Let $ (G, \cdot, \tau) $ and $ (H, \cdot', \tau') $ be Hausdorff topological groups, $ f: (G, \cdot, \tau) \rightarrow (H, \cdot ', \tau') $ is a continuous homomorphism. If $ U \in \mathcal{N}_{e_{G}} $ is such that $ \overline{U} $ is compact and $ f(U) \in \mathcal{N}_{e_H} $, prove that $ f $ is open.
I have managed to conclude, but I do not make use of some of the hypotheses and this raises doubts in my test, I would appreciate the comments.
Proof: Let $ V \subseteq G $ open, we will prove that $ f (V) \subseteq H $ is open, let $ b \in f (V) $ then $ f^{-1}(b) \subset V $, so $ V = f^{ -1} (b) U $ with $ U \in \mathcal{N}_{e_{H}} $, then there exists $ U '$ open such that $ f^{ -1} (b) U'\subseteq f^{- 1} (b) U \subset V $ and we obtain that: $ f (f^{- 1} (b) U') \subseteq f (f^{-1}(b) U) \subset f (V) $, $ bf (U') \subseteq bf(U) \subset f(V)$
Let us denote by $ V '= b f (U'), $ $ b \in V '= (V') ^ {\circ} $, like this: $ \forall b \in V \rightarrow \ \exists V' \in \mathcal{U}^{\circ}(b) \ : \ b \in V' \subset f(V)$ then $ f (V) $ is open in $ H $, since $ V $ is arbitrary we have that $ f $ is open.