theorem : Let $SL(n,\Bbb{C})$ be the group of matrices of complex entries and determinant $1$ then $SL(n,\Bbb{C})$ is a regular submanifold of $GL(n,\Bbb{C})$.
proof : $GL(n,\mathbb{C})$ is a Lie group and $SL(n,\mathbb{C})$ a closed subgroup. with Cartan's theorem $SL(n,\mathbb{C})$ is a regular Lie subgroup, in particular a regular submanifold.
Cartan's theorem : if $H$ is a closed subgroup of a Lie group G, then H is an embedded Lie group with the smooth structure (and hence the group topology) agreeing with the embedding.
why $SL(n,\mathbb{C})$ is a closed subgroup ? what is dimension of manifold $SL(n,\mathbb{C})$ ?
Consider the map $\det\colon GL(n,\Bbb C)\longrightarrow\Bbb C$. It is continuous and therefore $\det^{-1}\bigl(\{1\}\bigr)$ is closed. But $\det^{-1}\bigl(\{1\}\bigr)=SL(n,\Bbb C)$.
And its dimension is$$\dim\mathfrak{sl}(n,\Bbb C)=\dim\left\{M\in\Bbb C^{n\times n}\,\middle|\,\operatorname{tr}M=0\right\}=n^2-1.$$