# Confused with The Transversality Theorem when all manifolds are boundaryless

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

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.