Mixing
Equivalent definitions of adjunction morphisms
$begingroup$
I am struggling with the following exercise from Emily Riehl's Category Theory in Context, regarding adjunction morphisms: paraphrasing, let $F : C to D, G: D to C$ and $F' : C' to D', G' : D' to C'$ be functors such that $F dashv G$ and $F' dashv G'$. We now consider functors $H : C to C'$ and $K : D to D'$ such that the squares of adjunctions commute, that is so that $HG = G'K$ and $KF = F'H$. The exercise then asks to prove that the following are equivalent:
- if $eta, eta'$ are the units of the adjunctions, then $H eta = eta' H$, i.e. $Heta_c = eta'_{Hc}$.
- if $epsilon, epsilon'$ are the counits of the adjunctions, then $Kepsilon = epsilon'K$.
- the arrow compositions
$$
D(Fc,d) xrightarrow{simeq} C(c,Gd) xrightarrow{H} C(Hc,HGd) = C(Hc,G'Kd)
$$
and
$$
D(Fc,d) xrightarrow{K} D(KFc,Kd) = D(F'Hc,Kd) xrightarrow{simeq} C(Hc, G'Kd)
$$
are equal.
I've though about this for a fair amount of time, trying to use the adjunction relations between transposes/adjunts and the commutativity relations, with no luck: the only implication I have managed to prove is $(3) Rightarrow (1)$, via tracking $1_{Fc}$ along both arrows, which give $Heta_c$ and $eta'_{Hc}$ respectively.
Any hints?
category-theory adjoint-functors
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
$endgroup$
add a comment |
$begingroup$
I am struggling with the following exercise from Emily Riehl's Category Theory in Context, regarding adjunction morphisms: paraphrasing, let $F : C to D, G: D to C$ and $F' : C' to D', G' : D' to C'$ be functors such that $F dashv G$ and $F' dashv G'$. We now consider functors $H : C to C'$ and $K : D to D'$ such that the squares of adjunctions commute, that is so that $HG = G'K$ and $KF = F'H$. The exercise then asks to prove that the following are equivalent:
- if $eta, eta'$ are the units of the adjunctions, then $H eta = eta' H$, i.e. $Heta_c = eta'_{Hc}$.
- if $epsilon, epsilon'$ are the counits of the adjunctions, then $Kepsilon = epsilon'K$.
- the arrow compositions
$$
D(Fc,d) xrightarrow{simeq} C(c,Gd) xrightarrow{H} C(Hc,HGd) = C(Hc,G'Kd)
$$
and
$$
D(Fc,d) xrightarrow{K} D(KFc,Kd) = D(F'Hc,Kd) xrightarrow{simeq} C(Hc, G'Kd)
$$
are equal.
I've though about this for a fair amount of time, trying to use the adjunction relations between transposes/adjunts and the commutativity relations, with no luck: the only implication I have managed to prove is $(3) Rightarrow (1)$, via tracking $1_{Fc}$ along both arrows, which give $Heta_c$ and $eta'_{Hc}$ respectively.
Any hints?
category-theory adjoint-functors
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
$endgroup$
2
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44
add a comment |
$begingroup$
I am struggling with the following exercise from Emily Riehl's Category Theory in Context, regarding adjunction morphisms: paraphrasing, let $F : C to D, G: D to C$ and $F' : C' to D', G' : D' to C'$ be functors such that $F dashv G$ and $F' dashv G'$. We now consider functors $H : C to C'$ and $K : D to D'$ such that the squares of adjunctions commute, that is so that $HG = G'K$ and $KF = F'H$. The exercise then asks to prove that the following are equivalent:
- if $eta, eta'$ are the units of the adjunctions, then $H eta = eta' H$, i.e. $Heta_c = eta'_{Hc}$.
- if $epsilon, epsilon'$ are the counits of the adjunctions, then $Kepsilon = epsilon'K$.
- the arrow compositions
$$
D(Fc,d) xrightarrow{simeq} C(c,Gd) xrightarrow{H} C(Hc,HGd) = C(Hc,G'Kd)
$$
and
$$
D(Fc,d) xrightarrow{K} D(KFc,Kd) = D(F'Hc,Kd) xrightarrow{simeq} C(Hc, G'Kd)
$$
are equal.
I've though about this for a fair amount of time, trying to use the adjunction relations between transposes/adjunts and the commutativity relations, with no luck: the only implication I have managed to prove is $(3) Rightarrow (1)$, via tracking $1_{Fc}$ along both arrows, which give $Heta_c$ and $eta'_{Hc}$ respectively.
Any hints?
category-theory adjoint-functors
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
$endgroup$
I am struggling with the following exercise from Emily Riehl's Category Theory in Context, regarding adjunction morphisms: paraphrasing, let $F : C to D, G: D to C$ and $F' : C' to D', G' : D' to C'$ be functors such that $F dashv G$ and $F' dashv G'$. We now consider functors $H : C to C'$ and $K : D to D'$ such that the squares of adjunctions commute, that is so that $HG = G'K$ and $KF = F'H$. The exercise then asks to prove that the following are equivalent:
- if $eta, eta'$ are the units of the adjunctions, then $H eta = eta' H$, i.e. $Heta_c = eta'_{Hc}$.
- if $epsilon, epsilon'$ are the counits of the adjunctions, then $Kepsilon = epsilon'K$.
- the arrow compositions
$$
D(Fc,d) xrightarrow{simeq} C(c,Gd) xrightarrow{H} C(Hc,HGd) = C(Hc,G'Kd)
$$
and
$$
D(Fc,d) xrightarrow{K} D(KFc,Kd) = D(F'Hc,Kd) xrightarrow{simeq} C(Hc, G'Kd)
$$
are equal.
I've though about this for a fair amount of time, trying to use the adjunction relations between transposes/adjunts and the commutativity relations, with no luck: the only implication I have managed to prove is $(3) Rightarrow (1)$, via tracking $1_{Fc}$ along both arrows, which give $Heta_c$ and $eta'_{Hc}$ respectively.
Any hints?
category-theory adjoint-functors
category-theory adjoint-functors
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
edited Mar 5 at 12:51
Arnaud D.
16.2k52445
16.2k52445
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
asked Jan 29 at 13:18
Guido A.Guido A.
8,0951730
8,0951730
2
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44
add a comment |
2
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44
2
2
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint with another perspective:
An adjunction $Fdashv G$ with $F:Cto D$ can be represented as a category $mathcal F$ containing (disjoint isomorphic copies of) $C$ and $D$ with a further arrow $tildedelta:cto d$ for each $delta:Fcto d$.
Due to the adjunction, $C$ is a coreflective full subcategory, while $D$ is a reflective full subcategory of $mathcal F$.
Now the conditions imply that $H$ and $K$ induces a functor $mathcal H:mathcal Ftomathcal F'$, and both maps in 3. describe the action of this functor on the newly added homsets.
answered Jan 29 at 14:29
BerciBerci
61.8k23674
$endgroup$
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
add a comment |
$begingroup$
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) iff (3)$ one can similarly prove $(2) iff (3)$.
Having said that, $(1) iff (3)$ follows from the fact that the transpose of an arrow $f : Fc to d$ can be defined (or characterized) as the composition $eta_cGf$. Suppose that $(1)$ holds and let $f : Fc to d$ be an arrow. Now, commutativity of the diagram amounts to showing
$$
H(eta_cGf) = eta'_{Hc}G'Kf tag{$star$}
$$
since the transpose of $Kf : F'Hc = KFc to d$ is, via the same remark, $eta'_{Hc}G'Kf$. In effect,
$$
H(eta_cGf) = Heta_c HGf stackrel{(1)}{=} eta'_{Hc}HGf = eta'_{Hc}G'Kf
$$
as desired.
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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$begingroup$
Hint with another perspective:
An adjunction $Fdashv G$ with $F:Cto D$ can be represented as a category $mathcal F$ containing (disjoint isomorphic copies of) $C$ and $D$ with a further arrow $tildedelta:cto d$ for each $delta:Fcto d$.
Due to the adjunction, $C$ is a coreflective full subcategory, while $D$ is a reflective full subcategory of $mathcal F$.
Now the conditions imply that $H$ and $K$ induces a functor $mathcal H:mathcal Ftomathcal F'$, and both maps in 3. describe the action of this functor on the newly added homsets.
answered Jan 29 at 14:29
BerciBerci
61.8k23674
$endgroup$
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
add a comment |
$begingroup$
Hint with another perspective:
An adjunction $Fdashv G$ with $F:Cto D$ can be represented as a category $mathcal F$ containing (disjoint isomorphic copies of) $C$ and $D$ with a further arrow $tildedelta:cto d$ for each $delta:Fcto d$.
Due to the adjunction, $C$ is a coreflective full subcategory, while $D$ is a reflective full subcategory of $mathcal F$.
Now the conditions imply that $H$ and $K$ induces a functor $mathcal H:mathcal Ftomathcal F'$, and both maps in 3. describe the action of this functor on the newly added homsets.
answered Jan 29 at 14:29
BerciBerci
61.8k23674
$endgroup$
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
add a comment |
$begingroup$
Hint with another perspective:
An adjunction $Fdashv G$ with $F:Cto D$ can be represented as a category $mathcal F$ containing (disjoint isomorphic copies of) $C$ and $D$ with a further arrow $tildedelta:cto d$ for each $delta:Fcto d$.
Due to the adjunction, $C$ is a coreflective full subcategory, while $D$ is a reflective full subcategory of $mathcal F$.
Now the conditions imply that $H$ and $K$ induces a functor $mathcal H:mathcal Ftomathcal F'$, and both maps in 3. describe the action of this functor on the newly added homsets.
answered Jan 29 at 14:29
BerciBerci
61.8k23674
$endgroup$
Hint with another perspective:
An adjunction $Fdashv G$ with $F:Cto D$ can be represented as a category $mathcal F$ containing (disjoint isomorphic copies of) $C$ and $D$ with a further arrow $tildedelta:cto d$ for each $delta:Fcto d$.
Due to the adjunction, $C$ is a coreflective full subcategory, while $D$ is a reflective full subcategory of $mathcal F$.
Now the conditions imply that $H$ and $K$ induces a functor $mathcal H:mathcal Ftomathcal F'$, and both maps in 3. describe the action of this functor on the newly added homsets.
answered Jan 29 at 14:29
BerciBerci
61.8k23674
answered Jan 29 at 14:29
BerciBerci
61.8k23674
answered Jan 29 at 14:29
BerciBerci
61.8k23674
answered Jan 29 at 14:29
BerciBerci
61.8k23674
61.8k23674
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
add a comment |
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
$begingroup$
This is a bit different from what I was thinking, and also more advanced than my current abilities on the subject, as I have not studied (co) reflective subcategories yet. I will give this another read later on. Thanks (once again) for taking the time to answer :)
$endgroup$
– Guido A.
Jan 29 at 18:41
add a comment |
$begingroup$
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) iff (3)$ one can similarly prove $(2) iff (3)$.
Having said that, $(1) iff (3)$ follows from the fact that the transpose of an arrow $f : Fc to d$ can be defined (or characterized) as the composition $eta_cGf$. Suppose that $(1)$ holds and let $f : Fc to d$ be an arrow. Now, commutativity of the diagram amounts to showing
$$
H(eta_cGf) = eta'_{Hc}G'Kf tag{$star$}
$$
since the transpose of $Kf : F'Hc = KFc to d$ is, via the same remark, $eta'_{Hc}G'Kf$. In effect,
$$
H(eta_cGf) = Heta_c HGf stackrel{(1)}{=} eta'_{Hc}HGf = eta'_{Hc}G'Kf
$$
as desired.
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
$endgroup$
add a comment |
$begingroup$
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) iff (3)$ one can similarly prove $(2) iff (3)$.
Having said that, $(1) iff (3)$ follows from the fact that the transpose of an arrow $f : Fc to d$ can be defined (or characterized) as the composition $eta_cGf$. Suppose that $(1)$ holds and let $f : Fc to d$ be an arrow. Now, commutativity of the diagram amounts to showing
$$
H(eta_cGf) = eta'_{Hc}G'Kf tag{$star$}
$$
since the transpose of $Kf : F'Hc = KFc to d$ is, via the same remark, $eta'_{Hc}G'Kf$. In effect,
$$
H(eta_cGf) = Heta_c HGf stackrel{(1)}{=} eta'_{Hc}HGf = eta'_{Hc}G'Kf
$$
as desired.
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
$endgroup$
add a comment |
$begingroup$
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) iff (3)$ one can similarly prove $(2) iff (3)$.
Having said that, $(1) iff (3)$ follows from the fact that the transpose of an arrow $f : Fc to d$ can be defined (or characterized) as the composition $eta_cGf$. Suppose that $(1)$ holds and let $f : Fc to d$ be an arrow. Now, commutativity of the diagram amounts to showing
$$
H(eta_cGf) = eta'_{Hc}G'Kf tag{$star$}
$$
since the transpose of $Kf : F'Hc = KFc to d$ is, via the same remark, $eta'_{Hc}G'Kf$. In effect,
$$
H(eta_cGf) = Heta_c HGf stackrel{(1)}{=} eta'_{Hc}HGf = eta'_{Hc}G'Kf
$$
as desired.
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
$endgroup$
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) iff (3)$ one can similarly prove $(2) iff (3)$.
Having said that, $(1) iff (3)$ follows from the fact that the transpose of an arrow $f : Fc to d$ can be defined (or characterized) as the composition $eta_cGf$. Suppose that $(1)$ holds and let $f : Fc to d$ be an arrow. Now, commutativity of the diagram amounts to showing
$$
H(eta_cGf) = eta'_{Hc}G'Kf tag{$star$}
$$
since the transpose of $Kf : F'Hc = KFc to d$ is, via the same remark, $eta'_{Hc}G'Kf$. In effect,
$$
H(eta_cGf) = Heta_c HGf stackrel{(1)}{=} eta'_{Hc}HGf = eta'_{Hc}G'Kf
$$
as desired.
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
answered Jan 29 at 18:38
Guido A.Guido A.
8,0951730
8,0951730
add a comment |
add a comment |
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2
$begingroup$
Write out the isomorphisms of homsets explicitly in terms of units and counits.
$endgroup$
– Derek Elkins
Jan 29 at 13:23
$begingroup$
@DerekElkins Thanks for the idea, I think I was using (co) units as black boxes a bit too much. I believe I have managed to conclude the result via your hint, and would really appreciate a sanity check. Cheers.
$endgroup$
– Guido A.
Jan 29 at 18:44