Mixing
Notation: Definition of little $d$-disk operad $D_d$ for $d=infty$
$begingroup$
Let $D_d$ be the little $d$-disk operad as outlined in Fresse's book Homotopy of Operads and Grothendieck-Teichmuller Groups.
We have the sequence of inclusions of operads
$$D_1 to D_2 to cdots to D_n to cdots$$
The operad $D_infty$ was set as $colim_n D_n$.
I'm not sure about the notation $colim_n D_n$.
I'm assuming that it's the colimit of the diagram
$$D_1 to D_2 to cdots to D_n to cdots$$
Is it correct?
algebraic-topology category-theory limits-colimits operads
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
$endgroup$
add a comment |
$begingroup$
Let $D_d$ be the little $d$-disk operad as outlined in Fresse's book Homotopy of Operads and Grothendieck-Teichmuller Groups.
We have the sequence of inclusions of operads
$$D_1 to D_2 to cdots to D_n to cdots$$
The operad $D_infty$ was set as $colim_n D_n$.
I'm not sure about the notation $colim_n D_n$.
I'm assuming that it's the colimit of the diagram
$$D_1 to D_2 to cdots to D_n to cdots$$
Is it correct?
algebraic-topology category-theory limits-colimits operads
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
$endgroup$
$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26
add a comment |
$begingroup$
Let $D_d$ be the little $d$-disk operad as outlined in Fresse's book Homotopy of Operads and Grothendieck-Teichmuller Groups.
We have the sequence of inclusions of operads
$$D_1 to D_2 to cdots to D_n to cdots$$
The operad $D_infty$ was set as $colim_n D_n$.
I'm not sure about the notation $colim_n D_n$.
I'm assuming that it's the colimit of the diagram
$$D_1 to D_2 to cdots to D_n to cdots$$
Is it correct?
algebraic-topology category-theory limits-colimits operads
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
$endgroup$
Let $D_d$ be the little $d$-disk operad as outlined in Fresse's book Homotopy of Operads and Grothendieck-Teichmuller Groups.
We have the sequence of inclusions of operads
$$D_1 to D_2 to cdots to D_n to cdots$$
The operad $D_infty$ was set as $colim_n D_n$.
I'm not sure about the notation $colim_n D_n$.
I'm assuming that it's the colimit of the diagram
$$D_1 to D_2 to cdots to D_n to cdots$$
Is it correct?
algebraic-topology category-theory limits-colimits operads
algebraic-topology category-theory limits-colimits operads
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
edited Feb 1 at 9:32
YuiTo Cheng
2,3694937
2,3694937
asked Feb 1 at 9:24
MathsMyMathsMy
491215
asked Feb 1 at 9:24
MathsMyMathsMy
491215
asked Feb 1 at 9:24
MathsMyMathsMy
491215
491215
$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26
add a comment |
$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26
$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26
$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it's the colimit of this diagram. Concretely, $D_infty(k)$ is a quotient of the disjoint union $bigsqcup_{n ge 1} D_n(k)$ under some equivalence relation. The equivalence relation is generated by the identification of $x in D_n(k)$ with its image in $D_{n+1}(k)$ under the inclusion map. Then you can check that the operadic structure maps are compatible with this equivalence relation (essentially because the inclusions $D_n to D_{n+1}$ are operad morphisms) and thus you get an operad structure on $D_infty$.
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes, it's the colimit of this diagram. Concretely, $D_infty(k)$ is a quotient of the disjoint union $bigsqcup_{n ge 1} D_n(k)$ under some equivalence relation. The equivalence relation is generated by the identification of $x in D_n(k)$ with its image in $D_{n+1}(k)$ under the inclusion map. Then you can check that the operadic structure maps are compatible with this equivalence relation (essentially because the inclusions $D_n to D_{n+1}$ are operad morphisms) and thus you get an operad structure on $D_infty$.
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
add a comment |
$begingroup$
Yes, it's the colimit of this diagram. Concretely, $D_infty(k)$ is a quotient of the disjoint union $bigsqcup_{n ge 1} D_n(k)$ under some equivalence relation. The equivalence relation is generated by the identification of $x in D_n(k)$ with its image in $D_{n+1}(k)$ under the inclusion map. Then you can check that the operadic structure maps are compatible with this equivalence relation (essentially because the inclusions $D_n to D_{n+1}$ are operad morphisms) and thus you get an operad structure on $D_infty$.
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
add a comment |
$begingroup$
Yes, it's the colimit of this diagram. Concretely, $D_infty(k)$ is a quotient of the disjoint union $bigsqcup_{n ge 1} D_n(k)$ under some equivalence relation. The equivalence relation is generated by the identification of $x in D_n(k)$ with its image in $D_{n+1}(k)$ under the inclusion map. Then you can check that the operadic structure maps are compatible with this equivalence relation (essentially because the inclusions $D_n to D_{n+1}$ are operad morphisms) and thus you get an operad structure on $D_infty$.
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
$endgroup$
Yes, it's the colimit of this diagram. Concretely, $D_infty(k)$ is a quotient of the disjoint union $bigsqcup_{n ge 1} D_n(k)$ under some equivalence relation. The equivalence relation is generated by the identification of $x in D_n(k)$ with its image in $D_{n+1}(k)$ under the inclusion map. Then you can check that the operadic structure maps are compatible with this equivalence relation (essentially because the inclusions $D_n to D_{n+1}$ are operad morphisms) and thus you get an operad structure on $D_infty$.
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
answered Feb 5 at 8:39
Najib IdrissiNajib Idrissi
41.9k473143
41.9k473143
add a comment |
add a comment |
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$begingroup$
That sounds about right.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:26