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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?

1

$begingroup$

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

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

1

$begingroup$

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

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

$endgroup$

share|cite|improve this question

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

share|cite|improve this question

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

asked 2 hours ago

Keshav SrinivasanKeshav Srinivasan

2,32221445

2 Answers
2

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oldest

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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

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

$endgroup$

add a comment | 

3

$begingroup$

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

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

$endgroup$

add a comment | 

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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

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

$endgroup$

add a comment | 

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

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

answered 1 hour ago

Thibaut BenjaminThibaut Benjamin

867

3

$begingroup$

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

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

$endgroup$

add a comment | 

3

$begingroup$

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

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443

answered 1 hour ago

Arnaud D.Arnaud D.

16k52443