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Let $f:M\rightarrow N$ be a smooth function between two smooth manifolds.

A $\textit{regular point}$ is a point $p\in M$ for which the differential $df_p$ is surjective.

What does the surjectivity condition for the differential mean intuitively? What is then so "regular" about this point?

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The point is regular in the sense that it is not critical. In fact, Sard's theorem tells us that the image of the set of critical points (the critical values) of $f$ has measure zero as a subset of $N$. – M.B. Aug 8 '12 at 11:06
up vote 5 down vote accepted

What is regular about a regular point $p$ is that the fibre $f^{-1}(f(p))$ is itself a manifold in a suitable neighbourhood of $p$, and that manifold has dimension $dim (M)-dim (N)$.

About the simplest example is obtained by taking $M=\mathbb R^2, N=\mathbb R$ and $f(p)=f(x,y)=y^2-x^3$.
We have $df_{(a,b)}(u,v)=-3a^2u+2bv$ for $p=(a,b)$ and this linear form is surjective (=non-zero) unless $p=(a,b)=0$.
As a consequence the fibers of $f$, aka the contour curves $y^2-x^3=c$ are smooth submanifolds (of dimension $1$) of $M=\mathbb R^2$ for $c\neq0$.
However the contour curve through $p=(0,0) $ is the subset $C\subset \mathbb R^2$ given by $y^2-x^3=0$, which is not a manifold in any neighbourhood of $(0,0)$.

Differential geometers generally recoil in horror before beasts like $C$, whereas algebraic geometers study them under the name of singular varieties.

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