 Solution

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.