In a lot of references, people mention that $\operatorname{SL}_n(k)$ is a normal subgroup of $\operatorname{GL}_n(k)$ and $\operatorname{GL}_n(k)/\operatorname{SL}_n(k) \cong k^\times$ via the determinant map.
But I wonder if there is anything wrong with the following similar map
$$\operatorname{GL}_n(k)\to\operatorname{SL}_n(k), A\mapsto \frac{1}{\det A}A$$
This map is well-defined since $\det A\neq 0$ when $A\in\operatorname{GL}_n(k)$. The multiplication is preserved since $\det AB=\det A \det B$. This map is clearly surjective. So it induces $\operatorname{GL}_n(k)/k^\times \cong\operatorname{SL}_n(k)$.
Is there anything wrong here? I'm just surprised that I cannot find any source mentionging this one.
As pointed out in the comments, the given map does not map into $SL_n(k)$, and in general these groups are not isomorphic. For instance, if we consider the case of Lie groups ($k=\mathbb{C}$ or $k=\mathbb{R}$) then there is at most one map with prescribed differential at the identity. Any isomorphism would have the identity as differential at the identity, like the natural map $SL_n(\mathbb{C}) \to GL_n(\mathbb{C})/k^*$. However, this map has a finite kernel, namely the $n$th roots of unity times the identity (if $k=\mathbb{R}$, take $n$ even and $\pm I$). By uniqueness of maps corresponding to identity differential, there is no isomorphism between the groups.