Let $f:M→N$ be a smooth function between two smooth manifolds. Then $p\in M$ is a critical point if $df_p$ is not surjective.
I feel very confused about this definition, even in the case where $M=N=\mathbb{R}$. I can understand at a critical point $p$, $df_p$ sends all tangent vectors at $p$ (all on the x-axis) to a zero tangent vector w.r.t. the y-axis.
But on a regular point, I feel $df_p$ is not surjective either, "since" the tangent direction is fixed at each point $f(p)$. Where does my intuition go wrong?
Best Answer
I think you're just confused about the meaning of surjective in this case. In particular, $df_p$ does not surject onto $\mathbb{R}^2$: the tangent space of $\mathbb{R}$ is one-dimensional at every point.
For $M = N = \mathbb{R}$, $df_p$ is a linear map from $T_p \mathbb{R} \to T_{f(p)} \mathbb{R}$. We can identify the tangent space of $\mathbb{R}$ at any point with $\mathbb{R}$ itself, so we can consider $df_p$ as a linear map from $\mathbb{R} \to \mathbb{R}$. $df_p$ is given by $df_p(v) = f'(p)v$ where $f'(p)$ is the usual one-variable derivative. Since its codomain is $1$-dimensional, $df_p$ is either surjective or the zero map, which corresponds exactly to whether $f'(p) = 0$ or $f'(p) \neq 0$. This follows from the rank-nullity theorem. We have $$ 1 = \dim(\mathbb{R}) = \dim(\ker(df_p)) + \dim(\text{img}(df_p)) $$ so either $\dim(\text{img}(df_p)) = 1$ or $\dim(\text{img}(df_p)) = 0$.