim_ns_nx=(s_n)x$. Thus, $M$ becomes an $\hat{A}$-module. It is left to see that $\hat{\mu}:\hat{A}\times M\to M$ is continuous. Let $[f_n]\in\hat{A}$ and $x\in M$. Denote $y=[f_n]x=\lim_nf_nx$. We will show that $\hat{\mu}^{-1}(N)$ is a neighborhood of $([f_n],x)$ if $N\subset M$ is a neighborhood of $y$. By Remark 10 here, there is an open neighborhood $V\subset M$ of $y$ with $\overline{V}\subset N$. Here's when we use that $M$ is first-countable. Let $(W_n)$ be a countable neighborhood basis of $M$ at zero, and $(U_n)$ be a neighborhood basis of $A$ at zero. We claim that there is $K$ and $m$ such that $(f_n+U_N)\times(x+W_N)\subset\mu^{-1}(V)$ for all $n\geq m$ and $N\geq K$. If not, then, for each $K$ and $m$, there is $n_m$ and $N_K$, $s_{mK}\in U_{N_K},t_{mK}\in W_{N_K}$ such that $\lim_m n_m=+\infty=\lim_KN_K$ and \begin{align*}
V&\not\ni(f_{n_m}+s_{mK})(x+t_{mK})\\
&=f_{n_m}x+f_{n_m}t_{mK}+s_{mK}x+s_{mK}t_{mK}.
\end{align*}
Since $s_{mK},t_{mK}\to 0$ whenever $K\to+\infty$, by Lemma 8, part 3, we have that the three last summands goes to zero whenever $K\to+\infty$. Hence, taking limit in $K$ in the last expression gives $V\not\ni f_{n_m}x$, since $V$ is open. Similarly, $V\not\ni\lim_mf_{n_m}=y$, a contradiction. Pick then $K$ such that
$$
\tag{3}\label{eq}
(f_m+U_K)\times(x+W_K)\subset\mu^{-1}(V)
$$ for all $m\gg 0$. Pick $m$ such that \eqref{eq} holds and we have $[f_n]\in\widehat{f_m+U_K}$ (see this post, proof of Lemma 7, Claim i). We claim that $\widehat{f_m+U_K}\times(x+W_N)\subset\hat{\mu}^{-1}(W)$. Indeed, for $([g_n],z)\in\widehat{f_m+U_K}\times(x+W_N)$, since $g_n\in f_m+U_K$ for all $n\gg 0$, $g_nz\in V$ for all $n\gg 0$, whence $[g_n]y=\lim_ng_ny\in\overline{V}\subset W$, and we win. $\square$
Lemma 10. Let $M$ be a topological module over a topological ring $A$. Suppose $A,M$ are first-countable. Then $\hat{M}$, along with the structure map $\hat{A}\times\hat{M}\to\hat{M}$ coming from Lemma 8, is a topological $\hat{A}$-module.
Proof. Since $\hat{M}$ is first-countable, it suffices to show that the $\hat{A}$-module structure on $\hat{M}$ obtained from Lemma 9 coincides with the one obtained via Lemma 8 (hence, $\hat{M}$ becomes a topological $\hat{A}$-module). This is if and only if $[f_nx_n]=\lim_m[f_mx_n]$. Let $V\subset M$ be a neighborhood of zero. It suffices to show that $f_mx_n-f_nx_n\in V$ for all $m,n\gg 0$ (on that case, it follows $[f_mx_n]-[f_nx_n]\in\hat{W}$ for all $m\gg 0$ and for any neighborhood $W\subset M$ of zero, since $f_mx_n-f_nx_n+s_n\in V+V$ for all $m,n\gg 0$ and $s_n\in M$ convergent to zero, with $V\subset M$ a neighborhood of zero such that $V+V\subset W$). Let $N\subset M$ be a neighborhood of zero with $N+N\subset V$. Let $U\subset A$, $W\subset M$ be neighborhoods of zero with $UW\subset N$. Let $k$ be such that $x_n\in x_k+W$ for all $n\geq k$. By Lemma 8, part 1, and shrinking $U$ if necessary, we can assume $Ux_k\subset N$. Then, for all $n,m\gg 0$,
\begin{align*}
(f_m-f_n)x_n&\subset Ux_n\\
&\subset U(x_k+W)\\
&\subset Ux_k+UW\\
&\subset N+N\\
&\subset V.&\square
\end{align*}
Lemma 11. Let $A$ be a first-countable topological ring and let $M,N$ be sequentially complete topological $A$-modules. With the canonical $\hat{A}$-module structures on $M,N$ coming from Lemma 9, every morphism of topological $A$-modules $M\to N$ is $\hat{A}$-linear.
Proof. Let $\psi:M\to N$ be a morphism of topological $A$-modules.
\begin{align*}
\psi([f_n]x)&=\psi(\lim_n f_nx)\\
&=\lim_n\psi(f_nx)\\
&=\lim_nf_n\psi(x)\\
&=[f_n]\psi(x).&\square
\end{align*}
The continuous and additive map $\phi:M\to\hat{M}$ is $A$-linear, as it is easy to verify from the definitions. On the other hand, $\hat{M}$ is sequentially complete (this is a property of the underlying topological abelian group).
Lemma 12 (Universal property of the completion of a first-countable topological module over a first-countable topological ring). Let $M$ be a topological module over a topological ring $A$, and suppose $M$ and $A$ are first-countable. The map $M\to\hat{M}$ is universal among $A$-module morphisms of $M$ into sequentially complete topological $A$-modules. Moreover, if $M\to N$ is a morphism of topological $A$-modules, then the unique factorization $\hat{M}\to N$ is $\hat{A}$-linear, where $N$ has the $\hat{A}$-module structure from Lemma 9.
Proof. Let $\psi:M\to N$ be a morphism of topological $A$-modules. By Lemma 9 from here, we have a unique additive and continuous factorization $\tilde{\psi}:\hat{M}\to N$. To see that $\tilde{\psi}$ is $A$-linear, one uses the definition of $\tilde{\psi}$, $A$-linearity of $\psi$ and continuity of $A\times N\to N$. The $\hat{A}$-linearity of $\tilde{\psi}$ is Lemma 11. $\square$
To answer the rest of the OP's questions: Let $A$ be a ring, $\fra\subset A$ be an ideal and $M$ be an $A$-module. We can consider analogous containments to \eqref{contain} for $M$,
$$
i(\fra M)\subset
\fra^n\hat{M}\subset
\hat{\fra}^n\hat{M}\subset
\widehat{\fra^n}\hat{M}\subset
\widehat{\fra^nM},
$$
(where $i:M\to\hat{M}$ is the canonical map), plus a new containment (the last one), which becomes an equality if $M=A$. The two middle containments are in general strict for they are in general strict in the particular case $M=A$. In the case $\fra$ is finitely-generated, the three last containments become equalities, for $\fra^n$ is then also finitely generated, whence $\fra^n\hat{M}=\widehat{\fra^nM}$ by the argument given here.
Lemma 12 shows that the correct filtration to topologize $\hat{M}$ (if we want it to satisfy the universal property) is $(\widehat{\fra^nM})$.
To answer OP's question regarding the completion functor for the $\fra$-adic topology: any $A$-linear map $f:M\to N$ is continuous for the $\fra$-adic topologies on $M,N$ ($f(\fra^nM)=\fra^nf(M)\subset\fra^nN$). Hence, by Lemma 12 there is a unique morphism of topological $\hat{A}$-modules $\hat{f}:\hat{M}\to\hat{N}$ making the square
$$\require{AMScd}
\begin{CD}
M @>f>> N \\
@VVV @VVV \\
\hat{M} @>\smash{\hat{f}}>> \hat{N}
\end{CD}
$$
commute.
An alternative proof idea of continuity of $\hat{f}$ may be found here.
Addendum
If $A$ is a ring, $M$ is an $A$-module and $\fra$ is an ideal, there is a slight notational ambiguity when writing $\widehat{\fra^nM}$. Namely, this can be interpreted either as the $\fra$-adic completion of $\fra^nM$ (which is a topological $\hat{A}$-module) or as the subset of $\hat{M}$ given by taking $S=\fra^nM$ in \eqref{subset}. Each of these bears an $\hat{A}$-module structure; the former by Lemmas 7 and 10, and the latter since it is a submodule of $\hat{M}$ (and any submodule of a topological module is a topological module with the subspace topology).
From the following lemma it follows that these two $\hat{A}$-modules are the same.
Given a topological module $M$ over a topological ring $A$ (with $A$, $M$ first-countable), for the next lemma, we'll denote $M^\c$ to the completion of $M$ (we reserve $\widehat{(-)}$ for the subset given in \eqref{subset}). Recall that a quotient module of a topological $A$-module is a topological $A$-module with the quotient topology (apply this to get continuity of $+:M/N\times M/N\to M/N$ and use the same idea to get get continuity of $\cdot:A\times M/N\to M/N$).
Lemma 13. Let $M$ be a topological module over a topological ring $A$, and suppose that $A$ and $M$ are first-countable. Let $N\subset M$ be an open $A$-submodule. Then:
- the set $\hat{N}\subset M^\c$ is an $A^\c$-submodule.
- we have a short exact sequence of topological $A^\c$-modules
$$
0\to N^\c\to M^\c\to M/N\to 0,
$$
- the image of $N^\c\to M^\c$ equals $\hat{N}$, and
- the morphism $N^\c\to\hat{N}$ is an isomorphism of topological $A^\c$-modules.
Proof. We leave the first part as an exercise. The short exact sequence comes from application of Lemma 12. The rest is by Lemma 11 from here. $\square$
With this topology, $\hat{f}$ is continuous if $f$ is continuous.
– Cantlog Oct 04 '13 at 16:12