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Let $h:X \to Y$ be a smooth map between manifolds with boundaries.

How does one characterize a regular value for $h$ ? I am sifting through Milnor's Topology from the Differentiable Viewpoint and nowhere does he address the subtleties -if there are any- regarding such regular values (admittedly this isn't the book's focus...), as opposed to the definition given in §2. A motivating example would be the following (c.f. §4) :

Let $f,g : M \to N$ be smooth maps between boundaryless manifolds of the same dimension, where $M$ is compact and $N$ is connected. Let $F:M \times [0,1] \to N$ be a smooth homotopy between $f$ and $g$ (here $M \times [0,1]$ is a smooth manifold with boundary $(M \times 0) \cup (M \times 1)$). I would like to show that if $y\in N$ is a regular value for $F$, then it is a regular value for $f$ and $g$.

My initial approach was to simply compute, after choosing some coordinates in $M$, the jacobian matrices of these functions, and reason with linear algebra. However the approach seemed flawed, which is why I am wondering if regular values for maps like $F$, and more generally for maps like $h$, behaved differently than the usual case of a map between boundaryless manifolds.

(Edit : This type of result is probably what I am looking for)

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The rank of the Jacobi matrix is well defined even if $f(a)=b$ and $a\in \partial X$, $b\in \partial Y$. Note that in coordinates, $f$ is locally expressed as a map from $H^n$ to $H^k$ where $H^n, H^k$ are the upper half-spaces in $\Bbb R^n, \Bbb R^k$. Milnors definition of smoothness (see the first page of the book) in $a\in \partial H^n$ is via existence of a smooth extension of $f$ to a map from an open neighborhood of $a$ in $\Bbb R^n$ into an open neighborhood of $f(a)$ in $\Bbb R^k$ (I'm identifying $a,b$ with their coordinates). So, the matrix of partial derivatives is well-defined.

For the second question, yes, if $y$ is a regular value of $F$, then it is also a regular value of $f,g$. To show this, note that Milnors definition of a homotopy is such that $F(t,x)$ is independent of $t$ for $t\leq \epsilon$ and $t\geq 1-\epsilon$. Therefore, the matrix of derivatives of $F$ in $x\in F^{-1}(0)\cap (M\times \{0\})$ looks like $$ \begin{pmatrix} f' & 0 \\ \end{pmatrix} $$ which has the same rank as the matrix $(f')$.

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