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Localization of a model category
Clash Royale CLAN TAG#URR8PPP
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Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: Xrightarrow Y$ and $g: Xrightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Zrightarrow Ycup_X Z $ is a weak equivalence in the localized model category $mathrmL_ fM$. Is it true that $mathrmL_ fM=mathrmL_ rM$ ?
at.algebraic-topology homotopy-theory model-categories
asked Aug 31 at 10:34
Amadeus
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Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: Xrightarrow Y$ and $g: Xrightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Zrightarrow Ycup_X Z $ is a weak equivalence in the localized model category $mathrmL_ fM$. Is it true that $mathrmL_ fM=mathrmL_ rM$ ?
at.algebraic-topology homotopy-theory model-categories
asked Aug 31 at 10:34
Amadeus
734
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: Xrightarrow Y$ and $g: Xrightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Zrightarrow Ycup_X Z $ is a weak equivalence in the localized model category $mathrmL_ fM$. Is it true that $mathrmL_ fM=mathrmL_ rM$ ?
at.algebraic-topology homotopy-theory model-categories
asked Aug 31 at 10:34
Amadeus
734
Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: Xrightarrow Y$ and $g: Xrightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Zrightarrow Ycup_X Z $ is a weak equivalence in the localized model category $mathrmL_ fM$. Is it true that $mathrmL_ fM=mathrmL_ rM$ ?
at.algebraic-topology homotopy-theory model-categories
asked Aug 31 at 10:34
Amadeus
734
asked Aug 31 at 10:34
Amadeus
734
asked Aug 31 at 10:34
Amadeus
734
asked Aug 31 at 10:34
Amadeus
734
734
add a comment |Â
add a comment |Â
1 Answer
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No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $Delta^1$ with its endpoints identified and let $f : Delta^1 to S^1$ be the obvious map. Let $g : Delta^1 to Delta^0$ be the unique map. Then $r : Delta^0 to Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_rM = M$, but $L_fM neq M$ since $S^1$ is not equivalent to $Delta^1$ in $M$.
answered Aug 31 at 10:51
Valery Isaev
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $Delta^1$ with its endpoints identified and let $f : Delta^1 to S^1$ be the obvious map. Let $g : Delta^1 to Delta^0$ be the unique map. Then $r : Delta^0 to Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_rM = M$, but $L_fM neq M$ since $S^1$ is not equivalent to $Delta^1$ in $M$.
answered Aug 31 at 10:51
Valery Isaev
1,8111822
add a comment |Â
up vote
5
down vote
accepted
No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $Delta^1$ with its endpoints identified and let $f : Delta^1 to S^1$ be the obvious map. Let $g : Delta^1 to Delta^0$ be the unique map. Then $r : Delta^0 to Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_rM = M$, but $L_fM neq M$ since $S^1$ is not equivalent to $Delta^1$ in $M$.
answered Aug 31 at 10:51
Valery Isaev
1,8111822
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $Delta^1$ with its endpoints identified and let $f : Delta^1 to S^1$ be the obvious map. Let $g : Delta^1 to Delta^0$ be the unique map. Then $r : Delta^0 to Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_rM = M$, but $L_fM neq M$ since $S^1$ is not equivalent to $Delta^1$ in $M$.
answered Aug 31 at 10:51
Valery Isaev
1,8111822
No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $Delta^1$ with its endpoints identified and let $f : Delta^1 to S^1$ be the obvious map. Let $g : Delta^1 to Delta^0$ be the unique map. Then $r : Delta^0 to Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_rM = M$, but $L_fM neq M$ since $S^1$ is not equivalent to $Delta^1$ in $M$.
answered Aug 31 at 10:51
Valery Isaev
1,8111822
answered Aug 31 at 10:51
Valery Isaev
1,8111822
answered Aug 31 at 10:51
Valery Isaev
1,8111822
answered Aug 31 at 10:51
Valery Isaev
1,8111822
1,8111822
add a comment |Â
add a comment |Â
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