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I'm having trouble in proving the following:

If $U$ is a neighborhood of $e$, there exists $V$ neighborhood of $e$ such that $V^{-1}V \subseteq U$

Can someone please verify if what I wrote makes sense? And if possible show the way to go further, because I'm stucked.

My attempt:

Let $G$ a topological group. (Definition: A topological group $G$ is a topological space with a group structure such that the applications $m: G \times G \to G$ and $r:G \to G$, defined by $m(x,y) = x.y$ and $r(x) = x^{-1}$ are continuous.)

$V^{-1}V = \{ x^{-1}y ; x,y \in V\}$

$U$ is a neighborhood of $e$, then is $V^{-1}V$ a neighborhood of $e$?

Since $r$ is homeomorphism and the right-translation $\delta_y:G \to G, \delta_y(x)=x.y$ is homeomorphism, the application $\delta_y \circ r :G \to G$ is homeomorphism. Let $x \in U$. Then $\delta_y \circ r(x) = x^{-1}y$ is in an open set, because of continuity. Why is it a neighborhood of $e$?

C.F.G
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1 Answers1

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The map $f\colon(x,y)\mapsto(x^{-1}y)$ is continuous at $(e,e)$.

Let $U$ be a neighborhood of $e$; then there exist neighborhoods $V_1$ and $V_2$ of $e$ such that $V_1\times V_2\subseteq f^{-1}(U)$.

Take $V=V_1\cap V_2$.

egreg
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