A smooth map regular at a point is regular near it

Question

If a smooth $f:M\rightarrow N$ where $M, N$ are manifolds with boundary is regular at some point, is it regular near this point?

Answer 1

Assume $f(p)$ is a regular value of $f$. Pick local coordinates $x^i$ on an open set $U$ around $p$ and $y^j$ on an open set $V$ around $f(p)$. Let $\partial_{x^i} = \frac{\partial}{\partial x^i}$ denote the basis of tangent vectors on gets at each point in $U$ and similarly for $\partial_{y^j}$. Set $m = \dim M$ and $n = \dim N$.

Then for each $q\in U$, the linear transformation $d_q f$ can be written as a matrix with respect to the two bases $\{\partial_{x^i}\}$ and $\{\partial_{x^j}\}$. This matrix has dimension $n\times m$, and, because $f$ is smooth, the entries of $d_q f$ vary smoothly as a function of $a$.

Now, by assumption, at $p$, $d_p f$ is surjective, which means the matrix associated to it has rank $n$. In particular, there must be $n$ columns $c_i(p) ,...c_n(p)$ of $d_p f$ which form a basis of the tangent space $T_{f(p)}N$.

Now, form the matrix $C(p)$ with columns $c_i(p)$. Because the $c_i(p)$ form a basis, the determinant $\det(C(p))\neq 0$.

Consider the family of matrices $C(q)$ obtained by letting $p$ vary. Since the entries of $d_q f$ vary smoothly, and $C(q)$ is built out of the columns of $d_q f$, the entries of $C(q)$ also vary smoothly. Thus, by continuity of $\det$, there is an open subset $U_0\subseteq U$ for which $\det(C(q))\neq 0$ for all $q\in U_0$.

But this means that the columns of $C(q)$ form a basis of $T_{f(p)}N$ for all $q\in U_0$ which, in turn, means that $d_q f$ has rank $n$ for all $q\in U_0$, which, in turn means that $d_q f$ is surjective for all $q\in U_0$.