The complement of the special linear group

Idea. Show that $\{\det>1\}$ and $\{\det<1\}$ are each connected; they are clearly disjoint open sets covering the set.

Step 0: ingredients. $\{\det>1\}$ and $\{\det<1\}$ are open (preimages of open intervals under the continuous determinant), disjoint, and non-empty ($2^{1/n}I$ and $0$). By Problem 9, $\det^{-1}(1)=SL_n$ and $\det^{-1}(-1)$ are path connected.

Step 1: $\{\det>1\}$ is path connected. Let $\det A=d>1$ and $A=d^{1/n}A_1$ with $A_1\in SL_n$.
- Take a path $A_1(s)$ in $SL_n$ from $A_1$ to $I$. Then $d^{1/n}A_1(s)$ has determinant $d$ for all $s$, and runs from $A$ to $d^{1/n}I$.
- Then move along $c\,I$, $c$ between $d^{1/n}$ and $2^{1/n}$ (monotonically). Its determinant $c^n$ lies between $d$ and $2$, hence is $>1$.

So $A$ is joined within $\{\det>1\}$ to the fixed matrix $2^{1/n}I$.

Step 2: $\{\det<1\}$ is path connected. We show that every $A$ with $\det A<1$ is joined to $0$ inside $\{\det<1\}$.
- *$\det A=0$.* The ray $tA$ has $\det=0<1$.
- *$\det A=d\in(0,1)$.* As in step 1, $A$ is joined, at constant determinant $d$, to $d^{1/n}I$; then $c\mapsto cI$ with $c$ decreasing from $d^{1/n}$ to $0$ has $\det=c^n\le d<1$.
- *$\det A=d<0$.* $A=|d|^{1/n}A_1$ with $\det A_1=-1$; by connectedness of $\det^{-1}(-1)$, $A$ is joined at constant determinant $d$ to $|d|^{1/n}D$ with $D=\operatorname{diag}(-1,1,\dots,1)$. Then $c\mapsto cD$ for $c\searrow0$ has $\det=-c^n\le0<1$.

Step 3: conclusion. The set $\{\det\ne1\}$ is the disjoint union of two non-empty, open, path connected sets; these are its two components.

Remark. For $n=1$ the statement is also true (it is just $\mathbb{R}\setminus\{1\}$), but the proof above uses $n\ge2$ only through the connectedness of $SL_n$. Compare with Problem 22's contrast: the complement of $SL_n$ is *disconnected*, whereas the complement of $O(n)$ is connected (Problem 14), because $SL_n$ is a hypersurface (codimension $1$) and $O(n)$ is not.