Regular level set theorem and $SL(n,\mathbb{R})$ - explanation In Tu's book on manifolds, he gives this example as an application of the regular level set theorem.


It seems that, in order to prove that $1$ is a regular value for $\det$, he proves that $A$ has maximal rank. Shouldn't he prove that $\det_{*,p}$ has maximal rank for all points in $SL$ instead? After all, that is what is meant by $1$ being a regular value for all points in $SL$.
 A: As @Bass says in the comments, Tu is not proving that $A \in \mathrm{SL}(n,\Bbb{R})$ has maximal rank (this anyway follows from the fact that $A$ is invertible).
Tu properly proves that $\det \colon \mathrm{GL}(n,\Bbb{R}) \to \Bbb{R}$ has $1$ as a regular value by showing that for each $A$ in $\det^{-1}(1) = \mathrm{SL}(n,\Bbb{R})$, the map $d(\det)_A \colon T_A(\mathrm{GL}(n,\Bbb{R})) \to T_1(\Bbb{R})$ is surjective. He does this by choosing coordinate charts around $A$ and $1$ and computing the matrix of this linear map with respect to the induced bases.
In gory detail, here is what is going on in the proof. Let $\psi \colon \mathrm{GL}(n,\Bbb{R}) \to \Bbb{R}^{n^2}$ be the coordinate chart given by $$(a_{ij})_{1 \leq i,j \leq n} \mapsto (a_{11},\dotsc,a_{1n},a_{21},\dotsc,a_{2n},\dotsc,a_{n1},\dotsc,a_{nn})$$ and let $\psi(\mathrm{GL}(n,\Bbb{R})) =: U \subset \Bbb{R}^{n^2}$. Let $\varphi \colon \Bbb{R} \to \Bbb{R}$ be the coordinate chart given by $\varphi(a) = a$, and let $V := \varphi(\Bbb{R}) = \Bbb{R}$. Let $f \colon \mathrm{GL}(n,\Bbb{R}) \to \Bbb{R}$ be the map $f(A) = \det(A)$, and let $\tilde{f} := \varphi \circ f \circ \psi^{-1} \colon U \to V$. Then, given an $n^2$-tuple $\mathbf{a}$ in $U$ of the form $\mathbf{a} = (a_{11},\dotsc,a_{1n},a_{21},\dotsc,a_{2n},\dotsc,a_{n1},\dotsc,a_{nn})$, we have $\tilde{f}(\mathbf{a}) = \det[(a_{ij})_{1 \leq i,j \leq n}]$.
So, fix $A = (a_{ij})_{1\leq i,j \leq n} \in \mathrm{SL}(n,\Bbb{R}) = f^{-1}(1)$, and let $\psi(A) = \mathbf{a} \in U$. The matrix of the linear map $df_A \colon T_A(\mathrm{GL}(n,\Bbb{R})) \to T_1(\Bbb{R})$ with respect to the pair of bases induced by the local coordinates $\psi$ and $\varphi$ is the Jacobian of the map $\tilde{f} \colon U \to V$ at the point $\mathbf{a}$.
The entries of $\mathrm{Jac}(\tilde{f})|_\mathbf{a}$ are given by the partial derivatives of $\tilde{f}$ with respect to each of the coordinates (evaluated at $\mathbf{a}$). This is what Tu computes and expresses in terms of the minors of $A$.
Since $T_1(\Bbb{R}) \cong \Bbb{R}$, which is a one-dimensional vector space, $df_A$ is surjective if and only if it is nonzero. Equivalently, $df_A$ is not surjective if and only if $\mathrm{Jac}(\tilde{f})|_\mathbf{a}$ is the zero matrix. But this can happen only when $A$ is not invertible. Since every $A \in \mathrm{SL}(n,\Bbb{R})$ is invertible, this can never happen. So, $df_A$ is surjective.
Thus, $1$ is a regular value of $f$.
