Let $G$ be a first-countable abelian topological group. Its completion $\hat{G}$, set-theoretically, is the family of Cauchy sequences in $G$ modulo the equivalence relation $(x_n)\sim (y_n)$ iff $x_n-y_n\to 0$. It comes equipped with a topology that turns it into a topological abelian group (see here) plus a natural map $\phi:G\to\hat{G}$ that is a continuous abelian group homomorphism (see here). Moreover, $\hat{G}$ is Hausdorff (see Remark 4.2 from last link) and every Cauchy sequence in $\hat{G}$ converges (see here).
Now I'm wondering about the universal property of $\phi:G\to\hat{G}$.
Let us call a topological abelian group “complete” if it is Hausdorff and every Cauchy sequence converges. (One can show that a first-countable topological abelian group is complete iff the canonical morphism into its completion is an isomorphism; see Proposition 7 here.)
I know $\phi$ is universal among continuous abelian group homomorphisms of $G$ into first-countable complete topological abelian groups: the proof I know is by leveraging Corollary 8, part 2, here.
My question is:
Is $\phi$ universal among topological abelian group morphisms from $G$ into all complete topological abelian groups?
I get stuck at proving continuity of the factorization: Suppose $\psi:G\to H$ is a morphism of abelian topological groups and that $H$ is complete. Given a Cauchy sequence $(x_n)$, by continuity, $(\psi(x_n))$ is Cauchy in $H$. Hence we can define \begin{align*} \hat{\psi}:\hat{G}&\to H\\ [x_n]&\mapsto\lim_n\psi(x_n) \end{align*} Note that $(\psi(x_n))$ has a unique limit on $H$ for $H$ is Hausdorff. It is not difficult to show that this definition for $\hat{\psi}$ is independent of the choice of representative in $[x_n]$. Note that by continuity of $+_H$, the function $\hat{\psi}$ is additive.
This is my approach to proving continuity of $\hat{\psi}$: recall that it suffices to show continuity at zero (see this). In turn, since $\hat{G}$ is first-countable, it suffices to show sequential continuity at zero. Suppose $x_n\in\hat{G}$ converges to zero, where $x_n=[x_{nm}]$. This means that for every neighborhoood $U\subset G$ of the identity, $x_n\in\hat{U}$ for all $n\gg 0$. That is, $x_{nm}+s_m\in U$ for all $n,m\gg 0$ and all $s_m\in G$ converging to zero. Now, given an open neighborhood $V\subset H$ of zero, we want to show that for all $n\gg 0$, $$ V\ni\hat{\psi}(x_n)=\lim_m\psi(x_{nm}). $$
But how one does do this? I don't know how to continue from here.
