Why is epimorphism not defined as follows?
$begingroup$
Epimorphism in category theory, can be seen as a generalization of "surjective function" in Set.
But we could also have characterized surjective functions in Set as follows:
Definition (*). $f:Ato B$ is surjective if for all $Z$, and all $g:Zto B$, there is a $g_f:Zto A$ such that $g=fcirc g_f$.
In Set this definition characterized surjective functions, just as the standard definition for epimorphisms does. But in general categories, this definition is independent from the epimorphism definition (neither implies the other).
Why choose the standard epimorphism definition over (*), or over some other definition that matches in Set, as a generalization of surjective function? What makes the epimorphism definition desirable?
EDIT: Similarly for monomorphism: We can characterize injective functions in Set as:
Definition (&). $f:Ato B$ is injective, if for all $Z$ and all $g:Ato Z$, there is a $g_f:Bto Z$ such that $g=g_fcirc f$.
category-theory definition
$endgroup$
|
show 3 more comments
$begingroup$
Epimorphism in category theory, can be seen as a generalization of "surjective function" in Set.
But we could also have characterized surjective functions in Set as follows:
Definition (*). $f:Ato B$ is surjective if for all $Z$, and all $g:Zto B$, there is a $g_f:Zto A$ such that $g=fcirc g_f$.
In Set this definition characterized surjective functions, just as the standard definition for epimorphisms does. But in general categories, this definition is independent from the epimorphism definition (neither implies the other).
Why choose the standard epimorphism definition over (*), or over some other definition that matches in Set, as a generalization of surjective function? What makes the epimorphism definition desirable?
EDIT: Similarly for monomorphism: We can characterize injective functions in Set as:
Definition (&). $f:Ato B$ is injective, if for all $Z$ and all $g:Ato Z$, there is a $g_f:Bto Z$ such that $g=g_fcirc f$.
category-theory definition
$endgroup$
$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
1
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago
|
show 3 more comments
$begingroup$
Epimorphism in category theory, can be seen as a generalization of "surjective function" in Set.
But we could also have characterized surjective functions in Set as follows:
Definition (*). $f:Ato B$ is surjective if for all $Z$, and all $g:Zto B$, there is a $g_f:Zto A$ such that $g=fcirc g_f$.
In Set this definition characterized surjective functions, just as the standard definition for epimorphisms does. But in general categories, this definition is independent from the epimorphism definition (neither implies the other).
Why choose the standard epimorphism definition over (*), or over some other definition that matches in Set, as a generalization of surjective function? What makes the epimorphism definition desirable?
EDIT: Similarly for monomorphism: We can characterize injective functions in Set as:
Definition (&). $f:Ato B$ is injective, if for all $Z$ and all $g:Ato Z$, there is a $g_f:Bto Z$ such that $g=g_fcirc f$.
category-theory definition
$endgroup$
Epimorphism in category theory, can be seen as a generalization of "surjective function" in Set.
But we could also have characterized surjective functions in Set as follows:
Definition (*). $f:Ato B$ is surjective if for all $Z$, and all $g:Zto B$, there is a $g_f:Zto A$ such that $g=fcirc g_f$.
In Set this definition characterized surjective functions, just as the standard definition for epimorphisms does. But in general categories, this definition is independent from the epimorphism definition (neither implies the other).
Why choose the standard epimorphism definition over (*), or over some other definition that matches in Set, as a generalization of surjective function? What makes the epimorphism definition desirable?
EDIT: Similarly for monomorphism: We can characterize injective functions in Set as:
Definition (&). $f:Ato B$ is injective, if for all $Z$ and all $g:Ato Z$, there is a $g_f:Bto Z$ such that $g=g_fcirc f$.
category-theory definition
category-theory definition
edited 2 hours ago
J W
2,0631628
2,0631628
asked 9 hours ago
user56834user56834
3,39821252
3,39821252
$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
1
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago
|
show 3 more comments
$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
1
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago
$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
1
1
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
You definition corresponds to the concept of split epimorphism, which is stronger of that epimorphism.
However, when the category has a projective generator, then epimorphisms can be characterized with a similar concept to being surjective:
In a category with a projective generator $Z$, a morphism $f:Xto Y$ is an epimorphisms if and only if for all $y:Zto Y$ there exists $x:Zto X$ such that $y=xf$ (morphisms composition in diagramatic order).
Recall that an object $Z$ is a generator if for each pair of distinct parallel
morphisms $f,g:Xto Y$ there exists a morphisms $x:Zto X$ such that $xfneq xg$.
An object $Z$ is projective if and only if for each epimorphism $e:Xto Y$ and each morphism $y:Zto Y$ there exists a morphism $x:Zto X$ such that $y=xe$.
This condition is fullfilled, for example:
- in the category of sets by taking ${varnothing}$ as projective generator;
- in the cateogry of modules over a ring taking the ring itself as projective generator;
- in the category of groups taking $Bbb Z$ as projective generator.
The only if part follows since $Z$ is projective.
For the if part follows arguing by contradiction: if $f$ is not an epimorphism, then there exists a pair of distinct parallel arrows $u,v:Yto W$ such that $fu=fv$.

Since $Z$ is a generator, there exists $y:Zto Y$ such that $yuneq yv$.
Let $x:Zto X$ such that $xf=y$.
Then
$$yu=xfu=xfv=yv$$
a contradiction.
In that case, epimorphisms are also pullback-stable (see here).
$endgroup$
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
Take a look at concrete categories, like groups and topological spaces for instance. In those categories, you can define a notion of surjectivity (resp. injectivity) by using the one from set, because you have a notion of underlying set-map.
Surely, any generalization of surjectivity (resp. injectivity) you have should be satisfied by those maps as well.
However, there are (tons, actually) some maps that have surjective (resp. injective) underlying set-maps, but that do not satisfy your condition: there are epimorphisms (monomorphisms) that are not split (your definition is that of a split epimorphism/monomorphism). Therefore, even if your condition is an interesting one, it can't be used to generalize surjectivity (resp.injectivity)
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You definition corresponds to the concept of split epimorphism, which is stronger of that epimorphism.
However, when the category has a projective generator, then epimorphisms can be characterized with a similar concept to being surjective:
In a category with a projective generator $Z$, a morphism $f:Xto Y$ is an epimorphisms if and only if for all $y:Zto Y$ there exists $x:Zto X$ such that $y=xf$ (morphisms composition in diagramatic order).
Recall that an object $Z$ is a generator if for each pair of distinct parallel
morphisms $f,g:Xto Y$ there exists a morphisms $x:Zto X$ such that $xfneq xg$.
An object $Z$ is projective if and only if for each epimorphism $e:Xto Y$ and each morphism $y:Zto Y$ there exists a morphism $x:Zto X$ such that $y=xe$.
This condition is fullfilled, for example:
- in the category of sets by taking ${varnothing}$ as projective generator;
- in the cateogry of modules over a ring taking the ring itself as projective generator;
- in the category of groups taking $Bbb Z$ as projective generator.
The only if part follows since $Z$ is projective.
For the if part follows arguing by contradiction: if $f$ is not an epimorphism, then there exists a pair of distinct parallel arrows $u,v:Yto W$ such that $fu=fv$.

Since $Z$ is a generator, there exists $y:Zto Y$ such that $yuneq yv$.
Let $x:Zto X$ such that $xf=y$.
Then
$$yu=xfu=xfv=yv$$
a contradiction.
In that case, epimorphisms are also pullback-stable (see here).
$endgroup$
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
You definition corresponds to the concept of split epimorphism, which is stronger of that epimorphism.
However, when the category has a projective generator, then epimorphisms can be characterized with a similar concept to being surjective:
In a category with a projective generator $Z$, a morphism $f:Xto Y$ is an epimorphisms if and only if for all $y:Zto Y$ there exists $x:Zto X$ such that $y=xf$ (morphisms composition in diagramatic order).
Recall that an object $Z$ is a generator if for each pair of distinct parallel
morphisms $f,g:Xto Y$ there exists a morphisms $x:Zto X$ such that $xfneq xg$.
An object $Z$ is projective if and only if for each epimorphism $e:Xto Y$ and each morphism $y:Zto Y$ there exists a morphism $x:Zto X$ such that $y=xe$.
This condition is fullfilled, for example:
- in the category of sets by taking ${varnothing}$ as projective generator;
- in the cateogry of modules over a ring taking the ring itself as projective generator;
- in the category of groups taking $Bbb Z$ as projective generator.
The only if part follows since $Z$ is projective.
For the if part follows arguing by contradiction: if $f$ is not an epimorphism, then there exists a pair of distinct parallel arrows $u,v:Yto W$ such that $fu=fv$.

Since $Z$ is a generator, there exists $y:Zto Y$ such that $yuneq yv$.
Let $x:Zto X$ such that $xf=y$.
Then
$$yu=xfu=xfv=yv$$
a contradiction.
In that case, epimorphisms are also pullback-stable (see here).
$endgroup$
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
You definition corresponds to the concept of split epimorphism, which is stronger of that epimorphism.
However, when the category has a projective generator, then epimorphisms can be characterized with a similar concept to being surjective:
In a category with a projective generator $Z$, a morphism $f:Xto Y$ is an epimorphisms if and only if for all $y:Zto Y$ there exists $x:Zto X$ such that $y=xf$ (morphisms composition in diagramatic order).
Recall that an object $Z$ is a generator if for each pair of distinct parallel
morphisms $f,g:Xto Y$ there exists a morphisms $x:Zto X$ such that $xfneq xg$.
An object $Z$ is projective if and only if for each epimorphism $e:Xto Y$ and each morphism $y:Zto Y$ there exists a morphism $x:Zto X$ such that $y=xe$.
This condition is fullfilled, for example:
- in the category of sets by taking ${varnothing}$ as projective generator;
- in the cateogry of modules over a ring taking the ring itself as projective generator;
- in the category of groups taking $Bbb Z$ as projective generator.
The only if part follows since $Z$ is projective.
For the if part follows arguing by contradiction: if $f$ is not an epimorphism, then there exists a pair of distinct parallel arrows $u,v:Yto W$ such that $fu=fv$.

Since $Z$ is a generator, there exists $y:Zto Y$ such that $yuneq yv$.
Let $x:Zto X$ such that $xf=y$.
Then
$$yu=xfu=xfv=yv$$
a contradiction.
In that case, epimorphisms are also pullback-stable (see here).
$endgroup$
You definition corresponds to the concept of split epimorphism, which is stronger of that epimorphism.
However, when the category has a projective generator, then epimorphisms can be characterized with a similar concept to being surjective:
In a category with a projective generator $Z$, a morphism $f:Xto Y$ is an epimorphisms if and only if for all $y:Zto Y$ there exists $x:Zto X$ such that $y=xf$ (morphisms composition in diagramatic order).
Recall that an object $Z$ is a generator if for each pair of distinct parallel
morphisms $f,g:Xto Y$ there exists a morphisms $x:Zto X$ such that $xfneq xg$.
An object $Z$ is projective if and only if for each epimorphism $e:Xto Y$ and each morphism $y:Zto Y$ there exists a morphism $x:Zto X$ such that $y=xe$.
This condition is fullfilled, for example:
- in the category of sets by taking ${varnothing}$ as projective generator;
- in the cateogry of modules over a ring taking the ring itself as projective generator;
- in the category of groups taking $Bbb Z$ as projective generator.
The only if part follows since $Z$ is projective.
For the if part follows arguing by contradiction: if $f$ is not an epimorphism, then there exists a pair of distinct parallel arrows $u,v:Yto W$ such that $fu=fv$.

Since $Z$ is a generator, there exists $y:Zto Y$ such that $yuneq yv$.
Let $x:Zto X$ such that $xf=y$.
Then
$$yu=xfu=xfv=yv$$
a contradiction.
In that case, epimorphisms are also pullback-stable (see here).
edited 7 hours ago
answered 9 hours ago
Fabio LucchiniFabio Lucchini
8,79311426
8,79311426
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
You might want to clarify for the OP and other readers that you write compositions in diagrammatic order.
$endgroup$
– Arnaud D.
8 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
$begingroup$
@ArnaudD. Fixed, thank'you
$endgroup$
– Fabio Lucchini
7 hours ago
3
3
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
It's worth mentioning that split epimorphism is a much stronger concept than epimorphism. Split epimorphisms are absolute, meaning they are preserved by all functors. Also, the existence of a splitting for all surjective functions in $mathbf{Set}$ is equivalent to the Axiom of Choice. Indeed, "all epimorphisms split" is usually used as the categorical formulation of the Axiom of Choice. There are many other kinds of epimorphisms between (plain) epimorphisms and split epimorphisms.
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
Take a look at concrete categories, like groups and topological spaces for instance. In those categories, you can define a notion of surjectivity (resp. injectivity) by using the one from set, because you have a notion of underlying set-map.
Surely, any generalization of surjectivity (resp. injectivity) you have should be satisfied by those maps as well.
However, there are (tons, actually) some maps that have surjective (resp. injective) underlying set-maps, but that do not satisfy your condition: there are epimorphisms (monomorphisms) that are not split (your definition is that of a split epimorphism/monomorphism). Therefore, even if your condition is an interesting one, it can't be used to generalize surjectivity (resp.injectivity)
$endgroup$
add a comment |
$begingroup$
Take a look at concrete categories, like groups and topological spaces for instance. In those categories, you can define a notion of surjectivity (resp. injectivity) by using the one from set, because you have a notion of underlying set-map.
Surely, any generalization of surjectivity (resp. injectivity) you have should be satisfied by those maps as well.
However, there are (tons, actually) some maps that have surjective (resp. injective) underlying set-maps, but that do not satisfy your condition: there are epimorphisms (monomorphisms) that are not split (your definition is that of a split epimorphism/monomorphism). Therefore, even if your condition is an interesting one, it can't be used to generalize surjectivity (resp.injectivity)
$endgroup$
add a comment |
$begingroup$
Take a look at concrete categories, like groups and topological spaces for instance. In those categories, you can define a notion of surjectivity (resp. injectivity) by using the one from set, because you have a notion of underlying set-map.
Surely, any generalization of surjectivity (resp. injectivity) you have should be satisfied by those maps as well.
However, there are (tons, actually) some maps that have surjective (resp. injective) underlying set-maps, but that do not satisfy your condition: there are epimorphisms (monomorphisms) that are not split (your definition is that of a split epimorphism/monomorphism). Therefore, even if your condition is an interesting one, it can't be used to generalize surjectivity (resp.injectivity)
$endgroup$
Take a look at concrete categories, like groups and topological spaces for instance. In those categories, you can define a notion of surjectivity (resp. injectivity) by using the one from set, because you have a notion of underlying set-map.
Surely, any generalization of surjectivity (resp. injectivity) you have should be satisfied by those maps as well.
However, there are (tons, actually) some maps that have surjective (resp. injective) underlying set-maps, but that do not satisfy your condition: there are epimorphisms (monomorphisms) that are not split (your definition is that of a split epimorphism/monomorphism). Therefore, even if your condition is an interesting one, it can't be used to generalize surjectivity (resp.injectivity)
answered 8 hours ago
MaxMax
15.1k11143
15.1k11143
add a comment |
add a comment |
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$begingroup$
It's completely categorical?
$endgroup$
– Lord Shark the Unknown
9 hours ago
$begingroup$
@LordSharktheUnknown, my definitions are categorical as well?
$endgroup$
– user56834
9 hours ago
1
$begingroup$
The existence of $g_f$ for every $f:Zto B$ is equivalent to existence of a section $g:Bto A$.
$endgroup$
– Julian Rosen
9 hours ago
$begingroup$
@JulianRosen, your type signatures are different from mine so I'm not sure what you're saying.
$endgroup$
– user56834
9 hours ago
$begingroup$
@LeeMosher, my definition is not set-theoretic though........... My definition is purely categorical, and is a generalization of surjective functions JUST like the definition of epimorphism is a categorical generalization of surjective functions.
$endgroup$
– user56834
9 hours ago