# Critical Points of Functions on Manifolds and Elementary Analysis

I am trying to show that the definition of a critical point of a function between manifolds reduces to the usual definition of critical point encountered in elementary analysis. I believe what I have is correct but would appreciate a critique of the argument.

Specifically:

By definition, a smooth function $f:M \rightarrow N$ between manifolds $M$ and $N$ has a critical point at $p \in M$ iff the differential of $f$ at $p$, $$f_{*,p}:T_pM \rightarrow T_{f(p)}N$$ is not surjective.

I claim that a point $p \in R$ is a critical point of a differentiable function $$f:\mathbb{R}\rightarrow \mathbb{R}$$ iff $f^{\prime}(p) = 0$

Proof of Claim: The differential of a real-valued function $f := f(t)$ at the point $p$ is the linear map $$f_{*,p}:\mathbb{R}\rightarrow \mathbb{R}$$ determined by $$f_{*,p}(x) = f^{\prime}(p)\cdot x$$ for all $x \in \mathbb{R}$. Since $f_{*,p}$ is a linear map between spaces of the same dimension $f_{*,p}$ is injective iff it is surjective and the rank/nullity theorem reads

$$\text{dim}(\text{Im}(f_{*,p})) + \text{dim}(\text{Ker}(f_{*,p})) = \text{dim}(\mathbb{R}) = 1$$

Now, if $p$ is a critical point, $f_{*,p}$ is not surjective and it cannot be injective either since $f_{*,p}$ is a linear map between spaces of the same finite dimension. But, since it is not injective the kernel must be nontrivial, i.e., $\text{dim}(\text{Ker} (f_{*,p})) \neq 0$. But, since the kernel is a subspace of $\mathbb{R}$ its dimension can be no greater than $1$. Therefore, the dimension of the kernel must be $1$ and by the rank/nullity theorem the dimension of the image must be $0$. The only function that has a zero dimensional image is the $0$-function itself. Therefore, $f_{*,p}(x) = 0 \implies f^{\prime}(p) = 0$

On the other hand, if $f_{*,p}$ is the $0$-function, then the rank-nulltility theorem can be used again to conclude that $f_{*,p}$ is not surjective.

QED?

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Yes. But a bit "overblown": isn't it obvious to you, that a linear Map $\mathbb R \rightarrow \mathbb R$ (which is always multiplication by a real number $a$) is surjective iff it is injective iff this number $a\neq 0$? –  Blah Feb 6 '12 at 23:25
Well, if it were "obvious" to me, do you think I would have gone to the trouble to write it all out? "Obvious", "clearly" and other such adjectives are absolutely meaningless words in the context of a mathematical proof. A fact follows directly from something or it doesn't. The fact that you state about a map I suppose could be proved without resorting to the rank/nullity theorem but it is not "obvious" to me until I prove it one way or the other. –  ItsNotObvious Mar 1 '12 at 16:15