Does every functor from Set to Set preserve products?












2












$begingroup$


In general, not all functors preserve products. But my question is, is it at least true that all functors from Set to Set preserve products?



If not, does anyone know of a counterexample?






category-theory examples-counterexamples products functors







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    (Most) constant functors don't preserve products.
    $endgroup$
    – Arnaud D.
    11 hours ago










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    9 hours ago
















2












$begingroup$


In general, not all functors preserve products. But my question is, is it at least true that all functors from Set to Set preserve products?



If not, does anyone know of a counterexample?






category-theory examples-counterexamples products functors







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    (Most) constant functors don't preserve products.
    $endgroup$
    – Arnaud D.
    11 hours ago










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    9 hours ago














2












2








2





$begingroup$


In general, not all functors preserve products. But my question is, is it at least true that all functors from Set to Set preserve products?



If not, does anyone know of a counterexample?






category-theory examples-counterexamples products functors







share|cite|improve this question









$endgroup$




In general, not all functors preserve products. But my question is, is it at least true that all functors from Set to Set preserve products?



If not, does anyone know of a counterexample?






category-theory examples-counterexamples products functors




category-theory examples-counterexamples products functors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 11 hours ago









Keshav SrinivasanKeshav Srinivasan

2,33021445




2,33021445








  • 4




    $begingroup$
    (Most) constant functors don't preserve products.
    $endgroup$
    – Arnaud D.
    11 hours ago










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    9 hours ago














  • 4




    $begingroup$
    (Most) constant functors don't preserve products.
    $endgroup$
    – Arnaud D.
    11 hours ago










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    9 hours ago








4




4




$begingroup$
(Most) constant functors don't preserve products.
$endgroup$
– Arnaud D.
11 hours ago




$begingroup$
(Most) constant functors don't preserve products.
$endgroup$
– Arnaud D.
11 hours ago












$begingroup$
By "product", do you mean Cartesian product?
$endgroup$
– Acccumulation
9 hours ago




$begingroup$
By "product", do you mean Cartesian product?
$endgroup$
– Acccumulation
9 hours ago










2 Answers
2






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10












$begingroup$

There are a lot of counterexamples, actually.



For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve products, since the two projections $Stimes Sto S$ are not equal.



The functor $Xmapsto Stimes X$ does not preserve products either, because the function $$Stimes Xtimes Yto Stimes Xtimes Stimes Y:(s,x,y)mapsto (s,x,s,y)$$
is never surjective.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    If you consider the powerset functor $mathcal{P}$, which takes any $A$ to its powerset $mathcal{P}A$, and any function $f : A to B$ to the function $mathcal{P}f : mathcal{P}A to mathcal{P}B$, defined by the direct image, for $Xsubset A$, $(mathcal{P}f)(X) = f(X) subset B$. This defines a functor from Set to Set.



    Now you can check that this functor doesn't preserve products, since, for instance you have :
    $mathcal{P}{0} = {emptyset,{0}}$, so taking any non-empty finite set $A$, you have $mathcal{P}(Atimes{0}) = mathcal{P}(A) neq mathcal{P}(A)timesmathcal{P}({0})$. You can check that they are indeed different by looking at their cardinal :
    $|mathcal{P}(A)timesmathcal{P}({0})| = 2 |mathcal{P}(A)| $






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      10












      $begingroup$

      There are a lot of counterexamples, actually.



      For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve products, since the two projections $Stimes Sto S$ are not equal.



      The functor $Xmapsto Stimes X$ does not preserve products either, because the function $$Stimes Xtimes Yto Stimes Xtimes Stimes Y:(s,x,y)mapsto (s,x,s,y)$$
      is never surjective.






      share|cite|improve this answer









      $endgroup$


















        10












        $begingroup$

        There are a lot of counterexamples, actually.



        For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve products, since the two projections $Stimes Sto S$ are not equal.



        The functor $Xmapsto Stimes X$ does not preserve products either, because the function $$Stimes Xtimes Yto Stimes Xtimes Stimes Y:(s,x,y)mapsto (s,x,s,y)$$
        is never surjective.






        share|cite|improve this answer









        $endgroup$
















          10












          10








          10





          $begingroup$

          There are a lot of counterexamples, actually.



          For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve products, since the two projections $Stimes Sto S$ are not equal.



          The functor $Xmapsto Stimes X$ does not preserve products either, because the function $$Stimes Xtimes Yto Stimes Xtimes Stimes Y:(s,x,y)mapsto (s,x,s,y)$$
          is never surjective.






          share|cite|improve this answer









          $endgroup$



          There are a lot of counterexamples, actually.



          For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve products, since the two projections $Stimes Sto S$ are not equal.



          The functor $Xmapsto Stimes X$ does not preserve products either, because the function $$Stimes Xtimes Yto Stimes Xtimes Stimes Y:(s,x,y)mapsto (s,x,s,y)$$
          is never surjective.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 11 hours ago









          Arnaud D.Arnaud D.

          16.1k52444




          16.1k52444























              7












              $begingroup$

              If you consider the powerset functor $mathcal{P}$, which takes any $A$ to its powerset $mathcal{P}A$, and any function $f : A to B$ to the function $mathcal{P}f : mathcal{P}A to mathcal{P}B$, defined by the direct image, for $Xsubset A$, $(mathcal{P}f)(X) = f(X) subset B$. This defines a functor from Set to Set.



              Now you can check that this functor doesn't preserve products, since, for instance you have :
              $mathcal{P}{0} = {emptyset,{0}}$, so taking any non-empty finite set $A$, you have $mathcal{P}(Atimes{0}) = mathcal{P}(A) neq mathcal{P}(A)timesmathcal{P}({0})$. You can check that they are indeed different by looking at their cardinal :
              $|mathcal{P}(A)timesmathcal{P}({0})| = 2 |mathcal{P}(A)| $






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                If you consider the powerset functor $mathcal{P}$, which takes any $A$ to its powerset $mathcal{P}A$, and any function $f : A to B$ to the function $mathcal{P}f : mathcal{P}A to mathcal{P}B$, defined by the direct image, for $Xsubset A$, $(mathcal{P}f)(X) = f(X) subset B$. This defines a functor from Set to Set.



                Now you can check that this functor doesn't preserve products, since, for instance you have :
                $mathcal{P}{0} = {emptyset,{0}}$, so taking any non-empty finite set $A$, you have $mathcal{P}(Atimes{0}) = mathcal{P}(A) neq mathcal{P}(A)timesmathcal{P}({0})$. You can check that they are indeed different by looking at their cardinal :
                $|mathcal{P}(A)timesmathcal{P}({0})| = 2 |mathcal{P}(A)| $






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  If you consider the powerset functor $mathcal{P}$, which takes any $A$ to its powerset $mathcal{P}A$, and any function $f : A to B$ to the function $mathcal{P}f : mathcal{P}A to mathcal{P}B$, defined by the direct image, for $Xsubset A$, $(mathcal{P}f)(X) = f(X) subset B$. This defines a functor from Set to Set.



                  Now you can check that this functor doesn't preserve products, since, for instance you have :
                  $mathcal{P}{0} = {emptyset,{0}}$, so taking any non-empty finite set $A$, you have $mathcal{P}(Atimes{0}) = mathcal{P}(A) neq mathcal{P}(A)timesmathcal{P}({0})$. You can check that they are indeed different by looking at their cardinal :
                  $|mathcal{P}(A)timesmathcal{P}({0})| = 2 |mathcal{P}(A)| $






                  share|cite|improve this answer









                  $endgroup$



                  If you consider the powerset functor $mathcal{P}$, which takes any $A$ to its powerset $mathcal{P}A$, and any function $f : A to B$ to the function $mathcal{P}f : mathcal{P}A to mathcal{P}B$, defined by the direct image, for $Xsubset A$, $(mathcal{P}f)(X) = f(X) subset B$. This defines a functor from Set to Set.



                  Now you can check that this functor doesn't preserve products, since, for instance you have :
                  $mathcal{P}{0} = {emptyset,{0}}$, so taking any non-empty finite set $A$, you have $mathcal{P}(Atimes{0}) = mathcal{P}(A) neq mathcal{P}(A)timesmathcal{P}({0})$. You can check that they are indeed different by looking at their cardinal :
                  $|mathcal{P}(A)timesmathcal{P}({0})| = 2 |mathcal{P}(A)| $







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 11 hours ago









                  Thibaut BenjaminThibaut Benjamin

                  1167




                  1167






























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