Oredering of the vertices of simplicial complex
$begingroup$
In a proof of the theorem that states that singular and simplical homology groups of triangulable spaces are isomorphic there is the sentence (here $K$ is a simplicial complex):
Choose a partial ordering of the vertices of $K$ that induces a linear ordering on the vertices of each simplex of $K$.
Can we explicitly define such ordering or is the existence of such ordering a consequence of "something"?
algebraic-topology simplicial-complex
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
$endgroup$
add a comment |
$begingroup$
In a proof of the theorem that states that singular and simplical homology groups of triangulable spaces are isomorphic there is the sentence (here $K$ is a simplicial complex):
Choose a partial ordering of the vertices of $K$ that induces a linear ordering on the vertices of each simplex of $K$.
Can we explicitly define such ordering or is the existence of such ordering a consequence of "something"?
algebraic-topology simplicial-complex
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
$endgroup$
add a comment |
$begingroup$
In a proof of the theorem that states that singular and simplical homology groups of triangulable spaces are isomorphic there is the sentence (here $K$ is a simplicial complex):
Choose a partial ordering of the vertices of $K$ that induces a linear ordering on the vertices of each simplex of $K$.
Can we explicitly define such ordering or is the existence of such ordering a consequence of "something"?
algebraic-topology simplicial-complex
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
$endgroup$
In a proof of the theorem that states that singular and simplical homology groups of triangulable spaces are isomorphic there is the sentence (here $K$ is a simplicial complex):
Choose a partial ordering of the vertices of $K$ that induces a linear ordering on the vertices of each simplex of $K$.
Can we explicitly define such ordering or is the existence of such ordering a consequence of "something"?
algebraic-topology simplicial-complex
algebraic-topology simplicial-complex
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
edited Dec 10 '18 at 21:07
Eric Wofsey
184k13213339
184k13213339
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
asked Sep 4 '16 at 13:00
Madara UchihaMadara Uchiha
1798
1798
add a comment |
add a comment |
1 Answer
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$begingroup$
Literally just take any linear ordering of the set of vertices of $K$. This ordering, when restricted to the vertices of each simplex, will still be a linear ordering.
(The proof only actually needs the fact that the ordering is linear when restricted to the vertices of each simplex, which is why that is all it assumes. But the easiest way to verify the existence of such a partial ordering is to consider an ordering that is actually linear on all the vertices.)
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Literally just take any linear ordering of the set of vertices of $K$. This ordering, when restricted to the vertices of each simplex, will still be a linear ordering.
(The proof only actually needs the fact that the ordering is linear when restricted to the vertices of each simplex, which is why that is all it assumes. But the easiest way to verify the existence of such a partial ordering is to consider an ordering that is actually linear on all the vertices.)
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
$endgroup$
add a comment |
$begingroup$
Literally just take any linear ordering of the set of vertices of $K$. This ordering, when restricted to the vertices of each simplex, will still be a linear ordering.
(The proof only actually needs the fact that the ordering is linear when restricted to the vertices of each simplex, which is why that is all it assumes. But the easiest way to verify the existence of such a partial ordering is to consider an ordering that is actually linear on all the vertices.)
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
$endgroup$
add a comment |
$begingroup$
Literally just take any linear ordering of the set of vertices of $K$. This ordering, when restricted to the vertices of each simplex, will still be a linear ordering.
(The proof only actually needs the fact that the ordering is linear when restricted to the vertices of each simplex, which is why that is all it assumes. But the easiest way to verify the existence of such a partial ordering is to consider an ordering that is actually linear on all the vertices.)
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
$endgroup$
Literally just take any linear ordering of the set of vertices of $K$. This ordering, when restricted to the vertices of each simplex, will still be a linear ordering.
(The proof only actually needs the fact that the ordering is linear when restricted to the vertices of each simplex, which is why that is all it assumes. But the easiest way to verify the existence of such a partial ordering is to consider an ordering that is actually linear on all the vertices.)
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
answered Dec 10 '18 at 21:06
Eric WofseyEric Wofsey
184k13213339
184k13213339
add a comment |
add a comment |
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