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hom-functor [2014/06/26 14:30] nikolaj |
hom-functor [2014/11/01 16:48] nikolaj |
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| simply because their values lie in the field $\mathbb K$, which already has addition. And as a side note, if we restrict ourselves to the linear functionals, then this functional space becomes the so called dual vector space. | simply because their values lie in the field $\mathbb K$, which already has addition. And as a side note, if we restrict ourselves to the linear functionals, then this functional space becomes the so called dual vector space. | ||
| - | Now back to the general case. Instead of working with the category ${\bf C}$, one can work with the category of set valued covariant functors ${\bf Set}^{{\bf C}^\mathrm{op}}$ called [[Presheaf category]]). One substitutes an object $A$ with the contravariant functor $\mathrm{Hom}_{\bf C}(-,A)$ (=[[Yoneda embedding]]) and the arrows actually are the same/isomorphic to the old ones (=Yoneda lemma). | + | Now back to the general case. Instead of working with the category ${\bf C}$, one can work with the category of set valued covariant functors ${\bf Set}^{{\bf C}^\mathrm{op}}$ (called [[Presheaf category]]). One substitutes an object $A$ with the contravariant functor $\mathrm{Hom}_{\bf C}(-,A)$ (=[[Yoneda embedding]]) and the arrows actually are the same/isomorphic to the old ones (=Yoneda lemma). |
| The advantage of this is that that new category ${\bf Set}^{{\bf C}^\mathrm{op}}$ has more objects than ${\bf C}$, namely some non-representable functors $F$. (Representable means $F$ is isomorphic to some hom-functor anyway.) For example, the category ${\bf C}$ might not have products $\times$, but because ${\bf Set}$ has products, the category ${\bf Set}^{{\bf C}^\mathrm{op}}$ always has them. | The advantage of this is that that new category ${\bf Set}^{{\bf C}^\mathrm{op}}$ has more objects than ${\bf C}$, namely some non-representable functors $F$. (Representable means $F$ is isomorphic to some hom-functor anyway.) For example, the category ${\bf C}$ might not have products $\times$, but because ${\bf Set}$ has products, the category ${\bf Set}^{{\bf C}^\mathrm{op}}$ always has them. | ||
