Universal property of completion of a field at an absolute value

Question

Prove that the completion $\hat k$ of a field $k$ at one of its absolute values $|\ |$ satisfies the following universal property: every topological field embedding of $k$ into a complete field $k'$ extends uniquely to an embedding of $\hat k$ into $k'$ that is an isomorphism if and only if $k$ is dense in $k'$.

I think $(\implies)$ is easy.
I want to prove the other way implication. I define $\hat i : \hat k \rightarrow k'$ as $ \hat i([(x_n)])=\text {lim} \ (i(x_n))$.
The inverse of the above map can be defined as follows:
For any element, there is a sequence converging to that element in $k$. Send this element to equivalence class of that sequence.
As this is an isomorphism of topological fields I need to show $\hat i$ is a homeomorphism.
I am stuck here.

I have made the following attempt. Please let me know imperfections or unwarranted assumptions etc.

Assume that topology on $k'$ comes from an absolute value and embedding is continuous.
$$\hat i ^{-1}(B_{k'}(x, \epsilon))=\{ [(x_n) ] \ | \ \text{lim} \ i(x_n) \in B_{k'}(x, \epsilon) \} $$ $i$ being continuous, $y \in i^{-1}B_{k'}(x, \epsilon), \ \exists \delta \ : |z-y| < \delta \implies z\in i^{-1}B_{k'}(x', \epsilon') \subset B_{k'}(x, \epsilon) $
So if $|[(x_n)]-[(y_n)]| < \delta $ then after some $N$, $|x_n-y_n| < \delta $ I am having difficulty proceeding hereafter.

Answer 1

Recall that the topology of a field with an absolute value $|\cdot|$ is defined by the metric (a.k.a. distance) $d$ such that $d(x,y)=|x-y|$. This metric is translation invariant, i.e., $d(x+z, y+z)=d(x,y)$. Instead of fields with absolute values, let us investigate commutative groups with translation-invariant metric (NCP). We have the following proposition.

Lemma: A continuous group homomorphism between two NCPs maps cauchy sequences to cauchy sequences.

Proof: Let $i:(g, d)\to(g', d')$ be a continuous group homomorphism. Let $(x_n)$ be a cauchy sequence in $g$.

Let $B_{g'}(0, \epsilon)$ be the open sphere in $g'$ of radius $\epsilon>0$ centered at $0$. Since $i^{-1}\left(B_{g'}(0, \epsilon)\right)$ is open in $g$, it contains $B_{g}(0, \beta(\epsilon))$ for some $\beta(\epsilon)\gt0$. We can assume $\beta$ is a non-increasing function since, if not, we can reset $\beta(\epsilon)=\min_{m\in\mathbb N, m\ge \frac1\epsilon}\beta(\frac 1m)$ for all $\epsilon$.

Given $\epsilon>0$, since $(x_n)$ is a cauchy sequence, we can find $n_0=n_0(\epsilon)$ such that for all $n_1, n_2\ge n_0$, $d(x_{n_1}, x_{n_2})\lt\beta(\epsilon)$. Since $d(x_{n_1}, x_{n_2})=d(x_{n_1}-x_{n_2},0)$, $\ (x_{n_1}-x_{n_2})\in B_g(0,\beta(\epsilon))$.

$$d'\left(i(x_{n_1}), i(x_{n_2})\right)= d'\left(i(x_{n_1})-i(x_{n_2}),0\right)= d'\left(i(x_{n_1}-x_{n_2}),0\right)\lt \epsilon, $$ which means $(i(x_n))$ is a cauchy sequence in $g'.$ $\blacksquare$.

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Let us prove the following proposition.

Let $(g,d)$ be an NCP. Let $(\hat g, d)$ be the completion of $g$. Let $(g',d)$ be another NCP. (We abuse $d$ to mean the metric on difference spaces. Its meaning should be clear depending on the context.) $g'$ is complete. Let $i:g\to g'$ be a continuous group homomorphism. We have the following.
(gI). There is a unique way to extend $i$ to a continuous map from $\hat g$ to $g'$.
(gII). If $i$ is, in fact, a topological group embedding and $i(g)$ is dense in $g'$, that extension of $i$ must be a topological group isomorphism.

Proof for (gI). Let $[(x_n)]$ be an element in in $\hat g$, i.e., $x_n$ is a cauchy sequence in $g$. The lemma says $(i(x_n))$ is a cauchy sequence in $g'$. Since $g'$ is complete, $\lim\, (i(x_n))$ is well-defined.

Suppose $\hat i : \hat g \to g'$ is a continuous map that extends $i$. Note that $x_1, x_2, \cdots$ converges to $[(x_n)]$ in $\hat g$. Since a continuous map between metric space must map convergent sequences to convergent sequences, $ \hat i([(x_n)])$ must be $\lim\, (i(x_n))$. On the other hand, we can verify that the map from $\hat g\to g'$ defined by $[(x_n)]\to\lim\, (i(x_n))$ is well-defined and continuous. So, we have proved there is a unique continuous map that extends $i$ to $\hat g$, which is $\hat i$ that is defined by $ \hat i([(x_n)])=\lim\, (i(x_n))$. $\blacksquare$.

Proof for (gII). In the proof for (gI), we have shown that the extension is $\hat i$ defined by $\hat i([(x_n)])=\lim\, (i(x_n))$ for $[(x_n)]\in \hat g$.

It is routine to verify that $\hat i$ is a group morphism.

Suppose $[(x_n)]$ and $[(y_n)]$ are two points in $\hat g$ such that $\hat i([(x_n)])=\hat i[(y_n)])\in g'$. Then $\hat i\left([(x_n-y_n)]\right)=\hat i\left([(x_n)]-[(y_n)]\right)=0$, which means $\lim\ (x_n-y_n) = 0$, i.e., $[(x_n)]$ and $[(y_n)]$ are the same points. So $\hat i$ is injective. $\hat i: \hat g\to \hat i(\hat g)$ is a group isomorphism.

We have shown that $\hat i$ is bijective from $\hat g$ to $\hat i(\hat g)$. Since $\hat i$ is continuous, to show $\hat i: \hat g\to \hat i(\hat g)$ is a homeomorphism, we just need to show $\left(\hat i\right)^{-1}$ is continuous.

Suppose $(i(\hat{y_n}))$ is a cauchy sequence in $i(\hat g)$, where $\hat{y_n}=[(y_{n,1}, y_{n,2}, \cdots)]\in \hat g$ for some $y_{n,m}\in g$. For each $n$, select $\beta(n)>0$ such that $d(y_{n,\beta(n)}, \hat{y_n})\lt \frac 1n$. Verify that $(\hat i(y_{n,\beta(n)}))$ is cauchy sequence in $i(\hat g)$. That means, $(i(y_{n,\beta(n)}))$ is a cauchy sequence in $i(g)$,

Note that the assumption $i: g\to i(g)$ is a homeomorphism implies that $i^{-1}:i(g)\to g$ maps cauchy sequences to cauchy sequences. So, $(y_{n,\beta(n)})=(i^{-1}(i(y_{n,\beta(n)}))$ is a cauchy sequence in $g$. Let $\hat y=[(y_{n,\beta(n)})]\in \hat g$. We can verify that that $\lim \hat{y_n} = \hat{y}$, i.e., $(\hat y_n)$ is a cauchy sequence. Since $(\hat y_n)=\left(i^{-1}(i(\hat y_n))\right)$, we have shown $\left(\hat i\right)^{-1}$ maps all cauchy sequences to cauchy sequences. Since both $\hat g$ and $\hat i(\hat g)$ are metric spaces, $\left(\hat i\right)^{-1}$ is continuous.

So $\hat i: \hat g\to \hat i(\hat g)$ is a homeomorphism and, hence, a topological group isomorphism.

Since $\hat i$ is a topological group isomorphism between metric spaces, the fact that $\hat g$ is complete implies $\hat i(\hat g)$ is complete. Since $\hat i(\hat g)\supseteq i(g)$ is dense in $g'$, $\hat i(\hat g)=g'$. So $\hat i$ is a topological group isomorphism. $\blacksquare$.

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The proposition (gI) and (gII) can be applied to the additive groups of (complete) fields with absolute values. It becomes easy and routine to prove the following clearer and stronger version of the direction "$\impliedby$" in the question.

Let $k$ be a field with absolute value $|\cdot|$ and $k'$ be a complete field with absolute value $|\cdot|'$. space. Let $i:k\to k'$ be a field embedding that is continuous. Then we have the following.
(I). There is a unique way to extend $i$ to a continuous map from $\hat k$ to $k'$.
(II). If $i$ is, in fact, a topological field embedding and $i(k)$ is dense in $k'$, then that extension of $i$ must be a topological field isomorphism.

Answer 2

$\def\c{\mathrm{c}}$The universal property is Lemma 12 below and it works for good first-countable Hausdorff topological fields. For the meaning of “good,” see Definition 6 (cf. also Corollary 11). In particular, fields with an absolute value are good (Lemma 7).

A field with a topology that turns it into a topological ring either has the trivial topology or is Hausdorff, see this comment. For this reason, in the statements of the upcoming results we will always require all such rings to be Hausdorff.

Exercise 1. Let $G$ be a topological abelian group. Prove that a Cauchy sequence in $G$ converges if and only if it has a converging subsequence. (Hint: Imitate the proof seen in Analysis I for $G=\mathbb{R}$.)

Given an abelian group $G$ (written additively) with a topology (that may or may not be compatible with the group structure on $G$), recall that we say that a sequence $(x_n)$ in $G$ is Cauchy if for each neighborhood $U\subset G$ of zero there is $n_0$ such that $x_n-x_m\in U$ for all $n,m\geq n_0$.

For a ring $A$, we denote $A^*$ to the group of units.

Lemma 2. Let $A$ be a first-countable Hausdorff topological ring. Let $(x_n)$ be a sequence in $A$. Consider the following statements:

1. There is $n_0\geq 0$ such that $x_n\in A^*$ for all $n\geq n_0$ and for any such $n_0$ we have that $(x_n)_{n\geq n_0}$ is Cauchy in the multiplicative group $(A^*,\cdot)$.
2. $(x_n)$ is Cauchy in the topological additive group $(A,+)$ and not convergent to zero.

Then 1$\Rightarrow$2.

Proof. Let $n_0$ be like in the statement. Let $U\subset A$ be a neighborhood of zero. Since the function $A\times A\to A$, $(a,b)\mapsto a-b$ is continuous, there is a neighborhood $V\subset A$ of $1$ such that $V-V\subset U$. By hypothesis and taking a bigger $n_0$ if necessary, we have $x_nx_{n_0}^{-1}\in V$ for all $n\geq n_0$. Thus, for all $n,m\geq n_0$, we have $x_nx_{n_0}^{-1}-x_mx_{n_0}^{-1}\in V-V\subset U$. That is, $(x_nx_{n_0}^{-1})_n$ is Cauchy in $(A,+)$. Since multiplication by a unit is an automorphism of $(A,+)$ as a topological abelian group, we get that $(x_n)$ is Cauchy in $(A,+)$. Lastly, we see that $x_n\not\to 0$. Since $A$ is Hausdorff and $0\neq 1$, there is a closed neighborhood $C$ of $1$ not containing $0$. There is $n_0$ such that $x_nx_m^{-1}\in C$ for all $n,m\geq n_0$. In particular, $ x_n\in Cx_{n_0}$ for all $n\geq n_0$. Hence, $x_n$ cannot converge to zero, because otherwise $0=\lim x_n\in\overline{Cx_{n_0}}=\overline{C}x_{n_0}=Cx_{n_0}\not\ni 0$, where we have used that multiplication by a unit in $A$ is an automorphism of topological abelian groups. $\square$

We say that an abelian group $G$ with a topology (compatible or not with the group structure) is sequentially complete if it is Hausdorff and every Cauchy sequence in $G$ converges.

Corollary 3. Let $K$ be a first-countable Hausdorff topological ring which is a field such that $(K,+)$ is sequentially complete. Then $(K^*,\cdot)$ is sequentially complete.

Proof. Let $(x_n)$ be Cauchy in $(K,\cdot)$. By Lemma 2, it is Cauchy in $(K,+)$ and not convergent to zero. Hence, there is $x\in K$ such that $x_n\to x\neq 0$. Since $K$ is a field, $x\in K^*$; i.e., $(x_n)$ converges in $K^*$. $\square$

Lemma 4. Let $A$ be a topological ring, let $W\subset A$ be a neighborhood of zero and suppose that $(y_n)$ is a Cauchy sequence in $A$. Then there is a neighborhood $U\subset A$ of zero and $n_0$ such that $Uy_n\subset W$ for all $n\geq n_0$.

Proof. Let $V\subset A$ be a neighborhood of zero such that $VV+V\subset W$ (which exists by continuity of $(a,b,c)\mapsto ab+c$, with $a,b,c\in A$). Let $n_0$ be such that $y_n-y_m\in V$ for all $n,m\geq n_0$. By Lemma 3 from here, there is a neighborhood $U\subset A$ of zero with $Uy_{n_0}\subset V$. We can assume $U\subset V$ (replace $U$ by $U\cap V$). Thus, for $n\geq n_0$, \begin{align*} Uy_n&=U(y_n-y_{n_0}+y_{n_0})\\ &\subset U(y_n-y_{n_0})+Uy_{n_0}\\ &\subset VV+V\\ &\subset W.&\square \end{align*}

Lemma 5. Let $K$ be a first-countable Hausdorff topological field. The following are equivalent:

• For all Cauchy sequences $(x_n)$ in $(K,+)$ not converging to zero we have that $(x_n^{-1})_{n\geq n_0}$ is a Cauchy sequence in $(K,+)$, for any natural number $n_0$ such that $x_n\neq 0$ for all $n\geq n_0$ (such an $n_0$ always exists by Exercise 1).
• For all sequences $(x_n)$ in $K$ and taking $A=K$, one has 2$\Rightarrow$1 in the properties of the statement of Lemma 2 (hence, they are equivalent for $K$).

Proof. We see first that the second point implies the first one. Let $(x_n)$ be Cauchy in $(K,+)$ and not converging to zero. Then $(x_n)$ satisfies property 1 from the statement of Lemma 2. Pick any $n_0$ such that $x_n\neq 0$ for all $n\geq n_0$ (which exists by Exercise 1). Since $(K,\cdot)$ is a topological abelian group, $(x_n^{-1})_{n\geq n_0}$ is Cauchy in $(K,\cdot)$. By Lemma 2, $(x_n^{-1})_{n\geq n_0}$ is Cauchy in $(K,+)$.

Conversely, suppose the first point holds. Let $(x_n)$ be a Cauchy sequence in $(K,+)$ not converging to zero. By Exercise 1, there are finitely many $n$ such that $x_n=0$. Let $n_0$ be some natural number such that $x_n\neq 0$ for all $n\geq n_0$. We may suppose $n_0=1$ and show that $(x_n)$ is Cauchy in $(K,\cdot)$. We have that $(x^{-1}_n)$ is Cauchy in $(K,+)$ by hypothesis. Let $W\subset K$ be a neighborhood of zero. Let $V\subset K$ be a neighborhood of zero such that $V+V+V\subset W$. By Lemma 4, there is $n_0$ and a neighborhood $U\subset K$ of zero such that for all $n\geq n_0$, $V$ contains both $x_nU$ and $x_n^{-1}U$. Shrinking $U$ if necessary, we can assume $UU\subset V$. Replacing $n_0$ by a bigger number if necessary, we may assume that for all $n\geq n_0$, $U$ contains $x_n-x_m$ and $x_n^{-1}-x_m^{-1}$. Thus, for all $n,m\geq n_0$, \begin{align*} x_nx_m^{-1}&\in(x_{n_0}+U)(x_{n_0}^{-1}+U)\\ &\subset 1+x_{n_0}U+Ux_{n_0}^{-1}+UU\\ &\subset 1+V+V+V\\ &\subset 1+W. \end{align*} Hence $(x_n)$ is Cauchy in $(K,\cdot)$. $\square$

Definition 6. A first-countable Hausdorff topological field satisfying the equivalent conditions of Lemma 5 is said to be good.

(I must say: I made this terminology up.)

A first-countable Hausdorff topological field is good if and only if its completion is a topological field (Corollary 11). The “goodness” condition can be stated even when $K$ is not first-countable, but for that one requires the language of Cauchy filters; the general goodness condition is also equivalent to the completion (in the non-first-countable case) being a topological field, see Bourbaki, General Topology, Ch. III, §6, no. 8, Proposition 7.

There are first-countable Hausdorff topological fields that aren't good, see ibid., Exercise 6.26. However this does not happen in the cases algebraists mostly care about:

Lemma 7. Let $K$ be a field. If $K$ satisfies any of the following:

1. $K$ is a first-countable Hausdorff topological field and $(K,+)$ is sequentially complete.
2. $K$ is equipped with an absolute value $|\cdot|:K\to\mathbb{R}_{\geq 0}$.
3. $K$ is equipped with a valuation $v:K\to G\cup\{+\infty\}$ (where $G$ is a totally ordered abelian group) and the topology on $K$ induced by $v$ is first-countable (e.g., if $v$ is discrete, i.e., $\Gamma\cong\mathbb{Z}$).

Then $K$ is a good first-countable Hausdorff topological field.

On the second case, because of the scope of OP's question, we should say that the absolute value on $K$ uniquely extends to an absolute value on its completion and its induced topology coincides with that of the completion, see Bourbaki, General Topology, Ch. IX, §3, no. 2, Proposition 6. On the third case, even when the topology on $K$ induced by $v$ is not first-countable, $K$ is still “good” is the more general sense of Cauchy filters, see Bourbaki, Commutative Algebra, Ch. VI, §5, no. 3, Proposition 5 (a) (in ibid., (b)-(f) states that $v$ uniquely extends to a valuation $\hat{v}$ on the completion of $K$ and that $\hat{v}$ satisfies all the nice properties one would desire it to have).

Proof. 1. If $(x_n)$ is Cauchy in $(K,+)$ and not converging to zero, then there is $x\in K$ such that $x_n\to x\neq 0$. By Exercise 1, there is $n_0$ such that $x_n\neq 0$ for all $n\geq n_0$. Since $K$ is a topological field, $x_n^{-1}\to x^{-1}$. Hence, $(x_n^{-1})_{n\geq n_0}$ is convergent; thus, Cauchy in $(K,+)$.

2. An absolute value induces a metric on $K$ and hence a topology that is is first-countable Hausdorff. Let $(x_n)$ be a Cauchy sequence in $(K,+)$ not converging to zero. Replacing $(x_n)$ by some tail if necessary, we may assume $x_n\neq 0$ for all $n$. Again, replacing $(x_n)$ by some tail if necessary, we may assume that there is $\alpha\in\mathbb{R}$ with $\alpha>0$ such that $|x_n|>\alpha$ for all $n$. Let $\varepsilon>0$. There is $n_0$ such that $|x_n-x_m|<\varepsilon\alpha^2$ for all $n,m\geq n_0$. Since $x^{-1}-y^{-1}=-x^{-1}(x-y)y^{-1}$ for $x,y\in K^*$, we deduce $$ |x^{-1}_n-x_m^{-1}|=\frac{|x_n-y_n|}{|x_n||x_m|}\leq\varepsilon $$ for all $n,m\geq n_0$. Thus $(x_n^{-1})$ is Cauchy in $(K,+)$.

3. We recall that the topology that $v$ induces on $K$ has as an open neighborhood basis at zero the sets $V_\alpha=v^{-1}((\alpha,+\infty])$, where $\alpha\in G$. This topology turns $K$ into a Hausdorff topological field (the proof of this fact may be read in Bourbaki, Commutative Algebra, Ch. VI, §5, no. 1, Proposition 5). Since we are assuming that this topology is first-countable, it is left to show that $K$ is good: let $(x_n)$ be a Cauchy sequence in $(K,+)$ not convergent to zero. Then there is $\beta\in G$ and $n_0$ such that $v(x_n)<\beta$ for all $n\geq n_0$ (in particular, $x_n\neq 0$ for $n\geq n_0$). Let $\alpha\in G$. Since $(x_n)$ is Cauchy in $(K,+)$, there is $n_1$ such that $v(x_n-x_m)>\max(\alpha+2\beta,\beta)$ for $n,m\geq n_1$. By properties of valuations, this in turn implies $v(x_n^{-1}-x_m^{-1})>\alpha$ for $n\geq \max(n_0,n_1)=N$ (ibid., no. 1, Lemma 1), whence $(x_n^{-1})_{n\geq N}$ is Cauchy in $(K,+)$. $\square$

Lemma 8. Let $K$ be a first-countable Hausdorff topological field. Suppose that $\hat{K}$ is a topological field. Then $K$ is good.

Proof. Let $(x_n)$ be a Cauchy sequence in $K$ not convergent to zero. Then $[x_n]\in\hat{K}$ is not zero. We can assume $x_n\neq 0$ for all $n\geq 0$. Since $\phi(x_n)\to[x_n]$ (see Remark 4.2 here) and $\hat{K}$ is a topological field, $[x_n]$ is invertible and $\phi(x_n^{-1})=\phi(x_n)^{-1}\to[x_n]^{-1}$. Hence, $(\phi(x_n^{-1}))$ is convergent in $\hat{K}$; thus, Cauchy. This implies that $(x_n^{-1})$ is Cauchy in $K$ (see proof of second lemma here). $\square$

Lemma 9. Let $K$ be a good first-countable Hausdorff topological field. The ring $\hat{K}$ is a field. Thus, $(\hat{K}^*,\cdot)$ is sequentially complete.

Thus, in the ring $K'$ of Cauchy sequences in $K$, the ideal $K_0$ of converging to zero sequences in $K$ is maximal, for $\hat{K}=K'/K_0$.

Proof. The second assertion follows from the first one plus Corollary 3. We have that $0$ and $1$ are distinct elements in $\hat{K}$, so $\hat{K}$ is not trivial. Let $[x_n]\in\hat{K}\setminus\{0\}$. We will show that $[x_n]$ is invertible (whence $\hat{K}$ is a field). Since there is a finite amount of $x_n$'s that are zero, we may replace $(x_n)$ by some tail where every element is non-zero (see Exercise 6 here). Since $(x_n)$ is Cauchy in $(K,+)$ and not convergent to zero, it is Cauchy in $(K,\cdot)$ as $K$ is good. Since the latter is an abelian topological group, $(x_n^{-1})$ is Cauchy in $(K,\cdot)$. By Lemma 4, $(x_n^{-1})$ is Cauchy in $(K,+)$. Hence $[x_n^{-1}]\in\hat{K}$ and we have $[x_n][x_n^{-1}]=[x_nx_n^{-1}]=[1]=1$, whence $[x_n]$ is invertible in $\hat{K}$ and $[x_n]^{-1}=[x_n^{-1}]$. $\square$

For a topological abelian group $G$—written additively or multiplicatively—(resp., for a topological ring $A$) with first-countable topology, we temporarily denote $G^\c$ (resp., $A^\c$) to its group (resp., ring) completion.

Let $K$ be a good first-countable Hausdorff topological field. By Lemma 5 from here, we know that $K^\c$ is a topological ring. In particular, $(K^\c)^*$ is a group with a topology (the one inherited from $K^\c$) such that the multiplication map is continuous (although we don't know yet whether $x\mapsto x^{-1}$ is continuous in $(K^\c)^*$). By Lemma 9, we get a canonical continuous group homomorphism $$ \tag{1}\label{can} (K^*)^\c\to(K^\c)^* $$ obtained from Lemma 9 here. Indeed, after close inspection of the proof of the linked lemma, one sees that only continuity of the product in $H$ is required to construct the unique continuous group homomorphism $\hat{G}\to H$. In other words: the morphism $G\to\hat{G}$ is actually universal among continuous homomorphisms of $G$ into groups $H$ with a topology for which the multiplication map is continuous (but not necessarily $h\in H\mapsto -h\in H$).

The map \eqref{can} sends the class of a Cauchy sequence $(x_n)$ in $(K^*,\cdot)$ to $\lim_n\phi(x_n)$ on the RHS, where $\phi:K\to K^\c$ is the canonical morphism. By Lemma 2, $(x_n)$ is Cauchy in $(K,+)$, so that $\lim_n\phi(x_n)=[x_n]$ (Remark 4.2 from here). In other words, the map \eqref{can} sends the class of a Cauchy sequence in the LHS to the class of the same sequence on the RHS.

Lemma 10. Let $K$ be a good first-countable Hausdorff topological field. The map \eqref{can} is an isomorphism of abelian groups that is also a homeomorphism.

Hence, since the LHS of \eqref{can} is a topological abelian group, so is its RHS (i.e., $x\mapsto x^{-1}$ is continuous in $(K^\c)^*$). Thus, the completion of a good first-countable Hausdorff topological field is a topological field. Along with Lemma 8, this shows:

Corollary 11. The completion of a first-countable Hausdorff topological field $K$ is a topological field if and only if $K$ is good.

Proof. The morphism is injective: if the class of a Cauchy sequence $(z_n)$ on the LHS becomes the identity on the RHS, then $z_n\to 1$ in $K$, so $z_n\to 1$ in $K^*$. Thus it was the identity on the LHS. It is surjective: Let $[z_n]\in (K^\c)^*$. By Exercise 6 from here, we may suppose $z_n\neq 0$ for all $n$. Since $(z_n)$ is Cauchy in $(K,+)$ and does not converge to zero, it is Cauchy in $(K^*,\cdot)$ as $K$ is good. Thus, the class of $(z_n)$ on the LHS is an inverse image.

We know that the map is continuous. It is left to show that it is open. For each open set $W\subset K^*$, denote $$ \tilde{W}=\{x\in (K^*)^\c\mid \forall (z_n)\in x,\text{ it holds }z_n\in W,\forall n\gg 0\}. $$ It suffices to show that \eqref{can} maps $\tilde{W}$ to $\hat{W}$, where the latter set is defined in equation $(1)$ from here.

Let $[z_n]\in\hat{W}$ and note that $[z_n]\neq [0]$ for $0\not\in W$. Since $K$ is good, $(z_n)$ is Cauchy in $(K^*,\cdot)$. Let $v_n$ be a sequence in $K$ converging to one. Then $\hat{W}\ni[z_n]\cdot 1=[z_n][v_n]=[z_nv_n]$. In particular, $z_nv_n\in W$ for all $n\gg 0$. This means that the class of $(z_n)$ in $(K^*)^\c$ is in $\tilde{W}$.

Conversely, let $(z_n)$ be a Cauchy sequence in $(K^*,\cdot)$ whose class in $(K^*)^\c$ lies in $\tilde{W}$. In particular, $(z_n)$ is Cauchy in $(K,+)$, by Lemma 2. Let $(U_n)$ and $(V_n)$ be countable neighborhood bases of $K$ respectively at zero and at one. We claim that there is $N$ such that $z_nV_N+U_N\subset W$ for all $n\geq N$. Suppose not. Then, for each $N$, there are $n_N\geq N$, $v_N\in V_N$ and $s_N\in U_N$ such that $W\not\ni z_{n_N}v_N+s_N=z_{nN}(v_N+z_{n_N}^{-1}s_N)$. Since $v_N+z_{n_N}^{-1}s_N\to 1$ ($(z_n^{-1})$ is Cauchy in $(K,\cdot)$ and hence in $(K,+)$; now use Lemma 3.3 from here), this implies the class of $(z_{n_N})$ in $(K^*)^\c$ (same as that of $(z_n)$) is not in $\tilde{W}$. Let $N$ be then such an integer and let $s_n\in K$ be a sequence converging to zero. Then, for all $n\gg 0$, \begin{align*} z_n+s_n &=z_n+s_n+z_ns_n-z_ns_n\\ &=z_n(1+s_n)+(1-z_n)s_n\\ &\in z_nV_N+U_N\\ &\subset W. \end{align*} That is, $[z_n]\in\hat{W}$. $\square$

Lemma 12 (Universal property of the completion of a good first-countable Hausdorff topological field). Let $K$ be a good first-countable Hausdorff topological field. The morphism $K\to\hat{K}$ is universal among morphisms of $K$ into sequentially complete topological fields.

Proof. By Corollary 11, $\hat{K}$ is a topological field. Now apply Lemma 6 from here. $\square$

We address the second part of OP's question: suppose $(K,|\cdot|_K)$ is a field with an absolute value that is isometrically embedded on a field $(L,|\cdot|_L)$, and assume that the latter is complete. Write $\psi:K\to L$ for the inclusion. We get a canonical continuous factorization $\tilde{\psi}:\hat{K}\to L$. Suppose $\overline{\psi(K)}=L$. We show that $\tilde{\psi}$ is an isomorphism. First we see surjectivity: Let $\ell\in L$. There is a sequence $x_n\in K$ with $\psi(x_n)\to\ell$. Hence, $(\psi(x_n))$ is Cauchy, so $(x_n)$ is also Cauchy. We have $\tilde{\psi}([x_n])=\lim \psi(x_n)=\ell$. Next, it suffices to show that $\hat{\psi}$ is an isometry. Let $[x_n]\in\hat{K}$. Then $|\tilde{\psi}([x_n])|_L =|\lim\psi(x_n)|_L =\lim|\psi(x_n)|_L =\lim|x_n|_K =|[x_n]|_{\hat{K}}$.