Mixing
Is a left homotopy inverse of a quasi-isomorphism automatically a right homotopy inverse?
$begingroup$
Let $R$ be an associative ring with unit. Let $E$ and $F$ be two chain complexes of $R$-modules and $phi: Eoverset{sim}{to} F$ be a quasi-isomorphism between them, i.e. $phi$ induces ismorphisms on cohomology modules at each degree.
We call $psi: Fto E$ a left homotopy inverse of $phi$ if there exist a degree $-1$ map $eta: Eto E$ such that $psicirc phi-text{id}_E=d eta+eta d$. Similarly we call $psi: Fto E$ a right homotopy inverse of $phi$ if there exist a degree $-1$ map $tau: Fto F$ such that $phicirc psi-text{id}_F=d tau+tau d$.
My question is: if we know $phi: Eoverset{sim}{to} F$ is a quasi-isomorphism and $psi: Fto E$ is a left homotopy inverse of $phi$, does it automatically mean that $psi$ is also a right homotopy inverse of $phi$? or vice versa?
abstract-algebra homological-algebra homotopy-theory
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
$endgroup$
add a comment |
$begingroup$
Let $R$ be an associative ring with unit. Let $E$ and $F$ be two chain complexes of $R$-modules and $phi: Eoverset{sim}{to} F$ be a quasi-isomorphism between them, i.e. $phi$ induces ismorphisms on cohomology modules at each degree.
We call $psi: Fto E$ a left homotopy inverse of $phi$ if there exist a degree $-1$ map $eta: Eto E$ such that $psicirc phi-text{id}_E=d eta+eta d$. Similarly we call $psi: Fto E$ a right homotopy inverse of $phi$ if there exist a degree $-1$ map $tau: Fto F$ such that $phicirc psi-text{id}_F=d tau+tau d$.
My question is: if we know $phi: Eoverset{sim}{to} F$ is a quasi-isomorphism and $psi: Fto E$ is a left homotopy inverse of $phi$, does it automatically mean that $psi$ is also a right homotopy inverse of $phi$? or vice versa?
abstract-algebra homological-algebra homotopy-theory
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
$endgroup$
$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45
add a comment |
$begingroup$
Let $R$ be an associative ring with unit. Let $E$ and $F$ be two chain complexes of $R$-modules and $phi: Eoverset{sim}{to} F$ be a quasi-isomorphism between them, i.e. $phi$ induces ismorphisms on cohomology modules at each degree.
We call $psi: Fto E$ a left homotopy inverse of $phi$ if there exist a degree $-1$ map $eta: Eto E$ such that $psicirc phi-text{id}_E=d eta+eta d$. Similarly we call $psi: Fto E$ a right homotopy inverse of $phi$ if there exist a degree $-1$ map $tau: Fto F$ such that $phicirc psi-text{id}_F=d tau+tau d$.
My question is: if we know $phi: Eoverset{sim}{to} F$ is a quasi-isomorphism and $psi: Fto E$ is a left homotopy inverse of $phi$, does it automatically mean that $psi$ is also a right homotopy inverse of $phi$? or vice versa?
abstract-algebra homological-algebra homotopy-theory
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
$endgroup$
Let $R$ be an associative ring with unit. Let $E$ and $F$ be two chain complexes of $R$-modules and $phi: Eoverset{sim}{to} F$ be a quasi-isomorphism between them, i.e. $phi$ induces ismorphisms on cohomology modules at each degree.
We call $psi: Fto E$ a left homotopy inverse of $phi$ if there exist a degree $-1$ map $eta: Eto E$ such that $psicirc phi-text{id}_E=d eta+eta d$. Similarly we call $psi: Fto E$ a right homotopy inverse of $phi$ if there exist a degree $-1$ map $tau: Fto F$ such that $phicirc psi-text{id}_F=d tau+tau d$.
My question is: if we know $phi: Eoverset{sim}{to} F$ is a quasi-isomorphism and $psi: Fto E$ is a left homotopy inverse of $phi$, does it automatically mean that $psi$ is also a right homotopy inverse of $phi$? or vice versa?
abstract-algebra homological-algebra homotopy-theory
abstract-algebra homological-algebra homotopy-theory
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
edited Jan 12 at 1:45
Zhaoting Wei
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
asked Jan 11 at 20:44
Zhaoting WeiZhaoting Wei
36417
36417
$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45
add a comment |
$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45
$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45
add a comment |
1 Answer
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$begingroup$
The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)
Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0to Ito Rto R/Ito 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)
Then the unique map $f:Eto F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $Fto E$, we have that $fcirc g= id_F$ so $f$ is a left homotopy inverse of $g$ but $gcirc f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.
answered Jan 11 at 20:54
MaxMax
14.3k11142
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1 Answer
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active
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1 Answer
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$begingroup$
The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)
Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0to Ito Rto R/Ito 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)
Then the unique map $f:Eto F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $Fto E$, we have that $fcirc g= id_F$ so $f$ is a left homotopy inverse of $g$ but $gcirc f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.
answered Jan 11 at 20:54
MaxMax
14.3k11142
$endgroup$
add a comment |
$begingroup$
The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)
Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0to Ito Rto R/Ito 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)
Then the unique map $f:Eto F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $Fto E$, we have that $fcirc g= id_F$ so $f$ is a left homotopy inverse of $g$ but $gcirc f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.
answered Jan 11 at 20:54
MaxMax
14.3k11142
$endgroup$
add a comment |
$begingroup$
The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)
Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0to Ito Rto R/Ito 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)
Then the unique map $f:Eto F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $Fto E$, we have that $fcirc g= id_F$ so $f$ is a left homotopy inverse of $g$ but $gcirc f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.
answered Jan 11 at 20:54
MaxMax
14.3k11142
$endgroup$
The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)
Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0to Ito Rto R/Ito 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)
Then the unique map $f:Eto F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $Fto E$, we have that $fcirc g= id_F$ so $f$ is a left homotopy inverse of $g$ but $gcirc f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.
answered Jan 11 at 20:54
MaxMax
14.3k11142
answered Jan 11 at 20:54
MaxMax
14.3k11142
answered Jan 11 at 20:54
MaxMax
14.3k11142
answered Jan 11 at 20:54
MaxMax
14.3k11142
14.3k11142
add a comment |
add a comment |
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$begingroup$
Isn't it $deta + eta d$ and $dtau + tau d$ ?
$endgroup$
– Max
Jan 11 at 20:49
$begingroup$
@Max Yes I have modified it.
$endgroup$
– Zhaoting Wei
Jan 12 at 1:45