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What is the exact definition of submanifold of a manifold with boundary? For example, when $H$ is the half space of the plane and S is a cycle which intersects with the origin in the half plane. Then is S the submanifold of $M$?

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I do not know the exact answer; but I suppose it would be clear once you get straight the definitions of immersion and embedding. Then you could define immersed submanifold, and embedded submanifold, respectively. – George Sep 8 '11 at 8:41
up vote 4 down vote accepted

(See around page 189 of J.M. Lee's Introduction to Smooth Manifolds. I assume you are worried about smooth manifolds, since you tagged differential-topology.)

Recall that the tangent bundle of a smooth manifold with boundary is defined the same way as the tangent bundle of one without boundary in the interior, and on a boundary point $p\in\partial M$ we take $T_pM$ to be the span of $\{\partial_1, \ldots \partial_n\}$ ($\dim M = n$) in the boundary chart. So the tangent space at any point on a smooth $n$-manifold with boundary is an $n$-dimensional vector space.

Thus we can define immersions between manifolds with boundary: a smooth map $F:M\to N$ where $M, N$ are manifolds (possibly with boundary) is said to be an immersion if the differential $dF|_p:T_pM \to T_{F(p)}N$ is injective for every $p\in M$. Then an immersed submanifold of a manifold with boundary is simply $S\subset M$ such that $S$ can be given a smooth structure (possibly with boundary) such that the inclusion map $\iota:S\to M$ is an immersion. Similarly you can define embedded submanifolds.

In particular, for a smooth manifold with boundary $M$, the boundary $\partial M$ is a smooth submanifold of $M$. Similarly, under Lee's definition, your example is also a submanifold.

(Slightly off-topic:) Though honestly, instead of asking whether Blah is a Foobar, which really strikes me as a job for taxonomists, you should be thinking about the practical implications of the definitions. For example, while it is true still that for any embedded submanifold of a manifold with boundary that you still can find an open neighborhood of it that deformation retracts to the submanifold, it is not true that we can have the "tubular neighborhood theorem". (That is, it is not true that a sufficiently small open neighborhood of the embedded submanifold is diffeomorphic to its normal bundle; any neighborhood of $\partial M$ is a manifold with boundary, whereas the normal bundle of $\partial M$ is a manifold without boundary.)

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Just an addition. A manifold with boundary is not a differentiable retract of an open nbhd, but it is a continuous retract, and in fact the latter can be approximated (in the graph topology) by differentiable maps. This for instance allows to prove approximation theorems of continuous mappings by differentiable mappings in the presence of boundaries. – Jesus RS Feb 6 '15 at 16:08

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