Do subspaces of polyhedron give subcomplexes?
Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| to X$. If we have a subspace $A subseteq X$, is there a subcomplex $L$ of $K$ such that $h(|L|) = A$?
It's easy to see that $h^{-1}(A)$ is a subspace of $|K|$ (which is the geometric realization of the simplicial complex $K$), but I don't have an idea if there is a subcomplex of $K$ whose geometric realization is $h^{-1}(A)$.
If it's the case that we can't find a subcomplex of $K$ from subspaces of $X$, under what conditions can we determine if a subspace of $X$ has a corresponding subcomplex of $K$?
general-topology algebraic-topology simplicial-complex
asked Jan 3 at 22:06
Perturbative
4,13511450
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Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| to X$. If we have a subspace $A subseteq X$, is there a subcomplex $L$ of $K$ such that $h(|L|) = A$?
It's easy to see that $h^{-1}(A)$ is a subspace of $|K|$ (which is the geometric realization of the simplicial complex $K$), but I don't have an idea if there is a subcomplex of $K$ whose geometric realization is $h^{-1}(A)$.
If it's the case that we can't find a subcomplex of $K$ from subspaces of $X$, under what conditions can we determine if a subspace of $X$ has a corresponding subcomplex of $K$?
general-topology algebraic-topology simplicial-complex
asked Jan 3 at 22:06
Perturbative
4,13511450
add a comment |
Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| to X$. If we have a subspace $A subseteq X$, is there a subcomplex $L$ of $K$ such that $h(|L|) = A$?
It's easy to see that $h^{-1}(A)$ is a subspace of $|K|$ (which is the geometric realization of the simplicial complex $K$), but I don't have an idea if there is a subcomplex of $K$ whose geometric realization is $h^{-1}(A)$.
If it's the case that we can't find a subcomplex of $K$ from subspaces of $X$, under what conditions can we determine if a subspace of $X$ has a corresponding subcomplex of $K$?
general-topology algebraic-topology simplicial-complex
asked Jan 3 at 22:06
Perturbative
4,13511450
Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| to X$. If we have a subspace $A subseteq X$, is there a subcomplex $L$ of $K$ such that $h(|L|) = A$?
It's easy to see that $h^{-1}(A)$ is a subspace of $|K|$ (which is the geometric realization of the simplicial complex $K$), but I don't have an idea if there is a subcomplex of $K$ whose geometric realization is $h^{-1}(A)$.
If it's the case that we can't find a subcomplex of $K$ from subspaces of $X$, under what conditions can we determine if a subspace of $X$ has a corresponding subcomplex of $K$?
general-topology algebraic-topology simplicial-complex
general-topology algebraic-topology simplicial-complex
asked Jan 3 at 22:06
Perturbative
4,13511450
asked Jan 3 at 22:06
Perturbative
4,13511450
asked Jan 3 at 22:06
Perturbative
4,13511450
asked Jan 3 at 22:06
Perturbative
4,13511450
asked Jan 3 at 22:06
Perturbative
4,13511450
4,13511450
add a comment |
add a comment |
1 Answer
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Let us call $h : lvert K rvert to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A subset X$, is there a triangulation $h : lvert K rvert to X$
and a subcomplex $L subset K$ such that $h(lvert L rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.
answered Jan 3 at 23:16
Paul Frost
9,3762631
add a comment |
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1 Answer
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Let us call $h : lvert K rvert to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A subset X$, is there a triangulation $h : lvert K rvert to X$
and a subcomplex $L subset K$ such that $h(lvert L rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.
answered Jan 3 at 23:16
Paul Frost
9,3762631
add a comment |
Let us call $h : lvert K rvert to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A subset X$, is there a triangulation $h : lvert K rvert to X$
and a subcomplex $L subset K$ such that $h(lvert L rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.
answered Jan 3 at 23:16
Paul Frost
9,3762631
add a comment |
Let us call $h : lvert K rvert to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A subset X$, is there a triangulation $h : lvert K rvert to X$
and a subcomplex $L subset K$ such that $h(lvert L rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.
answered Jan 3 at 23:16
Paul Frost
9,3762631
Let us call $h : lvert K rvert to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A subset X$, is there a triangulation $h : lvert K rvert to X$
and a subcomplex $L subset K$ such that $h(lvert L rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.
answered Jan 3 at 23:16
Paul Frost
9,3762631
answered Jan 3 at 23:16
Paul Frost
9,3762631
answered Jan 3 at 23:16
Paul Frost
9,3762631
answered Jan 3 at 23:16
Paul Frost
9,3762631
9,3762631
add a comment |
add a comment |
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