# Why do Guillemin and Pollack avoid the term “submanifold with boundary”?

On p. 60 of the book "Differential Topology" by named authors they state:

Theorem. Let $$f$$ be a smooth map of a manifold $$X$$ with boundary onto a boundaryless manifold $$Y$$, and suppose that both $$f:X\to Y$$ and $$\partial f:X\to Y$$ are transversal with respect to a boundaryless submanifold $$Z$$ in $$Y$$. Then the premiage $$f^{-1}(Z)$$ is a manifold with boundary $$\partial\{f^{-1}(Z)\}=f^{-1}(Z)\cap \partial X$$ and the codimension of $$f^{-1}(Z)$$ in $$X$$ equals the codimension of $$Z$$ in $$Y$$.

Why don't they call $$f^{-1}(Z)$$ a submanifold of $$X$$ with boundary?

Would that be mathematically incorrect?

In the proof they show that around a boundary point, $$f^{-1}(Z)$$ is diffeomorphic to a submanifold of $$\mathbb{R}^k$$ intersected with the halfspace $$\mathbb{H}^k$$ under a chart for $$X$$, so I do not see why you wouldn't be able to call $$f^{-1}(Z)$$ a submanifold of $$X$$.

Same thing on p. 62:

Lemma. Suppose that S is a manifold without boundary and that $$\pi:S\to \mathbb{R}$$ is a smooth function with regular value $$0$$. Then the subset $$\{s\in S:\pi(s) \geq 0\}$$ is a manifold with boundary, and the boundary is $$\pi^{-1}(0)$$.

Why not call $$\pi^{-1}(0)$$ a submanifold of $$S$$ with boundary?

• You'll have to ask them. But in my experience of writing mathematics, it's not always worthwhile to pick a name for every mathematical concept. – Lee Mosher Jul 9 at 2:12