What do the open sets in $\mathbb R^4$ look like such that $ad - bc \neq 0$ J. M. Lee in his book on Smooth Manifolds (2nd Ed. p. 19) notes that the General linear group is an open subset of $\mathbb R^{n^2}$, because it's determinant is nonzero.  Even for figuring out what the elements NOT in $\mathbb R^4$ for the $ad - bc = 0$ is difficult for me to visualize.
With the change of variables $a=t+s$, $b=t-s$, $c=u+v$, and $d=u-v$, you find $$ad-bc=(t^2+u^2)-(s^2+v^2).$$
To get an idea, consider the first quadrant and let one axis represent $\sqrt{t^2+u^2}$, and the other, $\sqrt{s^2+v^2}$. These represent absolute value in the $(t,u)$ and $(s,v)$ planes respectively. The diagonal is where the determinant vanishes. Thus the set where the determinant is nonzero is split into two components.

There's still a bunch of details hiding behind the little word “thus” in the previous sentence, but that's the gist of one way to look at it.

It may help to note that, if you lock one of the variables $t$, $s$, $u$, $v$ to a fixed value, you get either a double cone (for the zero case) or a hyperboloid in the other three variables.
It has two components (I.e. two pieces), one piece for positive determinant and one for negative. The part you removed with determinant 0 has dimension 3 by algebraic geometry, so it has one dimension less than four-dimensional space.

So imagine some surface like a one-sheeted hyperboloid or a parabolic hyperboloid separraaing 3-d space into 2 pieces.