Doubts in the proof that a subgroup $H$ of a Lie group $G$ that is a submanifold of $G$ is (topologically) closed in $G$

Question

I am trying to understand the details in the proof of

Let $G$ be a Lie group, $H $ a subgroup. then

$H$ is a submanifold of $G \implies H$ is closed in $G$

proof:

Since $H$ is a submanifold of $G$, it is in particular a submanifold at the identity $e$. Then by definition of submanifold there exists an open neighborhood $U$ of $e$ and local coordinates $x_i$ on $U$ s.t $U\cap H=\{g \in U:x_{m+1}(g)=...x_n(g)=0 \}$.

$U\cap H= U \cap \overline H \tag 1$.

Take any $g \in \overline H$. Let's prove that $g \in H$.

$U $ is an open neighborhod of $e$ $\implies gU$ is an open neighborhood of $g \implies gU \cap H \neq \emptyset$. Let $h \in gU \cap H $. Consider the left translation map $l_{g^{-1}}: G \rightarrow G: x \mapsto g^{-1}x$ Then

$g^{-1}h \in U \cap \overline H H = U \cap \overline H = U \cap H \tag 2$.

Then $g^{-1}h \in H \implies g \in H$

I am scratching my head about the following things:

1. Why is $(1)$: $U\cap H= U \cap \overline H$ true? I tried to prove it by double inclusion :$U\cap H \subseteq U \cap \overline H$ is obvious. But I couldn't prove the other one.

2. In $(2)$, why is $g^{-1}h \in U \cap \overline H H = U \cap \overline H $ ?

$g^{-1}h \in U \cap \overline H H $ seems to follow from $g^{-1} \in \overline H$ , but I don't see how that can be true.

All I could say is that since $h \in gU \cap H \implies g^{-1}h= l_{g^{-1}}(h) \in l_{g^{-1}}(gU \cap H)= l_{g^{-1}}(gU)\cap l_{g^{-1}}(H)= U \cap g^{-1}H $, where the next to last equality is because $l_{g^{-1}}$ is bijective

The equality $U \cap \overline H H = U \cap \overline H$ seems to follow from $ \overline H H = \overline H$, but why is that true?

Answer 1

$(1)$ is true in the following sense:
Recall that an open set in $U$ is defined to be $V \cap U$ where $V$ is open in $G$. Then because $U$ is open in $G$, $V \cap U$ is also open in $G$. Suppose $g \in U \cap \overline{H}$, then any open neighborhood $V$ of $g$ in $G$ intersects $H$, and $V \cap U$ is also an open neighborhood of $g$ in $G$, so it intersects $H$. Since $V \cap U$ gives every open neighborhood of $g$ in $U$, this means that any open neighborhood of $g$ in $U$ intersects $U \cap H$. Equivalently, $g$ belongs to the closure of $U \cap H$ in $U$.

Because $U \cap H$ is given by the common zeroes of a set of continuous functions in $U$, it is closed in $U$ (not necessarily closed in $G$!!). As a result, $g \in U \cap H$, which gives the inclusion $U \cap \overline{H} \subseteq U \cap H$.

$(2)$ is true in the following sense:
For any open neighborhood $V$ of $g^{-1}$, $V^{-1} = \{ v^{-1} \mid v \in V \}$ is open and contains $g$ by the continuity of group inverse on $G$. So $g \in \overline{H} \implies V^{-1} \cap H \ne \emptyset \implies V \cap H^{-1} = V \cap H \ne \emptyset$ for any open neighborhood $V$ of $g^{-1}$, which means $g^{-1} \in \overline{H}$.

Similarly, by the continuity of group product on $G$, $\forall g,h \in G$, the preimage of any open neighborhood $W$ of $gh$ under the product map $G \times G \rightarrow G$ is open. Since $(g,h)$ lies in the preimage of $W$, it has an open neighborhood $U \times V$ inside the preimage where $U,V$ are open in $G$, causing $UV=\{ uv \mid u \in U, v \in V \} \subseteq W$. Suppose $g,h \in \overline{H}$, then for any open neighborhood $W$ of $gh$, there is some open neighborhood $U \ni g$ and $V \ni h$ such that $UV \subseteq W$. But we have $U \cap H \ne \emptyset, V \cap H \ne \emptyset \implies UV \cap H \ne \emptyset$, so $W \cap H \ne \emptyset$ and then $gh \in \overline{H}$. We then conclude that $\overline{H}$ is a supergroup of $H$ and we have $\overline{H}H=\overline{H}$.