Difference between revisions of "Contravariant functor/Definition"

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(Created page with "<noinclude> ==Definition== </noinclude> A ''covariant functor'', {{M|T:C\leadsto D}} (for categories {{M|C}} and {{M|D}}) is a pair of mappings{{rAIRM...")
 
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==Definition==
 
==Definition==
 
</noinclude>
 
</noinclude>
A ''covariant functor'', {{M|T:C\leadsto D}} (for [[category|categories]] {{M|C}} and {{M|D}}) is a pair of [[mapping|mappings]]{{rAIRMACCF}}:
+
A ''covariant functor'', {{M|S:C\leadsto D}} (for [[category|categories]] {{M|C}} and {{M|D}}) is a pair of [[mapping|mappings]]{{rAIRMACCF}}:
 
* {{M|1=S:\left\{Obj(C)Obj(D)XSX
\right. }}
 
* {{M|1=S:\left\{Obj(C)Obj(D)XSX
\right. }}
 
* {{M|1=S:\left\{Mor(C)Mor(D)fSf
\right. }}
 
* {{M|1=S:\left\{Mor(C)Mor(D)fSf
\right. }}

Latest revision as of 16:53, 2 February 2016

Definition

A covariant functor, S:CD (for categories C and D) is a pair of mappings[1]:

  • S:{Obj(C)Obj(D)XSX
  • S:{Mor(C)Mor(D)fSf
    • Note that if f:AB then Sf:BA

Which preserves only the identity morphism of each object - it reverses composition of morphisms, that is to say:

  • f,gMor(C)[Sgf=S(gf)=SfSg=SfSg] (I've added the s in to make it more obvious to the reader what is going on)
    • Where such composition makes sense. That is target(f)=source(g).
  • and AObj(C)[S1A=1SA]

Thus if f:XY and g:YZ are morphisms of C, then the following diagram commutes:

Thus the diagram just depicts the requirement that:

  • =Sgf=SfSg
  Note that the diagram is
similar to

References

  1. Jump up Algebra I: Rings, modules and categories - Carl Faith