problem to show that a map is continuous and open

Question

Consider the family $\hat G$ of all minimal Cauchy filters on a topological group $G$. For every $V\in\mathcal{N}_e$ put $$[V]=\{\mathscr{F}:\mathscr{F}\in\hat G,\;V\in \mathscr{F}\}\;.$$ Denote $\mathcal{M}=\{[V]:V\in\mathcal{N}_e\}$. $\mathcal{M}$ satisfies nbhd conditions on a topological groups (Pontryagin's conditions) and therefore there exists a group topology on $\hat G$. Now define the mapping $\alpha:G\to \hat G$ where for every $x\in G$, $\alpha(x)=\mathcal{N}_x$. I can show that for every open subset $V$ of $e$, $\alpha(V)=[V]\cap \alpha(G)$. but i have a problem to show that $\alpha:G\to \alpha(G)$ is open and continuous. thanks for any advice.

Answer 1

Let's show that for an open neighbourhood $V$ of $e$, the corresponding neighbourhood $[V]$ is also open. Let $\mathscr{F}\in [V]$ arbitrary, so $V \in \mathscr{F}$.

The filter basis

$$\mathscr{F}_0 = \{ F \cdot W : F \in \mathscr{F}, W \in \mathcal{N}_e \}$$

generates a Cauchy filter that is coarser than $\mathscr{F}$, and since $\mathscr{F}$ is a minimal Cauchy filter, it generates $\mathscr{F}$. Hence there is an $F \in \mathscr{F}$, and a $W\in\mathcal{N}_e$ with $F\cdot W \subset V$.

But that means $\mathscr{F}[W] \subset [V]$, so $[V]$ is a neighbourhood of $\mathscr{F}$. Since $\mathscr{F}$ was arbitrary, $[V]$ is open.

Then we see that $\alpha(V) = [V] \cap \alpha(G)$ is open in $\alpha(G)$, hence $\alpha$ is open at $e$, also $\alpha^{-1}([V]) = V$, so $\alpha$ is continuous at $e$. Since $\alpha$ is a homomorphism, $\alpha$ is continuous and open (the latter as a mapping to $\alpha(G)$, not in general to $\widehat{G}$, of course).