# Boundary conditions on Sard's theorem and transversality theorem

Sard's theorem. For any smooth map $f$ of a manifold $X$ with boundary into a boundaryless manifold $Y$, almost every point of $Y$ is regular value of both $f: X \rightarrow Y$ and $\partial f: \partial X \rightarrow Y$.

So removing the boundary condition, is this statement true (words in brackets are croseed out):

For any smooth map $f$ of a manifold $X$ {with boundary} into a boundaryless manifold $Y$, almost every point of $Y$ is regular value of {both} $f: X \rightarrow Y${ and $\partial f: \partial X \rightarrow Y$}.

In other words, is the condition $X$ with boundary applies to the later part of the theorem, $\partial f: \partial X \rightarrow Y$? So when we only need almost every point of $Y$ is regular value of $f: X \rightarrow Y$, we don't need to know if $X$ is with boundary.

Very similarly:

The transversality theorem. Suppose that $F: X \times S \rightarrow Y$ is a smooth map of manifolds, where only $X$ has boundary, and let $Z$ be any boundaryless submanifold of $Y$. If both $F$ and $\partial F$ are transversal to $Z$, then for almost every $s \in S$, both $f_s$ and $\partial f_s$ are transeversal to $Z$.

So removing the boundary condition, is this statement true?

The transversality theorem. Suppose that $F: X \times S \rightarrow Y$ is a smooth map of manifolds, where $Y$ is boundaryless, and let $Z$ be any boundaryless submanifold of $Y$. If $F$ is transversal to $Z$, then for almost every $s \in S$, $f_s$ is transeversal to $Z$.

In other words, is the condition $X$ with boundary applies to the later part of the theorem, $\partial f: \partial X \rightarrow Y$? Or only $X$ has boundary is necessary for the conclusion that for almost every $s \in S$, $f_s$ is transeversal to $Z$.

Thank you very much!

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