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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?

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  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Jul 9 at 2:12

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