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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 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?

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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 $\dim(\text{img}(df_p)) = 0$.

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  • $\begingroup$ Can I ask the reason why "Since its codomain is 1-dimensional, dfp is either surjective or the zero map"? $\endgroup$ – Vladimir Mar 13 '15 at 17:41
  • $\begingroup$ Suppose $df_p=1$. Why is this surjective? $\endgroup$ – Vladimir Mar 13 '15 at 17:43
  • $\begingroup$ Thanks! Yes, the space spanned by $df_p=1$ is $\mathbb{R}$, but that's not equivalent to $df_p$ being surjective? $\endgroup$ – Vladimir Mar 13 '15 at 17:46
  • $\begingroup$ What does $df_p = 1$ mean? $df_p$ is a linear map from $\mathbb{R} \to \mathbb{R}$, so it cannot be the constant $1$ map. If instead you mean that $f'(p) = 1$, so $df_p$ is multiplication by $1$, then $df_p$ is the identity map from $\mathbb{R} \to \mathbb{R}$, which is certainly surjective. $\endgroup$ – André 3000 Mar 13 '15 at 17:47
  • $\begingroup$ I think I'm confused about the definition of $df_p$. In my textbook, it seems $df_p$ is the derivative--I'm probably wrong on this, but I have this impression because it can be written as Jacobian matrix. $\endgroup$ – Vladimir Mar 13 '15 at 17:50
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It's quite correct, just that when all tangent vectors at p because it is fixed at each point but since the regular point,dfp wouldn't be surjective due to the point of fixation.

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    $\begingroup$ Sorry that I don't understand your explanation... If at a regular point the differential fails to be surjective, then there'd be something wrong with the definition?! $\endgroup$ – Vladimir Mar 13 '15 at 17:33
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in one dimension we have have a function $f:R\to R$. if $f'(a)\neq 0$ then $df_a (x)=f'(a)x$, i.e., a linear approximation of the function at the point $a$, whose domain is $R$ and codomain $R$. Notice that the image of the function is $R$ as well, therefore $df_a$ is surjective. But when $f'(a)=0$ we have that $df_a(x)=f'(a)\cdot x=0$, so the image is always zero being not equal to the codomain (being $R$). Conclusion: $f'(a)=0$ implies $df_a(x)$ not surjective (and is the only way of $df_a$ to be not surjective).

To prove the opposite, I appeal to the fact that the differential is a linear approximation, and the only way that a straight line not to be surjective is with null slope (because a small inclination projects images in the entire codomain, guaranteeing surjectivity), so to demand not surjectivity to $df_a(x)=f'(a)x$ (which is a straight line) is the same thing that to say $f'(a)=0$.

Final conclusion: Let $f:R\to R$ be a function of $C^\infty$. $a\in R$ is a critical point of $f$ if $df_a(x)$ is not surjective (that is, $f'(a)=0$).

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