Lemma 15.36.5 (Open mapping lemma). Let $u : N \to M$ be a continuous map of linearly topologized abelian groups. Assume that $N$ is complete, $M$ separated, and $N$ has a countable fundamental system of neighbourhoods of $0$. Then exactly one of the following holds

$u$ is open, or

for some open subgroup $N' \subset N$ the image $u(N')$ is nowhere dense in $M$.

**Proof.**
Let $N_ n$, $n \in \mathbf{N}$ be a fundamental system of neighbourhoods of $0$. We may assume that $N_{n + 1} \subset N_ n$. If (2) does not hold, then the closure $M_ n$ of $u(N_ n)$ is an open subgroup for $n = 1, 2, 3, \ldots $. Since $u$ is continuous, we see that $M_ n$, $n \in \mathbf{N}$ must be a fundamental system of open neighbourhoods of $0$ in $M$. Also, since $M_ n$ is the closure of $u(N_ n)$ we see that

\[ u(N_ n) + M_{n + 1} = M_ n \]

for all $n \geq 1$. Pick $x_1 \in M_1$. Then we can inductively choose $y_ i \in N_ i$ and $x_{i + 1} \in M_{i + 1}$ such that

\[ u(y_ i) + x_{i + 1} = x_ i \]

The element $y = y_1 + y_2 + y_3 + \ldots $ of $N$ exists because $N$ is complete. Whereupon we see that $x = u(y)$ because $M$ is separated. Thus $M_1 = u(N_1)$. In exactly the same way the reader shows that $M_ i = u(N_ i)$ for all $i \geq 2$ and we see that $u$ is open.
$\square$

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