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Does every functor from Set to Set preserve products?


when does a functor map products into products?Does the forgetful functor from $T/C$ to $C$ preserve non-trivial colimits?Does the functor treating a monoid as a category preserve colimits and limits?What (filtered) (homotopy) (co) limits does $pi_0:mathbf{sSets}tomathbf{Sets}$ preserve?Does free functor preserve monomorphism?Does the lax Gray tensor product preserve fully faithful 2-functors?Does the Hom-functor $H( _, A)$ take limits to colimits?Does the forgetful functor from Stone spaces to sets preserve colimits?When does a functor preserve or reflect strong monos / strong epis?Is additivity necessary for a left exact functor to preserve pullbacks?













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









$endgroup$








  • 2




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










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    1 hour 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









$endgroup$








  • 2




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










  • $begingroup$
    By "product", do you mean Cartesian product?
    $endgroup$
    – Acccumulation
    1 hour 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 3 hours ago









Keshav SrinivasanKeshav Srinivasan

2,32721445




2,32721445








  • 2




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










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














  • 2




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










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








2




2




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




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












$begingroup$
By "product", do you mean Cartesian product?
$endgroup$
– Acccumulation
1 hour ago




$begingroup$
By "product", do you mean Cartesian product?
$endgroup$
– Acccumulation
1 hour ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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









$endgroup$





















    4












    $begingroup$

    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









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      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









      $endgroup$


















        4












        $begingroup$

        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









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 3 hours ago









          Thibaut BenjaminThibaut Benjamin

          867




          867























              4












              $begingroup$

              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









              $endgroup$


















                4












                $begingroup$

                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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 3 hours ago









                  Arnaud D.Arnaud D.

                  16k52443




                  16k52443






























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