Homotopy type of smooth manifold with boundary
It seems very likely to me that a $n$-dimensional smooth manifold with boundary has the homotopy type of a $(n-1)$-dimensional CW complex. Is that true? Does the manifold need to be compact? What about $n$-dimensional open manifolds?
general-topology smooth-manifolds manifolds-with-boundary
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It seems very likely to me that a $n$-dimensional smooth manifold with boundary has the homotopy type of a $(n-1)$-dimensional CW complex. Is that true? Does the manifold need to be compact? What about $n$-dimensional open manifolds?
general-topology smooth-manifolds manifolds-with-boundary
It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
1
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
2
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45
add a comment |
It seems very likely to me that a $n$-dimensional smooth manifold with boundary has the homotopy type of a $(n-1)$-dimensional CW complex. Is that true? Does the manifold need to be compact? What about $n$-dimensional open manifolds?
general-topology smooth-manifolds manifolds-with-boundary
It seems very likely to me that a $n$-dimensional smooth manifold with boundary has the homotopy type of a $(n-1)$-dimensional CW complex. Is that true? Does the manifold need to be compact? What about $n$-dimensional open manifolds?
general-topology smooth-manifolds manifolds-with-boundary
general-topology smooth-manifolds manifolds-with-boundary
asked Dec 27 '18 at 17:58
Yeah
212
212
It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
1
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
2
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45
add a comment |
It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
1
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
2
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45
It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
1
1
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
2
2
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45
add a comment |
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It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$.
– Paul Frost
Dec 27 '18 at 18:25
1
@PaulFrost Yes, these obvious assumptions should be added to the question. But then?
– Yeah
Dec 27 '18 at 21:20
2
For a discussion in dimension 2, with allusions to higher dimensions, see here.
– Lee Mosher
Dec 27 '18 at 22:45