Mixing
partial zero morphisms
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I cannot understand the proof of Theorem 2 of this paper.
Definition: A morphism $alpha$ is coconstant if for any $f$ and $g$, $alpha circ f =alpha circ g$. It is constant if $f circ alpha = g circ alpha$. It is zero if constant and coconstant.
Definition: A quasi-zero structure basically means that you have zero morphisms but not in the entire category in the full subcategory of it.
Theorem: A category C which has an initial object 0 admits a quasi-zero structure if and only if for each object X in the category there exists at most one morphism X → 0. In this case, the zero morphisms are those morphisms which factor through the initial object. If the category C also has a terminal object 1, then the existence of a quasi-zero structure is equivalent to the unique morphism 0 → 1 being a monomorphism.
If-part of the first part is clear, for the only-if part I can show it is constant but cannot show it is coconstant to conclude it is zero!
Constant part:
$$
X xrightarrowalpha 0 xrightarrowbetaY xrightarrowf, gZ
$$
Therefore, $0_X, Y=beta circ alpha$ which is constant, $0 xrightarrow fcircbeta= gcircbetaZ $ becauase $0$ is initial.
Question: I cannot show $beta circ alpha$ is coconstant.
category-theory
asked Aug 2 at 12:26
Mobius Knot
12
 |Â
show 2 more comments
up vote
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I cannot understand the proof of Theorem 2 of this paper.
Definition: A morphism $alpha$ is coconstant if for any $f$ and $g$, $alpha circ f =alpha circ g$. It is constant if $f circ alpha = g circ alpha$. It is zero if constant and coconstant.
Definition: A quasi-zero structure basically means that you have zero morphisms but not in the entire category in the full subcategory of it.
Theorem: A category C which has an initial object 0 admits a quasi-zero structure if and only if for each object X in the category there exists at most one morphism X → 0. In this case, the zero morphisms are those morphisms which factor through the initial object. If the category C also has a terminal object 1, then the existence of a quasi-zero structure is equivalent to the unique morphism 0 → 1 being a monomorphism.
If-part of the first part is clear, for the only-if part I can show it is constant but cannot show it is coconstant to conclude it is zero!
Constant part:
$$
X xrightarrowalpha 0 xrightarrowbetaY xrightarrowf, gZ
$$
Therefore, $0_X, Y=beta circ alpha$ which is constant, $0 xrightarrow fcircbeta= gcircbetaZ $ becauase $0$ is initial.
Question: I cannot show $beta circ alpha$ is coconstant.
category-theory
asked Aug 2 at 12:26
Mobius Knot
12
How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
I edited my question.
– Mobius Knot
Aug 2 at 15:42
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17
 |Â
show 2 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I cannot understand the proof of Theorem 2 of this paper.
Definition: A morphism $alpha$ is coconstant if for any $f$ and $g$, $alpha circ f =alpha circ g$. It is constant if $f circ alpha = g circ alpha$. It is zero if constant and coconstant.
Definition: A quasi-zero structure basically means that you have zero morphisms but not in the entire category in the full subcategory of it.
Theorem: A category C which has an initial object 0 admits a quasi-zero structure if and only if for each object X in the category there exists at most one morphism X → 0. In this case, the zero morphisms are those morphisms which factor through the initial object. If the category C also has a terminal object 1, then the existence of a quasi-zero structure is equivalent to the unique morphism 0 → 1 being a monomorphism.
If-part of the first part is clear, for the only-if part I can show it is constant but cannot show it is coconstant to conclude it is zero!
Constant part:
$$
X xrightarrowalpha 0 xrightarrowbetaY xrightarrowf, gZ
$$
Therefore, $0_X, Y=beta circ alpha$ which is constant, $0 xrightarrow fcircbeta= gcircbetaZ $ becauase $0$ is initial.
Question: I cannot show $beta circ alpha$ is coconstant.
category-theory
asked Aug 2 at 12:26
Mobius Knot
12
I cannot understand the proof of Theorem 2 of this paper.
Definition: A morphism $alpha$ is coconstant if for any $f$ and $g$, $alpha circ f =alpha circ g$. It is constant if $f circ alpha = g circ alpha$. It is zero if constant and coconstant.
Definition: A quasi-zero structure basically means that you have zero morphisms but not in the entire category in the full subcategory of it.
Theorem: A category C which has an initial object 0 admits a quasi-zero structure if and only if for each object X in the category there exists at most one morphism X → 0. In this case, the zero morphisms are those morphisms which factor through the initial object. If the category C also has a terminal object 1, then the existence of a quasi-zero structure is equivalent to the unique morphism 0 → 1 being a monomorphism.
If-part of the first part is clear, for the only-if part I can show it is constant but cannot show it is coconstant to conclude it is zero!
Constant part:
$$
X xrightarrowalpha 0 xrightarrowbetaY xrightarrowf, gZ
$$
Therefore, $0_X, Y=beta circ alpha$ which is constant, $0 xrightarrow fcircbeta= gcircbetaZ $ becauase $0$ is initial.
Question: I cannot show $beta circ alpha$ is coconstant.
category-theory
asked Aug 2 at 12:26
Mobius Knot
12
edited Aug 3 at 12:53
asked Aug 2 at 12:26
Mobius Knot
12
asked Aug 2 at 12:26
Mobius Knot
12
asked Aug 2 at 12:26
Mobius Knot
12
12
How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
I edited my question.
– Mobius Knot
Aug 2 at 15:42
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17
 |Â
show 2 more comments
How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
I edited my question.
– Mobius Knot
Aug 2 at 15:42
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17
How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
I edited my question.
– Mobius Knot
Aug 2 at 15:42
I edited my question.
– Mobius Knot
Aug 2 at 15:42
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17
 |Â
show 2 more comments
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How is defined a quasi-zero structure?
– Fabio Lucchini
Aug 2 at 12:59
So if I understand correctly, your question is why $betacircalpha circ f$ will be the unique zero map $Zto Y$, for any $f:Zto Y$?
– Arnaud D.
Aug 2 at 13:00
I edited my question.
– Mobius Knot
Aug 2 at 15:42
if $u,v:Wto X$ are two parallel morphisms, then by assumption $alpha u = alpha v$ (since there is at most one morphism to $0$ from any object).
– Nex
Aug 3 at 12:15
You seem to have mixed constant and coconstant in either your definition or in your proof...
– Nex
Aug 3 at 12:17