Suppose $X$ is a manifold without boundary and $Y$ is a manifold.
Suppose there is a smooth function $f: X \rightarrow Y$ and we are given a $y \in Y$ such that $y$ is a regular value of $f$ and $f^{-1}(y) \neq \emptyset$.
Then $y$ is not in the boundary of $Y$.
I believe there would be some contradiction with the boundary of $X$ being void.
This is somewhat weird because i can't seem to find any connection between these hypothesis.
Any help would be appreciated.
Say that $X$ is of dimension $m$ and $Y$ is of dimension $n$. If $y$ is a regular value of $f$, then $f^{-1}(y)$ is a properly embedded submanifold of $X$ with dimension $m-n$. Call this submanifold $S$.
Given a point $p$ in $S$, this means that there are coordinate charts $(U, \varphi)$ in $X$ containing $p$ and $(V, \psi)$ in $Y$ containing $f(p)$ such that the local coordinate representation for $f$ is of the form: $$(\psi\circ f\circ\varphi^{-1})(x^1, \ldots, x^{m-n}, x^{m-n+1}, \ldots, x^{m})=(x^{m-n+1}, x^{m-n+2}, \ldots, x^m)$$ In particular, $\psi\circ f\circ\varphi^{-1}$ maps the last $n$ coordinates in $\varphi(U)$ diffeomorphically onto their image in $\psi(V)$. If $f(p)$ is a boundary point, then this is a diffeomorphism between an open $n$-ball and the image of a boundary chart in $\mathbb{R}^n$, a contradiction.