Right homotopy relation on the projective model structure of chain complexes
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In Mark Hovey's book $textit{Model Categories}$ it is said that in the projective model structure of chain complexes, the right homotopy relation is precisely the
chain homotopy relation. However, I've found the following counter example:
Suppose $mathbb{Q}$ is the rational numbers, and $R=mathbb{Q}[x]/(x^2)$.
Define $A$ to be the complex with $A_n=R$ for any $n$, and the differential $d$ is given by the multiplication by $x$. Define $f$ to be the identity map on $A$ and $g$ to be the zero map on $A$. We deduce that:
i) $f$ and $g$ are not chain homotopic.
This is because that if there exists maps $s_n:Rto R$ such that $ds_n+s_{n-1}d=f-g$, suppose $s_n(1)=a+bx$, then $s_{n-1}(x)=1-ax$, but this cannot be true since $s_{n-1}$ must be a $R$-module homomorphism and must maps $x$ to a multiple of $x$.
ii) $f$ and $g$ are right homotopic.
This is because $A$ is acyclic and $Axrightarrow{(mathrm{id},mathrm{id})} Aoplus Axrightarrow{mathrm{id}}Aoplus A$ gives a path object of $A$,
and $Axrightarrow{(mathrm{id},0)}Aoplus Axrightarrow{mathrm{id}}Aoplus A$ yields the homotopy.
Is my counter example correct?
model-categories
asked Nov 22 at 13:05
Frank Kong
111
add a comment |
up vote
2
down vote
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In Mark Hovey's book $textit{Model Categories}$ it is said that in the projective model structure of chain complexes, the right homotopy relation is precisely the
chain homotopy relation. However, I've found the following counter example:
Suppose $mathbb{Q}$ is the rational numbers, and $R=mathbb{Q}[x]/(x^2)$.
Define $A$ to be the complex with $A_n=R$ for any $n$, and the differential $d$ is given by the multiplication by $x$. Define $f$ to be the identity map on $A$ and $g$ to be the zero map on $A$. We deduce that:
i) $f$ and $g$ are not chain homotopic.
This is because that if there exists maps $s_n:Rto R$ such that $ds_n+s_{n-1}d=f-g$, suppose $s_n(1)=a+bx$, then $s_{n-1}(x)=1-ax$, but this cannot be true since $s_{n-1}$ must be a $R$-module homomorphism and must maps $x$ to a multiple of $x$.
ii) $f$ and $g$ are right homotopic.
This is because $A$ is acyclic and $Axrightarrow{(mathrm{id},mathrm{id})} Aoplus Axrightarrow{mathrm{id}}Aoplus A$ gives a path object of $A$,
and $Axrightarrow{(mathrm{id},0)}Aoplus Axrightarrow{mathrm{id}}Aoplus A$ yields the homotopy.
Is my counter example correct?
model-categories
asked Nov 22 at 13:05
Frank Kong
111
most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In Mark Hovey's book $textit{Model Categories}$ it is said that in the projective model structure of chain complexes, the right homotopy relation is precisely the
chain homotopy relation. However, I've found the following counter example:
Suppose $mathbb{Q}$ is the rational numbers, and $R=mathbb{Q}[x]/(x^2)$.
Define $A$ to be the complex with $A_n=R$ for any $n$, and the differential $d$ is given by the multiplication by $x$. Define $f$ to be the identity map on $A$ and $g$ to be the zero map on $A$. We deduce that:
i) $f$ and $g$ are not chain homotopic.
This is because that if there exists maps $s_n:Rto R$ such that $ds_n+s_{n-1}d=f-g$, suppose $s_n(1)=a+bx$, then $s_{n-1}(x)=1-ax$, but this cannot be true since $s_{n-1}$ must be a $R$-module homomorphism and must maps $x$ to a multiple of $x$.
ii) $f$ and $g$ are right homotopic.
This is because $A$ is acyclic and $Axrightarrow{(mathrm{id},mathrm{id})} Aoplus Axrightarrow{mathrm{id}}Aoplus A$ gives a path object of $A$,
and $Axrightarrow{(mathrm{id},0)}Aoplus Axrightarrow{mathrm{id}}Aoplus A$ yields the homotopy.
Is my counter example correct?
model-categories
asked Nov 22 at 13:05
Frank Kong
111
In Mark Hovey's book $textit{Model Categories}$ it is said that in the projective model structure of chain complexes, the right homotopy relation is precisely the
chain homotopy relation. However, I've found the following counter example:
Suppose $mathbb{Q}$ is the rational numbers, and $R=mathbb{Q}[x]/(x^2)$.
Define $A$ to be the complex with $A_n=R$ for any $n$, and the differential $d$ is given by the multiplication by $x$. Define $f$ to be the identity map on $A$ and $g$ to be the zero map on $A$. We deduce that:
i) $f$ and $g$ are not chain homotopic.
This is because that if there exists maps $s_n:Rto R$ such that $ds_n+s_{n-1}d=f-g$, suppose $s_n(1)=a+bx$, then $s_{n-1}(x)=1-ax$, but this cannot be true since $s_{n-1}$ must be a $R$-module homomorphism and must maps $x$ to a multiple of $x$.
ii) $f$ and $g$ are right homotopic.
This is because $A$ is acyclic and $Axrightarrow{(mathrm{id},mathrm{id})} Aoplus Axrightarrow{mathrm{id}}Aoplus A$ gives a path object of $A$,
and $Axrightarrow{(mathrm{id},0)}Aoplus Axrightarrow{mathrm{id}}Aoplus A$ yields the homotopy.
Is my counter example correct?
model-categories
model-categories
asked Nov 22 at 13:05
Frank Kong
111
asked Nov 22 at 13:05
Frank Kong
111
asked Nov 22 at 13:05
Frank Kong
111
asked Nov 22 at 13:05
Frank Kong
111
asked Nov 22 at 13:05
Frank Kong
111
111
most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22
add a comment |
most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22
most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22
most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22
add a comment |
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most likely yes, however, you have the problem that your object is not cofibrant, hence your Example might work out as a counterexample if you use the full category, however, for defining the homotopy category, you first need to take a cofibrant replacement. (otherwise you also can't guarantee that right homotopy is an equivalence). Hence it is a counterexample, but a counterexample with no effect, and especially outside of the objects you are actually conisdering
– Enkidu
Nov 22 at 13:22