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hom-set_adjunction [2016/02/11 13:23] 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 a 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 $Y$). | ||
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| == Idea == | == Idea == | ||
| More generally, | More generally, | ||
| - | 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 $Y$. | + | view the left adjoint $F$ as A-"thickening" of ist argument ($X$) and view $G$ as the A-indexing's 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. | ||
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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 == | ||
