Mixing
Epimorphism in the functor category $[mathbf{C}^{op}, mathbf{Set}]$
up vote
4
down vote
favorite
Let $mathbf{C}$ be a small category and $alpha: F Rightarrow G$ be a morphism in the functor category $Func(mathbf{C}^{op}, mathbf{Set})$. How do I prove that $alpha$ is an epimorphism if and only if for all $A in mathbf{C}$ we have that $alpha_A : FA rightarrow GA$ is surjetive? Any tips?
category-theory
edited 2 days ago
Oskar
2,6961718
asked 2 days ago
user435800
1518
add a comment |
up vote
4
down vote
favorite
Let $mathbf{C}$ be a small category and $alpha: F Rightarrow G$ be a morphism in the functor category $Func(mathbf{C}^{op}, mathbf{Set})$. How do I prove that $alpha$ is an epimorphism if and only if for all $A in mathbf{C}$ we have that $alpha_A : FA rightarrow GA$ is surjetive? Any tips?
category-theory
edited 2 days ago
Oskar
2,6961718
asked 2 days ago
user435800
1518
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $mathbf{C}$ be a small category and $alpha: F Rightarrow G$ be a morphism in the functor category $Func(mathbf{C}^{op}, mathbf{Set})$. How do I prove that $alpha$ is an epimorphism if and only if for all $A in mathbf{C}$ we have that $alpha_A : FA rightarrow GA$ is surjetive? Any tips?
category-theory
edited 2 days ago
Oskar
2,6961718
asked 2 days ago
user435800
1518
Let $mathbf{C}$ be a small category and $alpha: F Rightarrow G$ be a morphism in the functor category $Func(mathbf{C}^{op}, mathbf{Set})$. How do I prove that $alpha$ is an epimorphism if and only if for all $A in mathbf{C}$ we have that $alpha_A : FA rightarrow GA$ is surjetive? Any tips?
category-theory
category-theory
edited 2 days ago
Oskar
2,6961718
asked 2 days ago
user435800
1518
edited 2 days ago
Oskar
2,6961718
asked 2 days ago
user435800
1518
edited 2 days ago
Oskar
2,6961718
edited 2 days ago
Oskar
2,6961718
edited 2 days ago
Oskar
2,6961718
2,6961718
asked 2 days ago
user435800
1518
asked 2 days ago
user435800
1518
asked 2 days ago
user435800
1518
1518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let $mathcal{A}$ and $mathcal{B}$ be categories, $mathcal{F},mathcal{G}colonmathcal{A}tomathcal{B}$ be functors, $alphacolonmathcal{F}tomathcal{G}$ be a natural transformation.
If $alpha(a)$ is an epimorphism for every $aintext{Obj}(A)$, then $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$. It is an easy exercise.
If $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$ and $mathcal{B}$ is finitely cocomplete, then $alpha(a)$ is an epimorphism in $mathcal{B}$ for every $aintext{Obj}(mathcal{A})$. For every object $aintext{Obj}(mathcal{A})$ denote by $Delta_acolonmathbf{1}tomathcal{A}$ such functor, that $Delta_a(0)=a$. Note, that if $mathcal{B}$ is finitely cocomplete, then the inverse image functor $mathcal{B}^{Delta_a}colonmathcal{B}^{mathcal{A}}tomathcal{B}^{mathbf{1}}$ is right exact. Therefore, the evaluation functor $text{ev}_acolonmathcal{B}^{mathcal{A}}tomathcal{B}$, such that $text{ev}_a(mathcal{T})=mathcal{T}(a)$ for every $mathcal{T}intext{Func}(mathcal{A},mathcal{B})$, which is isomorphic to $mathcal{B}^{Delta_a}$, preserves epimorphisms.
Then it is sufficient to note that $mathbf{Set}$ is finitely cocomplete.
Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.
answered 2 days ago
Oskar
2,6961718
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
|
show 2 more comments
up vote
2
down vote
Here is a direct calculational argument..
begin{align*}
quad& text{ each } α_A : F,A → G;A text{ is epic } \
≡ & color{green}{{text{ Definition of epic }}} \
& quad ∀ A • ∀ g,h • quad h ∘ α_A = g ∘ α_A ⇒ h = g \
≡ & color{green}{{text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }}} \
&quad ∀ A • ∀ η,ε • quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \
≡ &color{green}{{text{ Composition of natural transformations }}} \
&quad ∀ A • ∀ η,ε • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \
≡ & color{green}{{text{ Quantifier Nesting }}} & \
& quad ∀ η,ε • ∀ A • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\
Rightarrow & color{green}{{text{ Quantifier distributivity }}} \
& quad ∀ η,ε • left(∀ A •; (η ∘ α)_A = (ε ∘ α)_A right) ⇒ left(∀ A •; η_A = ε_Aright) \
≡ & color{green}{{text{ Extensionality }}} \
& quad ∀ η,ε •quad η ∘ α = ε ∘ α ⇒ η = ε \
≡ & color{green}{{text{ Definition of epic }}} \
& quad α text{ is epic }
end{align*}
Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.
answered yesterday
Musa Al-hassy
1,2171711
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $mathcal{A}$ and $mathcal{B}$ be categories, $mathcal{F},mathcal{G}colonmathcal{A}tomathcal{B}$ be functors, $alphacolonmathcal{F}tomathcal{G}$ be a natural transformation.
If $alpha(a)$ is an epimorphism for every $aintext{Obj}(A)$, then $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$. It is an easy exercise.
If $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$ and $mathcal{B}$ is finitely cocomplete, then $alpha(a)$ is an epimorphism in $mathcal{B}$ for every $aintext{Obj}(mathcal{A})$. For every object $aintext{Obj}(mathcal{A})$ denote by $Delta_acolonmathbf{1}tomathcal{A}$ such functor, that $Delta_a(0)=a$. Note, that if $mathcal{B}$ is finitely cocomplete, then the inverse image functor $mathcal{B}^{Delta_a}colonmathcal{B}^{mathcal{A}}tomathcal{B}^{mathbf{1}}$ is right exact. Therefore, the evaluation functor $text{ev}_acolonmathcal{B}^{mathcal{A}}tomathcal{B}$, such that $text{ev}_a(mathcal{T})=mathcal{T}(a)$ for every $mathcal{T}intext{Func}(mathcal{A},mathcal{B})$, which is isomorphic to $mathcal{B}^{Delta_a}$, preserves epimorphisms.
Then it is sufficient to note that $mathbf{Set}$ is finitely cocomplete.
Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.
answered 2 days ago
Oskar
2,6961718
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
|
show 2 more comments
up vote
3
down vote
accepted
Let $mathcal{A}$ and $mathcal{B}$ be categories, $mathcal{F},mathcal{G}colonmathcal{A}tomathcal{B}$ be functors, $alphacolonmathcal{F}tomathcal{G}$ be a natural transformation.
If $alpha(a)$ is an epimorphism for every $aintext{Obj}(A)$, then $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$. It is an easy exercise.
If $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$ and $mathcal{B}$ is finitely cocomplete, then $alpha(a)$ is an epimorphism in $mathcal{B}$ for every $aintext{Obj}(mathcal{A})$. For every object $aintext{Obj}(mathcal{A})$ denote by $Delta_acolonmathbf{1}tomathcal{A}$ such functor, that $Delta_a(0)=a$. Note, that if $mathcal{B}$ is finitely cocomplete, then the inverse image functor $mathcal{B}^{Delta_a}colonmathcal{B}^{mathcal{A}}tomathcal{B}^{mathbf{1}}$ is right exact. Therefore, the evaluation functor $text{ev}_acolonmathcal{B}^{mathcal{A}}tomathcal{B}$, such that $text{ev}_a(mathcal{T})=mathcal{T}(a)$ for every $mathcal{T}intext{Func}(mathcal{A},mathcal{B})$, which is isomorphic to $mathcal{B}^{Delta_a}$, preserves epimorphisms.
Then it is sufficient to note that $mathbf{Set}$ is finitely cocomplete.
Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.
answered 2 days ago
Oskar
2,6961718
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
|
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $mathcal{A}$ and $mathcal{B}$ be categories, $mathcal{F},mathcal{G}colonmathcal{A}tomathcal{B}$ be functors, $alphacolonmathcal{F}tomathcal{G}$ be a natural transformation.
If $alpha(a)$ is an epimorphism for every $aintext{Obj}(A)$, then $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$. It is an easy exercise.
If $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$ and $mathcal{B}$ is finitely cocomplete, then $alpha(a)$ is an epimorphism in $mathcal{B}$ for every $aintext{Obj}(mathcal{A})$. For every object $aintext{Obj}(mathcal{A})$ denote by $Delta_acolonmathbf{1}tomathcal{A}$ such functor, that $Delta_a(0)=a$. Note, that if $mathcal{B}$ is finitely cocomplete, then the inverse image functor $mathcal{B}^{Delta_a}colonmathcal{B}^{mathcal{A}}tomathcal{B}^{mathbf{1}}$ is right exact. Therefore, the evaluation functor $text{ev}_acolonmathcal{B}^{mathcal{A}}tomathcal{B}$, such that $text{ev}_a(mathcal{T})=mathcal{T}(a)$ for every $mathcal{T}intext{Func}(mathcal{A},mathcal{B})$, which is isomorphic to $mathcal{B}^{Delta_a}$, preserves epimorphisms.
Then it is sufficient to note that $mathbf{Set}$ is finitely cocomplete.
Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.
answered 2 days ago
Oskar
2,6961718
Let $mathcal{A}$ and $mathcal{B}$ be categories, $mathcal{F},mathcal{G}colonmathcal{A}tomathcal{B}$ be functors, $alphacolonmathcal{F}tomathcal{G}$ be a natural transformation.
If $alpha(a)$ is an epimorphism for every $aintext{Obj}(A)$, then $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$. It is an easy exercise.
If $alpha$ is an epimorphism in $text{Func}(mathcal{A},mathcal{B})$ and $mathcal{B}$ is finitely cocomplete, then $alpha(a)$ is an epimorphism in $mathcal{B}$ for every $aintext{Obj}(mathcal{A})$. For every object $aintext{Obj}(mathcal{A})$ denote by $Delta_acolonmathbf{1}tomathcal{A}$ such functor, that $Delta_a(0)=a$. Note, that if $mathcal{B}$ is finitely cocomplete, then the inverse image functor $mathcal{B}^{Delta_a}colonmathcal{B}^{mathcal{A}}tomathcal{B}^{mathbf{1}}$ is right exact. Therefore, the evaluation functor $text{ev}_acolonmathcal{B}^{mathcal{A}}tomathcal{B}$, such that $text{ev}_a(mathcal{T})=mathcal{T}(a)$ for every $mathcal{T}intext{Func}(mathcal{A},mathcal{B})$, which is isomorphic to $mathcal{B}^{Delta_a}$, preserves epimorphisms.
Then it is sufficient to note that $mathbf{Set}$ is finitely cocomplete.
Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.
answered 2 days ago
Oskar
2,6961718
edited 2 days ago
answered 2 days ago
Oskar
2,6961718
answered 2 days ago
Oskar
2,6961718
answered 2 days ago
Oskar
2,6961718
2,6961718
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
|
show 2 more comments
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
– user435800
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
@user435800 $mathcal{A}$ may be arbitraty.
– Oskar
2 days ago
1
1
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
@user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
– Oskar
2 days ago
1
1
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
@user435800 It is a consequence of the isomorphism $mathcal{B}cong mathcal{B}^{mathbf{1}}$. Functors from the trivial category to $mathcal{B}$ "are" objects of $mathcal{B}$. And it should be noted, that they are not isomorphic in a functor category (they have different codomains), but in the arrow category of $mathbf{Cat}$.
– Oskar
2 days ago
|
show 2 more comments
up vote
2
down vote
Here is a direct calculational argument..
begin{align*}
quad& text{ each } α_A : F,A → G;A text{ is epic } \
≡ & color{green}{{text{ Definition of epic }}} \
& quad ∀ A • ∀ g,h • quad h ∘ α_A = g ∘ α_A ⇒ h = g \
≡ & color{green}{{text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }}} \
&quad ∀ A • ∀ η,ε • quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \
≡ &color{green}{{text{ Composition of natural transformations }}} \
&quad ∀ A • ∀ η,ε • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \
≡ & color{green}{{text{ Quantifier Nesting }}} & \
& quad ∀ η,ε • ∀ A • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\
Rightarrow & color{green}{{text{ Quantifier distributivity }}} \
& quad ∀ η,ε • left(∀ A •; (η ∘ α)_A = (ε ∘ α)_A right) ⇒ left(∀ A •; η_A = ε_Aright) \
≡ & color{green}{{text{ Extensionality }}} \
& quad ∀ η,ε •quad η ∘ α = ε ∘ α ⇒ η = ε \
≡ & color{green}{{text{ Definition of epic }}} \
& quad α text{ is epic }
end{align*}
Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.
answered yesterday
Musa Al-hassy
1,2171711
add a comment |
up vote
2
down vote
Here is a direct calculational argument..
begin{align*}
quad& text{ each } α_A : F,A → G;A text{ is epic } \
≡ & color{green}{{text{ Definition of epic }}} \
& quad ∀ A • ∀ g,h • quad h ∘ α_A = g ∘ α_A ⇒ h = g \
≡ & color{green}{{text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }}} \
&quad ∀ A • ∀ η,ε • quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \
≡ &color{green}{{text{ Composition of natural transformations }}} \
&quad ∀ A • ∀ η,ε • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \
≡ & color{green}{{text{ Quantifier Nesting }}} & \
& quad ∀ η,ε • ∀ A • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\
Rightarrow & color{green}{{text{ Quantifier distributivity }}} \
& quad ∀ η,ε • left(∀ A •; (η ∘ α)_A = (ε ∘ α)_A right) ⇒ left(∀ A •; η_A = ε_Aright) \
≡ & color{green}{{text{ Extensionality }}} \
& quad ∀ η,ε •quad η ∘ α = ε ∘ α ⇒ η = ε \
≡ & color{green}{{text{ Definition of epic }}} \
& quad α text{ is epic }
end{align*}
Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.
answered yesterday
Musa Al-hassy
1,2171711
add a comment |
up vote
2
down vote
up vote
2
down vote
Here is a direct calculational argument..
begin{align*}
quad& text{ each } α_A : F,A → G;A text{ is epic } \
≡ & color{green}{{text{ Definition of epic }}} \
& quad ∀ A • ∀ g,h • quad h ∘ α_A = g ∘ α_A ⇒ h = g \
≡ & color{green}{{text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }}} \
&quad ∀ A • ∀ η,ε • quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \
≡ &color{green}{{text{ Composition of natural transformations }}} \
&quad ∀ A • ∀ η,ε • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \
≡ & color{green}{{text{ Quantifier Nesting }}} & \
& quad ∀ η,ε • ∀ A • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\
Rightarrow & color{green}{{text{ Quantifier distributivity }}} \
& quad ∀ η,ε • left(∀ A •; (η ∘ α)_A = (ε ∘ α)_A right) ⇒ left(∀ A •; η_A = ε_Aright) \
≡ & color{green}{{text{ Extensionality }}} \
& quad ∀ η,ε •quad η ∘ α = ε ∘ α ⇒ η = ε \
≡ & color{green}{{text{ Definition of epic }}} \
& quad α text{ is epic }
end{align*}
Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.
answered yesterday
Musa Al-hassy
1,2171711
Here is a direct calculational argument..
begin{align*}
quad& text{ each } α_A : F,A → G;A text{ is epic } \
≡ & color{green}{{text{ Definition of epic }}} \
& quad ∀ A • ∀ g,h • quad h ∘ α_A = g ∘ α_A ⇒ h = g \
≡ & color{green}{{text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }}} \
&quad ∀ A • ∀ η,ε • quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \
≡ &color{green}{{text{ Composition of natural transformations }}} \
&quad ∀ A • ∀ η,ε • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \
≡ & color{green}{{text{ Quantifier Nesting }}} & \
& quad ∀ η,ε • ∀ A • quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\
Rightarrow & color{green}{{text{ Quantifier distributivity }}} \
& quad ∀ η,ε • left(∀ A •; (η ∘ α)_A = (ε ∘ α)_A right) ⇒ left(∀ A •; η_A = ε_Aright) \
≡ & color{green}{{text{ Extensionality }}} \
& quad ∀ η,ε •quad η ∘ α = ε ∘ α ⇒ η = ε \
≡ & color{green}{{text{ Definition of epic }}} \
& quad α text{ is epic }
end{align*}
Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.
answered yesterday
Musa Al-hassy
1,2171711
answered yesterday
Musa Al-hassy
1,2171711
answered yesterday
Musa Al-hassy
1,2171711
answered yesterday
Musa Al-hassy
1,2171711
1,2171711
add a comment |
add a comment |
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