Why is $SL(n, \mathbb{R})$ the kernel of $\det : GL(n, \mathbb{R}) \mapsto\Bbb R^*$?

Question

The special linear group of invertible matrices is defined as the kernel of the determinant of the map:

$$\det:GL(n,\mathbb{R}) \mapsto \mathbb{R}^*$$

In my mind the kernel of a linear map is the set of vectors that are mapped to the zero vector. So the map above would contain all the matrices that have determinant zero (which doesn't make sense since the codomain of the function excludes zero)? But isn't the special linear group made of matrices with determinant 1?

Answer 1

But $\det$ is not a linear map. It is a group homomorphism. Its kernel is$$\left\{M\in GL(n,\Bbb R)\,\middle|\,\det(M)=1\right\},$$since $1$ is the identity element of $\bigl(\Bbb R\setminus\{0\},\times\bigr)$.

Answer 2

$\mathbb{R}^∗$ is the multiplicative group, so the identity is $1$ - in particular the kernel is the elements of the group homomorphism sent to it.

NB: the determinant map is not linear and indeed the group operation on $GL_2(\mathbb{R})$ is not abelian so it's clearly not a vector space in a naive way.