Are inversion and multiplicaton open?

Question

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.

Answer 1

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.

Answer 2

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.