Which of the following iss the image of a continuous surjection from $GL_2 (\Bbb R)\ $?
$(a)$ The subspace $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}$ of $\Bbb R^2.$
$(b)$ The complement of the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}$ of $\Bbb R^2.$
$(b)$ is false because if $f : X \longrightarrow Y$ is a continuous surjection then the number of connected components of $Y$ is fewer than than the number of connected components of $X.$ In this case $GL_2 (\Bbb R)$ has two connected components namely $\{A \in GL_2(\Bbb R)\ |\ \det A \gt 0 \}$ and $\{A \in GL_2(\Bbb R)\ |\ \det A \lt 0 \}.$ But on the other hand the complement of the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}$ has three connected components. Hence there does not exist any continuous surjective map from $GL_2(\Bbb R)$ onto the complement of the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}.$
I think $(a)$ is true. For instance we could have taken the map $A \mapsto \left (\det A, \frac {1} {\det A} \right )$ which will give us a continuous surjection from $GL_2 (\Bbb R)$ onto the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}.$ But I have a query here. We know that $GL_2 (\Bbb R)$ is an open subset of $M_2 (\Bbb R) \cong \Bbb R^4.$ Now the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}$ is closed. So if there were a continuous surjective map from $GL_2(\Bbb R)$ onto the space $\left \{\left (x, \frac {1} {x} \right )\ |\ x \neq 0 \right \}$ then by the closedness of the space it follows that $GL_2 (\Bbb R)$ is also closed (Since inverse image of a closed set under a continuous map is closed). Therefore $GL_2(\Bbb R)$ is clopen in $\Bbb R^4$ and since $\Bbb R^4$ is connected it follows that $GL_2 (\Bbb R) = \Bbb R^4 \cong M_2 (\Bbb R),$ which is clearly not true. So I definitely did some mistake here. Where did I do mistake? Can anybody please help me finding it out?
Thanks for your time.
Best Answer
The preimage of your space is closed in the topological space $GL_2(\mathbb{R})$, which is endowed with the induced topology. It does not mean that it is closed in $M_2(\mathbb{R})$. It simply means that it is the intersection of a closed susbet of $M_2(\mathbb{R})$ with the open subset $GL_2(\mathbb{R})$.