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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?

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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 $n \ge 1$).

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  • $\begingroup$ 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. $\endgroup$
    – Mark
    Sep 16, 2020 at 2:02

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