0

I can find a continuous surjection from $\text{GL}(2,\mathbb R)$ to $\left\{\left(x,\frac 1x\right)\big|x\neq 0,x \in \mathbb R\right\}$,i.e. $\left(\det A,\frac{1}{\det A}\right)$. What about the existence of a continuous surjection to its complement in $\mathbb R^2$? My guess is that such a function will not exist but I cannot think of any conclusive argument.

Sumanta
  • 9,777
sxccalmat1100
  • 460
  • 1
    Just to be clear, you’re asking for a surjection $GL_2(\mathbb{R}) \rightarrow \mathbb{R}^2 \backslash {(x,y),, xy=1}$? If so, note that the range has three connected components while the domain only has two. – Aphelli Feb 28 '21 at 14:20
  • Does a continuous surjection always preserve the number of connected components? – sxccalmat1100 Feb 28 '21 at 14:22
  • Not necessarily. But $GL_2(\mathbb{R})$ has two connected components, so its continuous image is the reunion of two connected subspaces – so it can’t have three connected components. However, there is a surjective map $GL_2(\mathbb{R}) \rightarrow SL_2(\mathbb{R})$ and the range is connected. – Aphelli Feb 28 '21 at 15:44

0 Answers0