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If $G$ is a topological group, are inversion $G \to G$ and multiplication $G\times G \to G$ open mappings? More concretely, I try to show that division of complex numbers $$\{(z,w) \in \mathbb{C}^2;\; w \neq 0\} \to \mathbb{C},\; (z,w) \mapsto \tfrac{z}{w}$$ is an open mapping. I want to use this to construct charts on $\mathbb{CP}^1 = (\mathbb{C}^2\setminus\{0\})/\mathbb{C}^{\times}$. I don't know where to begin.

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By the way there's another open question (no pun intended) I'm still very interested in, which is also related to this one. I hope it's okay to advertise a bit. –  k.stm Oct 3 '12 at 9:47

2 Answers 2

up vote 2 down vote accepted

Yes, open maps. The inversion $g\mapsto g^{-1}$ is in fact a homeomorphism, continuous by def. and its inverse is itself.

For the product, if $U,V\subseteq G$ open subsets, then the image of $(U,V)$ under multiplication is the complex product $$U\cdot V =\{u\cdot v\mid u\in U,v\in V\} =\bigcup_{u\in U}(u\cdot V)$$ is a union of open sets.

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Yes, thanks. And it's so easy, too! Wait a second, isn't $U \cdot V$ open, even if $V$ is not open? That argument should work for that case, too, am I right? Edit: NICE. –  k.stm Oct 3 '12 at 9:53
    
Exactly. For example $U\cdot\{1\} = U$. –  Berci Oct 3 '12 at 9:55

Take $O_1$ and $O_2$ two open subsets ot $G$. As the inversion $i$ is an homeorphism and an involution, $i(O_1)=i^{-1}(O_1)$ is open. Denote $m$ the multiplication map. Then $$m(0_1\times O_2)=\bigcup_{y\in O_2}O_1y$$ is open. We deal with the general case using the definition of product topology.

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Thanks, I'll accept Berci's answer for he was a bit faster. : - ) –  k.stm Oct 3 '12 at 9:56

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