Why is the general linear group a smooth manifold?

Question

First of all, any norm $||\cdot||$ on a finite dimensional vector space $V$ over $\mathbb{R}$ induces a metric $d(a,b):=||a-b||$. This metric then gives rise to a metric topology on $V$ and this topology makes $V$ a topological manifold. All good here.

The space of $n \times n$ matrices $M(n,\mathbb{R})$ with real entries is a finite dimensional real vector space. Now, the argument goes like this: $GL(n,\mathbb{R})$ is an open subset of $M(n,\mathbb{R})$ so it is also a topological manifold.

The thing is, how do you know $GL(n,\mathbb{R})$ is open? What norm are we using here?

Answer 1

It is the inverse image of $\mathbb{R}\setminus{\{0\}}$ (which is open) under the continuous map that maps the matrices in $Ml(n,\mathbb{R})$ to the determinant. And the inverse image is open (continuity).