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Why are finite cell complexes also finite as infinity-categories?
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A quasicategory ($infty$-category) $mathcalC$ is finite if there is a finite simplicial set $K$ and a categorical equivalence $KrightarrowmathcalC$.
On the other hand, a Kan complex (space) $X$ is finite if there is a finite simplicial set $K$ and a weak homotopy equivalence $Krightarrow X$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes.
Question: Now suppose $X$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $X$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof?
To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $K$ and a weak homotopy equivalence $f:Krightarrow S^1$. But $f$ is not a categorical equivalence. To build a finite model for the $infty$-category $S^1$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other.
higher-category-theory
asked 1 hour ago
John Berman
39919
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up vote
5
down vote
favorite
A quasicategory ($infty$-category) $mathcalC$ is finite if there is a finite simplicial set $K$ and a categorical equivalence $KrightarrowmathcalC$.
On the other hand, a Kan complex (space) $X$ is finite if there is a finite simplicial set $K$ and a weak homotopy equivalence $Krightarrow X$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes.
Question: Now suppose $X$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $X$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof?
To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $K$ and a weak homotopy equivalence $f:Krightarrow S^1$. But $f$ is not a categorical equivalence. To build a finite model for the $infty$-category $S^1$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other.
higher-category-theory
asked 1 hour ago
John Berman
39919
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
A quasicategory ($infty$-category) $mathcalC$ is finite if there is a finite simplicial set $K$ and a categorical equivalence $KrightarrowmathcalC$.
On the other hand, a Kan complex (space) $X$ is finite if there is a finite simplicial set $K$ and a weak homotopy equivalence $Krightarrow X$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes.
Question: Now suppose $X$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $X$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof?
To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $K$ and a weak homotopy equivalence $f:Krightarrow S^1$. But $f$ is not a categorical equivalence. To build a finite model for the $infty$-category $S^1$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other.
higher-category-theory
asked 1 hour ago
John Berman
39919
A quasicategory ($infty$-category) $mathcalC$ is finite if there is a finite simplicial set $K$ and a categorical equivalence $KrightarrowmathcalC$.
On the other hand, a Kan complex (space) $X$ is finite if there is a finite simplicial set $K$ and a weak homotopy equivalence $Krightarrow X$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes.
Question: Now suppose $X$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $X$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof?
To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $K$ and a weak homotopy equivalence $f:Krightarrow S^1$. But $f$ is not a categorical equivalence. To build a finite model for the $infty$-category $S^1$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other.
higher-category-theory
higher-category-theory
asked 1 hour ago
John Berman
39919
asked 1 hour ago
John Berman
39919
asked 1 hour ago
John Berman
39919
asked 1 hour ago
John Berman
39919
asked 1 hour ago
John Berman
39919
39919
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add a comment |Â
1 Answer
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Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K hookrightarrow K' rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g circ f => 1$ and $f circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' hookrightarrow Y rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $Lambda^0 [2] hookrightarrow Delta[2]$ and one by a $Lambda^2 [2] hookrightarrow Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f circ g => 1$ and $h circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.
answered 1 hour ago
Simon Henry
14k14480
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K hookrightarrow K' rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g circ f => 1$ and $f circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' hookrightarrow Y rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $Lambda^0 [2] hookrightarrow Delta[2]$ and one by a $Lambda^2 [2] hookrightarrow Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f circ g => 1$ and $h circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.
answered 1 hour ago
Simon Henry
14k14480
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
add a comment |Â
up vote
5
down vote
accepted
Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K hookrightarrow K' rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g circ f => 1$ and $f circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' hookrightarrow Y rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $Lambda^0 [2] hookrightarrow Delta[2]$ and one by a $Lambda^2 [2] hookrightarrow Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f circ g => 1$ and $h circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.
answered 1 hour ago
Simon Henry
14k14480
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K hookrightarrow K' rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g circ f => 1$ and $f circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' hookrightarrow Y rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $Lambda^0 [2] hookrightarrow Delta[2]$ and one by a $Lambda^2 [2] hookrightarrow Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f circ g => 1$ and $h circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.
answered 1 hour ago
Simon Henry
14k14480
Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K hookrightarrow K' rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g circ f => 1$ and $f circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' hookrightarrow Y rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $Lambda^0 [2] hookrightarrow Delta[2]$ and one by a $Lambda^2 [2] hookrightarrow Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f circ g => 1$ and $h circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.
answered 1 hour ago
Simon Henry
14k14480
edited 25 mins ago
answered 1 hour ago
Simon Henry
14k14480
answered 1 hour ago
Simon Henry
14k14480
answered 1 hour ago
Simon Henry
14k14480
14k14480
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
add a comment |Â
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
There is something I don't understand. Start with the simplicial set $Delta^1$ as a model for a contractible space. What are the pushouts of outer horn inclusions? It seems to me that you end up building $S^infty$, which is not finite.
– John Berman
37 mins ago
1
1
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
@JohnBerman : You are indeed right there was a small problem, I have clarified the precise construction that needs to be done at the end.
– Simon Henry
25 mins ago
add a comment |Â
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