I want to show that: $$ M := \{J \in M_{2n}(\mathbb{R}) \mid J^2 = -I \} $$ is the submanifold of $M_{2n}(\mathbb{R})$, and I am given a hint to use Theorem 1.
Theorem 1. Let $m,n,l \geq 0$. Suppose that $O$ is an open set of $\mathbb{R}^{n}$, and $O'$ is a open set of $\mathbb{R}^m$, and the mappings $F: O \to O'$ and $G: O' \to \mathbb{R}^l$ are of the class $C^{\infty}$. Suppose that the points $a_0 \in O$, $\,b_0 \in O'$, and $c_0 \in \mathbb{R}^{l}$ satisfy the following conditions: (i) $F(a_0) = b_0$, (ii) $GF(O) = c_0$, and (iii) the sequence of linear mappings $$ \mathbb{R}^n \xrightarrow{(JF)_{a_0}} \mathbb{R}^m \xrightarrow{(JG)_{b_0}} \mathbb{R}^l $$ is exact. Then, there is a open set $W \subseteq \mathbb{R}^m$ that satisfies $b_0 \in W \subseteq O'$, and we have $$ W \cap F(O) = W \cap G^{-1}(c_0) $$ and $W \cap F(O)$ will be a $C^{\infty}$-submanifold of $\mathbb{R}^m$.
However, I don't know how to use this theorem. In my opinion, I can consider $G$ as $\det$, and $\det^{-1} \{\pm{1}\}$ will be $M$, but I don't know what $J(\det)$ is.
Please help me!!!!! I am Japanese, so I am not good at $\LaTeX$ and English. There may be a lot of mistakes in my context, please forgive me.