Continuous Images of $GL_n(\mathbb{R})$ Which of the following is a continuous image of $GL_n(\mathbb{R})$?

*

*The real line $\mathbb{R}$


*The subspace $\{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ of $\mathbb{R}^2$


*$\mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$


*$\{(x_1,x_2,...,x_n):x_1^2+x_2^2+...+x_n^2 \le 1\}$ of $\mathbb{R}^n$
How do I proceed?
$GL_n(\mathbb{R})$ is not compact, not connected, so I cannot rule out some options using topological properties.
 A: Currently do not have enough reputation to comment, apologies.
How about just the trace function? One can always construct a matrix $M \in GL(n, \mathbb{R})$ such that $\mathrm{Tr}(M)=x, \forall x \in \mathbb{R}$
Trivial example, $M_{11} = x, M_{ij} = 1, i,j \neq (1,1)$.
A: The determinant $\det : GL_n(\mathbb R) \to \mathbb R$ is continuous and has image $\mathbb R \setminus \{0\} = (-\infty,0) \cup (0,\infty)$. But $(0,\infty)$ is homeomorphic to $\mathbb R$; for example $\ln : (0,\infty) \to \mathbb R$ is a homeomorphism.

*

*Yes. Let $f = \phi \circ \det$, where $\phi \mid_{(0,\infty)}$ is a homeomorphism and $\phi \mid_{(-\infty,0)}$ is any map. Lubin's comment takes $f = \ln \circ \lvert det \rvert$ which is a special case of this construction.


*Yes. $\phi : \mathbb R \setminus \{0\} \to \mathbb R^2, \phi(x) = (x,1/x)$, is continuous and has the desired image. Take $f = \phi \circ \det$. Note that $\phi$ is a homeomorphism onto its image.


*No. $X  = \mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ has three components, but $GL_n(\mathbb R)$ has only two. See How many connected components does $\mathrm{GL}_n(\mathbb R)$ have?


*Yes. Let $g : GL_n(\mathbb R) \to \mathbb  R^n$ be the map assigning to each matrix $M$ its first column vector. This is a continuous map with image $\mathbb  R^n \setminus \{0\}$. Now define
$$r : \mathbb R^n \setminus \{0\} \to \mathbb R^n, r(x) = \begin{cases} x & \lVert x \rVert \le 1 \\ (2- \lVert x \rVert)\frac{x}{\lVert x \rVert} & 1 \le \lVert x \rVert \le 2 \\ 0 & \lVert x \rVert \ge 2 \end{cases}$$
This a continuous map and $f = r \circ g$ has the desired image.
