# Sard's Theorem in one dimension

On page 76 of Guillemin and Pollaek (http://www.maths.adelaide.edu.au/pedram.hekmati/GP.pdf) it's stated that for any smooth map $$f:M \rightarrow N$$ from a manifold with boundary into a manifold without boundary, then almost every point $$y \in Y$$ is a regular point for both $$f:M \rightarrow N$$ and $$f_{|_{\partial{M}}}: \partial{M} \rightarrow N$$

But consider a very simple example of a smooth function $$f:[0,1] \rightarrow \mathbb{R}$$. Now, Sard's theorem guarantees that almost all $$y \in \mathbb{R}$$ is a regular value for both $$f:[0,1] \rightarrow \mathbb{R}$$ and $$f_{|_{\partial{[0,1]}}}: \partial{[0,1]} \rightarrow \mathbb{R}$$

But the boundary of $$[0,1]$$ is $$\partial{[0,1]}=\{0,1\}$$, a $$0-$$dimensional "smooth" manifold. Thinking about $$y$$ being a regular value of $$f_{|_{\partial{[0,1]}}}$$ would require us to check that for some $$x \in f_{|_{\partial{[0,1]}}}^{-1}(y) \in \{0,1\}$$,

$$Df(x) : T_{x}(\{0,1\}) \rightarrow \mathbb{R}$$

is surjective.

My question is, how is $$T_{x}(\{0,1\})$$ even defined? Normally, if $$\phi:U \subseteq \mathbb{R}^{n} \rightarrow V \cap M$$ is some diffeomorphism such that $$x \in V \cap M$$, then the tangent space would be defined as $$D\phi(v)(\mathbb{R}^{n})$$, where $$\phi(v)=x$$. It's clear that $$\{0,1\}$$ being a 0-dimensional smooth manifold means $$0 \in \{0,1\}$$ is "diffeomorphic" to $$\mathbb{R}^{0}=\{0\}$$ through a function $$\phi:\{0\} \rightarrow \{0\}$$.

What is the Jacobian of $$\phi:\{0\} \rightarrow \{0\}$$? What is $$T_{x}(\{0,1\})=D\phi(0)(\{0\})$$?

If $$D\phi(0)(\{0\})$$ is somehow defined to be $$0$$, then surjectivity can never hold. So it would have to be defined as some non-zero constant....but why?

First, you have the wrong quantifier. For $$y$$ to be a regular value requires you to check that FOR ALL $$x \in f_{|_{\partial{[0,1]}}}^{-1}(y) \in \{0,1\}$$, $$Df(x) : T_{x}(\{0,1\}) \rightarrow \mathbb{R}$$ is surjective.
If $$f_{|_{\partial{[0,1]}}}^{-1}(y)$$ happens to be empty, then the required condition for $$y$$ to be a regular value of the boundary restriction $$f \mid_{\partial{[0,1]}}$$ is vacuously true.
In the case where the set $$f_{|_{\partial{[0,1]}}}^{-1}(y)$$ is not empty, you just have to think sensibly about the concept of a $$0$$-dimensional vector space to be clear on what happens. First, that set is a subset of $$\partial[0,1]=\{0,1\}$$ which is a 0-dimensional manifold. The tangent space of $$\partial[0,1]$$ at each point of its two points is a $$0$$-dimensional vector space. The derivative of $$f|_{\partial{[0,1]}}$$ is certainly defined at each point in $$x \in \partial[0,1]$$, it is the unique homomorphism from the $$0$$-dimensional tangent space $$T_x \partial{[0,1]}$$ to the $$n$$-dimensional vector space $$T_{f(x)} \mathbb R^n$$. Since there is no surjective homomorphism from a $$0$$-dimensional tangent space to a positive dimensional tangent space, the values $$y_0=f(0)$$ and $$y_1=f(1)$$ are never regular values of $$f \mid_{\partial{[0,1]}}$$ (when $$n \ge 1$$).
• Could I go so far as to say that because $T_{x}(\partial{[0,1]})$ is required to be a $0$ dimensional vector space, and the set $\{0\}$ is the only 0 dimensional vectors space, then we must have $T_{x}(\partial{[0,1]})=\{0\}$. Thus, regardless of what $Df(x)$ is at some point $x \in \partial{[0,1]}$, $$Df(x): \{0\} \rightarrow \mathbb{R}$$ can never be surjective.