Epimorphism in the functor category $[mathbf{C}^{op}, mathbf{Set}]$











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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?










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    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?










    share|cite|improve this question


























      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?










      share|cite|improve this question















      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






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      edited Nov 19 at 19:12









      Oskar

      2,6961718




      2,6961718










      asked Nov 19 at 18:34









      H R

      1608




      1608






















          2 Answers
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          active

          oldest

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          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.






          share|cite|improve this answer























          • What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
            – H R
            Nov 19 at 19:21












          • @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
            – Oskar
            Nov 19 at 19:25










          • @user435800 $mathcal{A}$ may be arbitraty.
            – Oskar
            Nov 19 at 19:26






          • 1




            @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
            – Oskar
            Nov 19 at 19:36








          • 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
            Nov 19 at 19:42




















          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.






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
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            active

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            active

            oldest

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            active

            oldest

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            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.






            share|cite|improve this answer























            • What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
              – H R
              Nov 19 at 19:21












            • @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
              – Oskar
              Nov 19 at 19:25










            • @user435800 $mathcal{A}$ may be arbitraty.
              – Oskar
              Nov 19 at 19:26






            • 1




              @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
              – Oskar
              Nov 19 at 19:36








            • 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
              Nov 19 at 19:42

















            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.






            share|cite|improve this answer























            • What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
              – H R
              Nov 19 at 19:21












            • @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
              – Oskar
              Nov 19 at 19:25










            • @user435800 $mathcal{A}$ may be arbitraty.
              – Oskar
              Nov 19 at 19:26






            • 1




              @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
              – Oskar
              Nov 19 at 19:36








            • 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
              Nov 19 at 19:42















            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.






            share|cite|improve this answer














            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 19 at 19:31

























            answered Nov 19 at 19:04









            Oskar

            2,6961718




            2,6961718












            • What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
              – H R
              Nov 19 at 19:21












            • @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
              – Oskar
              Nov 19 at 19:25










            • @user435800 $mathcal{A}$ may be arbitraty.
              – Oskar
              Nov 19 at 19:26






            • 1




              @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
              – Oskar
              Nov 19 at 19:36








            • 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
              Nov 19 at 19:42




















            • What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
              – H R
              Nov 19 at 19:21












            • @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
              – Oskar
              Nov 19 at 19:25










            • @user435800 $mathcal{A}$ may be arbitraty.
              – Oskar
              Nov 19 at 19:26






            • 1




              @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
              – Oskar
              Nov 19 at 19:36








            • 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
              Nov 19 at 19:42


















            What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
            – H R
            Nov 19 at 19:21






            What do you mean by $mathcal{B}^{Delta_0}$? Does $mathcal{A}$ need to have zero object?
            – H R
            Nov 19 at 19:21














            @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
            – Oskar
            Nov 19 at 19:25




            @user435800 Inverse image functor. You can find its definition in my old answer: math.stackexchange.com/questions/1663325/…
            – Oskar
            Nov 19 at 19:25












            @user435800 $mathcal{A}$ may be arbitraty.
            – Oskar
            Nov 19 at 19:26




            @user435800 $mathcal{A}$ may be arbitraty.
            – Oskar
            Nov 19 at 19:26




            1




            1




            @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
            – Oskar
            Nov 19 at 19:36






            @user435800 Ah, sorry, it was a typo, thanks. Not $Delta_0$, but $Delta_a$.
            – Oskar
            Nov 19 at 19:36






            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
            Nov 19 at 19:42






            @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
            Nov 19 at 19:42












            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.






            share|cite|improve this answer

























              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.






              share|cite|improve this answer























                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.






                share|cite|improve this answer












                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.







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                answered Nov 19 at 23:10









                Musa Al-hassy

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