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Does every covariant functor on module category preserve inclusion?

0

$begingroup$

Does every covariant functor $F : ~_RmathcalM rightarrow mathcalC$ on module category $_RmathcalM$ preserve inclusion? I have proceeded in the following way.

Suppose $A subseteq B$ in $_RmathcalM$ then the natural inclusion map $i$ acts as an identity $1_A$ on $A$, then $F(i)$ must act as an identity on $F(A)$, therefore $F(A) subseteq F(B)$.
But clearly the functor $mathbbZ_n otimes_mathbbZ -$ doesnot preserve the inclusion $mathbbZ subseteq mathbbQ$. Am I overlooking something? Any help would be appreciated.

category-theory modules homological-algebra abelian-categories functors

share|cite|improve this question

edited Mar 26 at 9:32

jean_23

asked Mar 26 at 8:15

jean_23jean_23

84

$endgroup$

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  $begingroup$
  It doesn't make sense to say it "acts as an identity on $A$". Or more precisely, it does, but that has no reason to be preserved by a functor
  $endgroup$
  – Max
  Mar 26 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max But the inclusion map $i$ works as the identity map $1_A$ on $A$, and by the definition of functor $F(1_A) = 1_F(A)$ therefore $F(i) : F(A) rightarrow F(B)$ acts as identity map on $F(A)$. Isn't it?
  $endgroup$
  – jean_23
  Mar 26 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not post a question simultaneously on MathSE and MathOverflow. In this case, I think this site is more appropriate than MO, so I would advise you to delete the MO version and keep this question.
  $endgroup$
  – Arnaud D.
  Mar 26 at 9:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also this Math Meta post and this MO Meta post on the topic.
  $endgroup$
  – Arnaud D.
  Mar 26 at 9:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $F(1_A) = 1_F(A)$ is true but that doesn't imply "working on $A$ as the identity" is preserved under $F$, because "working on $A$ as the identity" is about evaluating certain arrows on certain elements, which makes no sense categorically. Try to write out a formal proof of what you are saying (that is, without using intuitive words such as "acts as the identity") and you'll see that you are unable to do so
  $endgroup$
  – Max
  Mar 26 at 9:58
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Does every covariant functor $F : ~_RmathcalM rightarrow mathcalC$ on module category $_RmathcalM$ preserve inclusion? I have proceeded in the following way.

Suppose $A subseteq B$ in $_RmathcalM$ then the natural inclusion map $i$ acts as an identity $1_A$ on $A$, then $F(i)$ must act as an identity on $F(A)$, therefore $F(A) subseteq F(B)$.
But clearly the functor $mathbbZ_n otimes_mathbbZ -$ doesnot preserve the inclusion $mathbbZ subseteq mathbbQ$. Am I overlooking something? Any help would be appreciated.

category-theory modules homological-algebra abelian-categories functors

share|cite|improve this question

edited Mar 26 at 9:32

jean_23

asked Mar 26 at 8:15

jean_23jean_23

84

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't make sense to say it "acts as an identity on $A$". Or more precisely, it does, but that has no reason to be preserved by a functor
  $endgroup$
  – Max
  Mar 26 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max But the inclusion map $i$ works as the identity map $1_A$ on $A$, and by the definition of functor $F(1_A) = 1_F(A)$ therefore $F(i) : F(A) rightarrow F(B)$ acts as identity map on $F(A)$. Isn't it?
  $endgroup$
  – jean_23
  Mar 26 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not post a question simultaneously on MathSE and MathOverflow. In this case, I think this site is more appropriate than MO, so I would advise you to delete the MO version and keep this question.
  $endgroup$
  – Arnaud D.
  Mar 26 at 9:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also this Math Meta post and this MO Meta post on the topic.
  $endgroup$
  – Arnaud D.
  Mar 26 at 9:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $F(1_A) = 1_F(A)$ is true but that doesn't imply "working on $A$ as the identity" is preserved under $F$, because "working on $A$ as the identity" is about evaluating certain arrows on certain elements, which makes no sense categorically. Try to write out a formal proof of what you are saying (that is, without using intuitive words such as "acts as the identity") and you'll see that you are unable to do so
  $endgroup$
  – Max
  Mar 26 at 9:58
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Does every covariant functor $F : ~_RmathcalM rightarrow mathcalC$ on module category $_RmathcalM$ preserve inclusion? I have proceeded in the following way.

Suppose $A subseteq B$ in $_RmathcalM$ then the natural inclusion map $i$ acts as an identity $1_A$ on $A$, then $F(i)$ must act as an identity on $F(A)$, therefore $F(A) subseteq F(B)$.
But clearly the functor $mathbbZ_n otimes_mathbbZ -$ doesnot preserve the inclusion $mathbbZ subseteq mathbbQ$. Am I overlooking something? Any help would be appreciated.

category-theory modules homological-algebra abelian-categories functors

share|cite|improve this question

edited Mar 26 at 9:32

jean_23

asked Mar 26 at 8:15

jean_23jean_23

84

$endgroup$

share|cite|improve this question

edited Mar 26 at 9:32

jean_23

asked Mar 26 at 8:15

jean_23jean_23

84

share|cite|improve this question

edited Mar 26 at 9:32

jean_23

asked Mar 26 at 8:15

jean_23jean_23

84

asked Mar 26 at 8:15

jean_23jean_23

84

asked Mar 26 at 8:15

jean_23jean_23

84