simplicial objects in a model category
$begingroup$
Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?
ct.category-theory homotopy-theory model-categories
asked 10 hours ago
ParisParis
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add a comment |
$begingroup$
Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?
ct.category-theory homotopy-theory model-categories
asked 10 hours ago
ParisParis
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?
ct.category-theory homotopy-theory model-categories
asked 10 hours ago
ParisParis
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?
ct.category-theory homotopy-theory model-categories
ct.category-theory homotopy-theory model-categories
asked 10 hours ago
ParisParis
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
ParisParis
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
ParisParis
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
ParisParis
233
asked 10 hours ago
ParisParis
233
233
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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No, it is not.
If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
$endgroup$
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
add a comment |
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1 Answer
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$begingroup$
No, it is not.
If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
$endgroup$
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
add a comment |
$begingroup$
No, it is not.
If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
$endgroup$
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
add a comment |
$begingroup$
No, it is not.
If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
$endgroup$
No, it is not.
If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
answered 8 hours ago
Mike ShulmanMike Shulman
36.8k483226
36.8k483226
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
add a comment |
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
$begingroup$
I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
$endgroup$
– Paris
8 hours ago
2
2
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
$begingroup$
@Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
$endgroup$
– Mike Shulman
5 hours ago
add a comment |
Paris is a new contributor. Be nice, and check out our Code of Conduct.
Paris is a new contributor. Be nice, and check out our Code of Conduct.
Paris is a new contributor. Be nice, and check out our Code of Conduct.
Paris is a new contributor. Be nice, and check out our Code of Conduct.
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