Consider the group $GL(n,\mathbb{R})$. I am interested in showing that the determinant map on this group has a constant rank.
Our professor justified this fact in the following way, let $A\in GL(n,\mathbb{R})$ and $a=det(A).$ Then, $$\operatorname{det} = l_a\circ \operatorname{det}\circ L_{A^{-1}}$$ where $l_a(x)=ax$ for $x\neq 0$ and $L_{A^{-1}}(X)=A^{-1}X$ are left translation on $\mathbb{R}^*$ and $GL(n,\mathbb{R})$ respectively.
Then using the chain rule one gets that $$\operatorname{rank} d_{X}\operatorname{det} = \operatorname{rank} d_{A^{-1}X}\operatorname{det}$$ and then choosing $A=X$ we can see that the rank is constant.
I am not sure how exactly is the chain rule being applied here explicitly to deduce that the rank of the two operators is the same at different points. Could one show me the computation explicitly if possible?
The chain rule on $$\operatorname{det} = l_a\circ \operatorname{det}\circ L_{A^{-1}}$$ implies that
$$ d\operatorname{det}_X = dl_a\circ d\operatorname{det}_{A^{-1}X } \circ d(L_{A^{-1}})_X.$$
Since both $\ell_a$, $L_{A^{-1}}$ are invertible (note $a\neq 0$) with smooth inverse, their differentials are bijective. Thus
\begin{align} \mathrm{rank} ( d\operatorname{det}_X) &= \mathrm{rank}( dl_a\circ d\operatorname{det}_{A^{-1}X } \circ d(L_{A^{-1}})_X) \\ &= \mathrm{rank}( dl_a\circ d\operatorname{det}_{A^{-1}X }) \\ &= \mathrm{rank}( d\operatorname{det}_{A^{-1}X }) \end{align}
Remark one can check directly that $\det$ has constant rank $1$ on $GL(n , \mathbb R)$: for any $A \in GL(n, \mathbb R)$, $$ d{\det} _A (A) = \frac{d}{dt}\bigg|_{t=0} \det (A + tA) = a \frac{d}{dt}\bigg|_{t=0} (1+t)^n = an \neq 0.$$