Mixing
Homotopy category of Chain complex - isomoprhism = quasi isomoprhism?
Let $Ho(Ch_Z)$ be the localization of the category of nonnegatively graded complexes of abelian groups, $Ch_Z$, wrt the quasi-isomorphisms.
If two objects are isomorphic, in this localization, then are they quasi isomoprhic? I do not think this is true. But why in page 43 of DS's notes, we have two objects in $A,B in Ho(Ch_Z)$ for which the author concludes
$$H_iA cong H_iB$$
I know this may be the case when $A,B$ are special objects. The problem is I cannot really tell if this is case in the notes.
algebraic-topology category-theory model-categories
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
|
show 2 more comments
Let $Ho(Ch_Z)$ be the localization of the category of nonnegatively graded complexes of abelian groups, $Ch_Z$, wrt the quasi-isomorphisms.
If two objects are isomorphic, in this localization, then are they quasi isomoprhic? I do not think this is true. But why in page 43 of DS's notes, we have two objects in $A,B in Ho(Ch_Z)$ for which the author concludes
$$H_iA cong H_iB$$
I know this may be the case when $A,B$ are special objects. The problem is I cannot really tell if this is case in the notes.
algebraic-topology category-theory model-categories
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
1
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
1
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
2
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40
|
show 2 more comments
Let $Ho(Ch_Z)$ be the localization of the category of nonnegatively graded complexes of abelian groups, $Ch_Z$, wrt the quasi-isomorphisms.
If two objects are isomorphic, in this localization, then are they quasi isomoprhic? I do not think this is true. But why in page 43 of DS's notes, we have two objects in $A,B in Ho(Ch_Z)$ for which the author concludes
$$H_iA cong H_iB$$
I know this may be the case when $A,B$ are special objects. The problem is I cannot really tell if this is case in the notes.
algebraic-topology category-theory model-categories
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
Let $Ho(Ch_Z)$ be the localization of the category of nonnegatively graded complexes of abelian groups, $Ch_Z$, wrt the quasi-isomorphisms.
If two objects are isomorphic, in this localization, then are they quasi isomoprhic? I do not think this is true. But why in page 43 of DS's notes, we have two objects in $A,B in Ho(Ch_Z)$ for which the author concludes
$$H_iA cong H_iB$$
I know this may be the case when $A,B$ are special objects. The problem is I cannot really tell if this is case in the notes.
algebraic-topology category-theory model-categories
algebraic-topology category-theory model-categories
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
asked Nov 22 '18 at 7:45
CL.CL.
2,1752822
2,1752822
Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
1
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
1
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
2
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40
|
show 2 more comments
Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
1
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
1
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
2
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40
Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
1
1
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
1
1
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
2
2
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40
|
show 2 more comments
1 Answer
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Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $mathbb{Z}$-modules):
$$ 0to 0to mathbb{Z}/2mathbb{Z} xrightarrow{0} mathbb{Z} xrightarrow{2} mathbb{Z} to 0 $$
and
$$0to mathbb{Z} xrightarrow{2} mathbb{Z} to 0to mathbb{Z}/2mathbb{Z}to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $mathbb{Z}/2mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to
$$0 to 0 to mathbb{Z}/2mathbb{Z} to 0 to mathbb{Z}/2mathbb{Z}to 0$$.
Regarding your question in the proof:
as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $mathcal{Ho}(mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
add a comment |
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Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $mathbb{Z}$-modules):
$$ 0to 0to mathbb{Z}/2mathbb{Z} xrightarrow{0} mathbb{Z} xrightarrow{2} mathbb{Z} to 0 $$
and
$$0to mathbb{Z} xrightarrow{2} mathbb{Z} to 0to mathbb{Z}/2mathbb{Z}to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $mathbb{Z}/2mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to
$$0 to 0 to mathbb{Z}/2mathbb{Z} to 0 to mathbb{Z}/2mathbb{Z}to 0$$.
Regarding your question in the proof:
as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $mathcal{Ho}(mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
add a comment |
Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $mathbb{Z}$-modules):
$$ 0to 0to mathbb{Z}/2mathbb{Z} xrightarrow{0} mathbb{Z} xrightarrow{2} mathbb{Z} to 0 $$
and
$$0to mathbb{Z} xrightarrow{2} mathbb{Z} to 0to mathbb{Z}/2mathbb{Z}to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $mathbb{Z}/2mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to
$$0 to 0 to mathbb{Z}/2mathbb{Z} to 0 to mathbb{Z}/2mathbb{Z}to 0$$.
Regarding your question in the proof:
as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $mathcal{Ho}(mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
add a comment |
Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $mathbb{Z}$-modules):
$$ 0to 0to mathbb{Z}/2mathbb{Z} xrightarrow{0} mathbb{Z} xrightarrow{2} mathbb{Z} to 0 $$
and
$$0to mathbb{Z} xrightarrow{2} mathbb{Z} to 0to mathbb{Z}/2mathbb{Z}to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $mathbb{Z}/2mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to
$$0 to 0 to mathbb{Z}/2mathbb{Z} to 0 to mathbb{Z}/2mathbb{Z}to 0$$.
Regarding your question in the proof:
as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $mathcal{Ho}(mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $mathbb{Z}$-modules):
$$ 0to 0to mathbb{Z}/2mathbb{Z} xrightarrow{0} mathbb{Z} xrightarrow{2} mathbb{Z} to 0 $$
and
$$0to mathbb{Z} xrightarrow{2} mathbb{Z} to 0to mathbb{Z}/2mathbb{Z}to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $mathbb{Z}/2mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to
$$0 to 0 to mathbb{Z}/2mathbb{Z} to 0 to mathbb{Z}/2mathbb{Z}to 0$$.
Regarding your question in the proof:
as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $mathcal{Ho}(mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
answered Nov 22 '18 at 8:37
EnkiduEnkidu
1,08819
1,08819
add a comment |
add a comment |
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Yes, it's true, pretty much by definition. Why don't you think it's true?
– Najib Idrissi
Nov 22 '18 at 8:01
I could not really find the problem in the notes you link, but maybe I am not familiar enough with the situation to see it. Anyway, note that homology factors through the localization with respect to quasi-isomorphisms since these induce isomorphisms on homology groups. Thus homology becomes a functor $Ho(Ch_Z) to Ab$ and so if two objects are isomorphic in $Ho(Ch_Z)$ they will have isomorphic homology. This does not imply the existence of a quasi-isomorphism between $A$ and $B$.
– Matthias Klupsch
Nov 22 '18 at 8:02
1
@NajibIdrissi : There could also be a zigzag sequence $A to C$ and $B to C$ of quasi-isomorphisms without the existence of a quasi-isomorphism $A to C$. I guess it depends on what 'quasi-isomorphic' is supposed to mean.
– Matthias Klupsch
Nov 22 '18 at 8:03
1
@NajibIdrissi I agree this is the "correct" definition of the notion of quasi-isomorphism for homotopy theoretic purposes, but I've almost always seen "quasi-isomorphism" defined to be a map $X to Y$ inducing isomorphisms on all homology groups.
– leibnewtz
Nov 22 '18 at 8:14
2
@leibnewtz You are confusing "quasi isomorphism" (a noun that denotes a certain class of maps) and "quasi isomorphic" (an adjective that applies to chain complexes).
– Najib Idrissi
Nov 22 '18 at 8:40