In Guillemin-Pollack's book Differential Topology, the Transversality theorem states that

The transversility Theorem. Suppose that $F:X \times S \to Y$ 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 transversal to $Z$

(Note that the notation $\partial f$ is just the restriction to the boundary $\partial X$)

However, I am wondering what if $X$ also has empty boundary? and how to define $\partial F$ in this case.

For example, I want to use transversility theorem to show the following problem:

Problem. Let X and Y be submanifolds of $\mathbb R^N$. Show that for almost every $a \in \mathbb R^N$, the translate $X+a$ intersects $Y$ transversally.

My sketch of proof: construct $F:X \times \mathbb R^N \to \mathbb R^N$ by $(x,a) \to x+a$. Now let $Z=Y$ in this case and "apply” the above theorem. However, I am not sure whether I could apply the theorem since here $X$ could be boundaryless.


1 Answer 1


The theorem you cite has a weaker version which makes no claim about the boundary of $X$.

Theorem: Suppose that $F:X\times S\to Y$ is a smooth map of manifolds and $Z$ is a submanifold of $Y$, all manifolds without boundary. If $F$ is transverse to $Z$ then for almost every $s\in S$ the map $f_s : x\mapsto F(x,s)$ is transverse to $Z$.

We can deduce this theorem as a corollary of your theorem, using facts about the empty set.

If $\partial X=\varnothing$ then $\partial F$ is the empty function, since its domain of definition of $\varnothing$. Now $\partial F$ is transverse to $Z$ if for every $y\in Z\cap\partial F(X)$ a certain condition is satisfied. But stop right there! The intersection is empty, so the transversality condition is vacuously true.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .