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hom-set_adjunction [2016/02/11 13:23]
nikolaj
hom-set_adjunction [2016/04/19 18:43] (current)
nikolaj
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 == An example in the category of sets == == An example in the category of sets ==
  
-Let both ${\bf C}$ and ${\bf D}$ be the category ${\bf Set}$, which has products and exponential objects. Fix some objects (sets) $A$ and $Y$. Many examples can be thought of as Variation ​of the pretty obvious relation+Let both ${\bf C}$ and ${\bf D}$ be the category ${\bf Set}$, which has products and exponential objects. Fix some objects (sets) $A$ and $Y$. Many examples can be thought of as variation ​of the pretty obvious relation
  
 $\mathrm{Hom}_{\bf Set}(*\times A,​Y)\cong\mathrm{Hom}_{\bf Set}(A,Y):= Y^A\cong\mathrm{Hom}_{\bf Set}(*,​Y^A)$ $\mathrm{Hom}_{\bf Set}(*\times A,​Y)\cong\mathrm{Hom}_{\bf Set}(A,Y):= Y^A\cong\mathrm{Hom}_{\bf Set}(*,​Y^A)$
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 $\mathrm{Hom}_{\bf Set}(X\times A,​Y)\cong\mathrm{Hom}_{\bf Set}(X,​Y^A)$ $\mathrm{Hom}_{\bf Set}(X\times A,​Y)\cong\mathrm{Hom}_{\bf Set}(X,​Y^A)$
  
-and this is hom-set adjunction ​+and this is hom-set adjunction ​
  
-$\mathrm{Hom}_{\bf Set}(FX,​Y)\cong\mathrm{Hom}_{\bf Set}(X,Y^A)$+$\mathrm{Hom}_{\bf Set}(FX,​Y)\cong\mathrm{Hom}_{\bf Set}(X,GY)$
  
 if we define the Action of $F$ on object via $FX:​=X\times A$ (Cartesian product) and let the action of $G$ on object be given by $GY:=Y^A$ (function space from $A$ to $Y$).  if we define the Action of $F$ on object via $FX:​=X\times A$ (Cartesian product) and let the action of $G$ on object be given by $GY:=Y^A$ (function space from $A$ to $Y$). 
  
-== Idea == +>== Idea == 
-More generally, ​ +>More generally, view the left adjoint $F$ as A-"​thickening"​ of ist argument ($X$), enabling to attack data, and view $G$ as the A-indexing'​s ​of aspects of it's argument $Y$, enabling to consider processes.  
-view the left adjoint $F$ as A-"​thickening"​ of ist argument ($X$) and view $G$ as the A-indexing of aspects of it's argument $Y$. +>If ${\bf C}\neq{\bf D}$, then viewing $G$ as indexing may be harder.
- +
-If ${\bf C}\neq{\bf D}$, then viewing $G$ as indexing may be harder.+
  
 == Currying == == Currying ==
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 $\left((X\land A)\implies Y\right)\leftrightarrow\left(X\implies(A\implies Y)\right)$ $\left((X\land A)\implies Y\right)\leftrightarrow\left(X\implies(A\implies Y)\right)$
  
-Here the A-"​thickening"​ side says you have more arugments to prove $Y$ to begin with, while the $A$-"indexing"​ side means you only demonstrate A-conditional truth of $Y$.+Here the A-"​thickening"​ side says you have more arugments to prove $Y$ to begin with, while the "$A$-indexing's" side means you only demonstrate A-conditional truth of $Y$.
  
 == Example from Algebra == == Example from Algebra ==
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