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?










share|cite|improve this question




























    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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 19:12









      Oskar

      2,6961718




      2,6961718










      asked Nov 19 at 18:34









      H R

      1608




      1608






















          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.






          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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005318%2fepimorphism-in-the-functor-category-mathbfcop-mathbfset%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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.






            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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 19 at 23:10









                Musa Al-hassy

                1,2971711




                1,2971711






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005318%2fepimorphism-in-the-functor-category-mathbfcop-mathbfset%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                    ComboBox Display Member on multiple fields

                    Is it possible to collect Nectar points via Trainline?