A nbd of identity $V$ such that $V V^{-1 }$ is contained in another nbd of identity

Question

Let $G$ be a Lie group and $D$ be a discret subgroup of center and $U$ be nbd of identity such that $U \cap D$ contains only identity. I want to find a nbd $V$ of identity such that $V V^{-1} \subset U.$ I have considered the map $G \times G \to G : (a,b) \mapsto ab^{-1}.$ This map is smooth, in particulare continous. So the set $ \{ (a,b) : a b^{-1} \in U \} $ is open. I have tried to construct required $V$ from $W$ considering projections but I am not able to figure out.

Answer 1

Given $U$ a neighborhood of the identity $e$ in $G$. We know $(a,b) \mapsto ab^{-1}$ is continuous at $(e,e)$. And $ee^{-1}=e$. So there exist neighborhoods $V_1, V_2$ of $e$ so that: $$ \forall a \in V_1, \forall b \in V_2,\qquad ab^{-1} \in U. $$ Let $V = V_1 \cap V_2$. So $V$ is a neighborhood of $e$. Then of course $V \subseteq V_1$ and $V \subseteq V_2$.

We claim $VV^{-1} \subseteq U$. Indeed, let $c \in VV^{-1}$. There exists $a,b \in V$ with $c=ab^{-1}$. Then $a \in V_1$ and $b \in V_2$, so $ab^{-1} \in U$.

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Perhaps first prove the following, it you do not know it already. (The "neighborhood" definition of continuity.)
Let $X,Y$ be topological spaces. Let$f : X \to Y$. Let $x_0 \in X$. Then The following are equivalent:
$\bullet\;$ $f$ is continuous at $x_0$
$\bullet\;$ for every neighborhood $U$ of $f(x_0)$ in $Y$, there exists a neighorhood $V$ of $x_0$ in $X$ with $f(V) \subseteq U$.