Mixing
Can you express this easy theorem in fancy categorical terms?
$begingroup$
Here is a theorem (of homological algebra):
Given $A rightarrow B rightarrow C$ in an abelian category
$mathcal{A}$.
If for all $D in mathcal{A}$ we have that $Hom(D,A) rightarrow
Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A
rightarrow B rightarrow C$ is an exact sequence.
This is a nice enough theorem that it feels like this can be expressed by saying that some certain representable functor is exact, or faithful, or a generator.
I tried using that an additive functor is faithful iff it sends nonexact sequences to nonexact sequences, but still haven't come up with anything.
category-theory homology-cohomology
asked Jan 23 at 13:26
SSFSSF
423111
$endgroup$
add a comment |
$begingroup$
Here is a theorem (of homological algebra):
Given $A rightarrow B rightarrow C$ in an abelian category
$mathcal{A}$.
If for all $D in mathcal{A}$ we have that $Hom(D,A) rightarrow
Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A
rightarrow B rightarrow C$ is an exact sequence.
This is a nice enough theorem that it feels like this can be expressed by saying that some certain representable functor is exact, or faithful, or a generator.
I tried using that an additive functor is faithful iff it sends nonexact sequences to nonexact sequences, but still haven't come up with anything.
category-theory homology-cohomology
asked Jan 23 at 13:26
SSFSSF
423111
$endgroup$
2
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
1
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
2
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37
add a comment |
$begingroup$
Here is a theorem (of homological algebra):
Given $A rightarrow B rightarrow C$ in an abelian category
$mathcal{A}$.
If for all $D in mathcal{A}$ we have that $Hom(D,A) rightarrow
Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A
rightarrow B rightarrow C$ is an exact sequence.
This is a nice enough theorem that it feels like this can be expressed by saying that some certain representable functor is exact, or faithful, or a generator.
I tried using that an additive functor is faithful iff it sends nonexact sequences to nonexact sequences, but still haven't come up with anything.
category-theory homology-cohomology
asked Jan 23 at 13:26
SSFSSF
423111
$endgroup$
Here is a theorem (of homological algebra):
Given $A rightarrow B rightarrow C$ in an abelian category
$mathcal{A}$.
If for all $D in mathcal{A}$ we have that $Hom(D,A) rightarrow
Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A
rightarrow B rightarrow C$ is an exact sequence.
This is a nice enough theorem that it feels like this can be expressed by saying that some certain representable functor is exact, or faithful, or a generator.
I tried using that an additive functor is faithful iff it sends nonexact sequences to nonexact sequences, but still haven't come up with anything.
category-theory homology-cohomology
category-theory homology-cohomology
asked Jan 23 at 13:26
SSFSSF
423111
asked Jan 23 at 13:26
SSFSSF
423111
asked Jan 23 at 13:26
SSFSSF
423111
asked Jan 23 at 13:26
SSFSSF
423111
asked Jan 23 at 13:26
SSFSSF
423111
423111
2
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
1
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
2
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37
add a comment |
2
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
1
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
2
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37
2
2
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
1
1
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
2
2
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm gonna have to go with @Malice on this one.
Recall that a functor $F$ is “representable”
precisely when $F ≅ mathsf{Hom}(D,-)$ for some object $D$.
Recall that a functor $F$ “reflects a property $P$”
precisely when $P(X) ;⇐; P(F, X)$ for all $X$.
( This’ the converse of property preservation. )
Recall that a family of functors consists of a functor $Fᵢ$
for each $i$ in some index set $I$.Recall that a family $Fᵢ$ “jointly reflects property $P$”
precisely when $P(X) ;⇐; (∀ i.; P(Fᵢ , x))$ for all $X$.
Now your phrase can be stated compactly,
“Representables jointly reflect exact ternary sequences”
Neato!
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
$endgroup$
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
add a comment |
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$begingroup$
I'm gonna have to go with @Malice on this one.
Recall that a functor $F$ is “representable”
precisely when $F ≅ mathsf{Hom}(D,-)$ for some object $D$.
Recall that a functor $F$ “reflects a property $P$”
precisely when $P(X) ;⇐; P(F, X)$ for all $X$.
( This’ the converse of property preservation. )
Recall that a family of functors consists of a functor $Fᵢ$
for each $i$ in some index set $I$.Recall that a family $Fᵢ$ “jointly reflects property $P$”
precisely when $P(X) ;⇐; (∀ i.; P(Fᵢ , x))$ for all $X$.
Now your phrase can be stated compactly,
“Representables jointly reflect exact ternary sequences”
Neato!
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
$endgroup$
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
add a comment |
$begingroup$
I'm gonna have to go with @Malice on this one.
Recall that a functor $F$ is “representable”
precisely when $F ≅ mathsf{Hom}(D,-)$ for some object $D$.
Recall that a functor $F$ “reflects a property $P$”
precisely when $P(X) ;⇐; P(F, X)$ for all $X$.
( This’ the converse of property preservation. )
Recall that a family of functors consists of a functor $Fᵢ$
for each $i$ in some index set $I$.Recall that a family $Fᵢ$ “jointly reflects property $P$”
precisely when $P(X) ;⇐; (∀ i.; P(Fᵢ , x))$ for all $X$.
Now your phrase can be stated compactly,
“Representables jointly reflect exact ternary sequences”
Neato!
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
$endgroup$
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
add a comment |
$begingroup$
I'm gonna have to go with @Malice on this one.
Recall that a functor $F$ is “representable”
precisely when $F ≅ mathsf{Hom}(D,-)$ for some object $D$.
Recall that a functor $F$ “reflects a property $P$”
precisely when $P(X) ;⇐; P(F, X)$ for all $X$.
( This’ the converse of property preservation. )
Recall that a family of functors consists of a functor $Fᵢ$
for each $i$ in some index set $I$.Recall that a family $Fᵢ$ “jointly reflects property $P$”
precisely when $P(X) ;⇐; (∀ i.; P(Fᵢ , x))$ for all $X$.
Now your phrase can be stated compactly,
“Representables jointly reflect exact ternary sequences”
Neato!
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
$endgroup$
I'm gonna have to go with @Malice on this one.
Recall that a functor $F$ is “representable”
precisely when $F ≅ mathsf{Hom}(D,-)$ for some object $D$.
Recall that a functor $F$ “reflects a property $P$”
precisely when $P(X) ;⇐; P(F, X)$ for all $X$.
( This’ the converse of property preservation. )
Recall that a family of functors consists of a functor $Fᵢ$
for each $i$ in some index set $I$.Recall that a family $Fᵢ$ “jointly reflects property $P$”
precisely when $P(X) ;⇐; (∀ i.; P(Fᵢ , x))$ for all $X$.
Now your phrase can be stated compactly,
“Representables jointly reflect exact ternary sequences”
Neato!
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
answered Jan 24 at 15:49
Musa Al-hassyMusa Al-hassy
1,3331711
1,3331711
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
add a comment |
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
$begingroup$
That's nice, I accept it. Have you seen "jointly reflects" defined anywhere in the literature before?
$endgroup$
– SSF
Jan 24 at 21:31
1
1
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
$begingroup$
I'm just extrapolating, but consider comparing with legitimate phrases such as "jointly monic" in which each member of the family must satisfy the monic premise in order for the monic conclusion to hold..
$endgroup$
– Musa Al-hassy
Jan 24 at 23:20
1
1
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
I actually googled this because I was sure I had just made it up on the spot, but "jointly reflects" has been used is basically this sense in the literature before. It looks like our sense of reasonable terminology is vindicated.
$endgroup$
– Malice Vidrine
Jan 24 at 23:39
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
$begingroup$
sweet! way to go MV!
$endgroup$
– Musa Al-hassy
Jan 25 at 15:36
add a comment |
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2
$begingroup$
"$mathsf{Hom}(D,-)$ reflects exact sequences of length 3"? ;-)
$endgroup$
– Musa Al-hassy
Jan 23 at 16:32
1
$begingroup$
You mean for a fixed choice of $D$? That can't be it, right? To say that for some fixed $D$, $Hom(D,-)$ reflects exacts sequences of length 3 would be the following statement: "Given $D$ and $A rightarrow B rightarrow C$ in an abelian category $mathcal{A}$, if $Hom(D,A) rightarrow Hom(D,B) rightarrow Hom(D,C)$ is an exact sequence, then $A rightarrow B rightarrow C$ is an exact sequence." And that's surely not a true statement. Whereas the theorem I gave is really a true statement, because it asks for the premise to hold over all $D$.
$endgroup$
– SSF
Jan 23 at 18:50
2
$begingroup$
"Representables jointly reflect exact sequences"? Something in that neighborhood?
$endgroup$
– Malice Vidrine
Jan 24 at 0:37