The sign of the determinant under squaring

Idea. $\det$ is continuous, so preimages of open sets are open; multiplicativity of $\det$ controls the sign of a square.

Step 1: openness. $GL_n^+=\det^{-1}((0,\infty))$ and $GL_n^-=\det^{-1}((-\infty,0))$ are preimages of open sets by the continuous map $\det$.

Step 2: squares have positive determinant. For $A\in GL_n$, $\det(A^2)=(\det A)^2>0$, so $A^2\in GL_n^+$.

Step 3: not onto for $n=2$. The matrix $N=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$ has $\det N=1$, so $N\in GL_2^+$. If $N=B^2$ with $B$ real, the eigenvalues of $B$ are two conjugate non-real numbers $\pm i$ or two real numbers; real $\mu$ would have $\mu^2=-1$, impossible, so they are $i,-i$, distinct. Then $B$ is diagonalisable over $\mathbb{C}$, so is $B^2$, with eigenvalues $-1,-1$: $B^2=-I\neq N$. Contradiction.

Remark.
- Combined with Problem 74: the image of squaring in $GL_n$ is a *proper* subset of $GL_n^+$ ($n\ge2$) it contains the open set of matrices with no negative eigenvalue (which have a real square root by the principal logarithm/power series), but it is not open-and-closed in $GL_n^+$ in an obvious way.
- Squaring on $GL_n^+$ is far from injective: $B$ and $-B$ have the same square, and for $n\ge2$ also $\operatorname{diag}(1,-1)$ squares to $I$.