Singularity of a smooth map

Remarks

Example: $f \colon \R^2 \to \R^2$, $f(x, y) = (x, y^2)$, has $df_{(x,y)} = \begin{pmatrix} 1 & 0 \\ 0 & 2y \end{pmatrix}$, of rank $1 < 2$ exactly when $y = 0$. So $\Sigma(f)$ is the line $y = 0$, along which $f$ folds the plane. Non-example: for $f \colon \R \to \R^2$, $f(t) = (t, t^2)$, the differential $df_t$ has matrix $(1, 2t)^T$ of rank $1 = \min(1, 2)$ at every $t$, so $f$ has no singular points. A point is regular if and only if $f$ is an immersion near it (when $m \le n$) or a submersion near it (when $m \ge n$). For $N = \R$ the singular points are the critical points of $f$, where $df_p = 0$. Some authors call every point with $df_p$ not surjective a critical point; this agrees with the definition above when $m \ge n$.