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Can vector space dualization be made into a covariant functor?

Discussion in 'Mathematics' started by goblin, Aug 1, 2020 at 8:03 PM.

  1. goblin

    goblin Guest

    The contravariant endofunctor $$\mathbf{Set}^{op} \rightarrow \mathbf{Set}$$ $$X \mapsto [X,2]$$

    can be made into a covariant endofunctor $$\mathbf{Set} \rightarrow \mathbf{Set}$$ $$X \mapsto \mathcal{P}(X)$$

    in such a way that:

    • these functors do the same thing to objects
    • they do the same thing to isomorphisms, excepting that $X \mapsto [X,2]$ flips the direction.

    This isn't too surprising, since $X \mapsto [X,2]$ can be seen as "completing" a set to a suplattice, and then forgetting the suplattice structure. It's probably fair to say that the covariant structure on this functor comes essentially comes from this fact (though perhaps there is a better way at looking at it).

    Anyway, I'm wondering if we can do something similar to the dual-vectorspace functor

    $$\mathbb{R}\mathbf{Mod}^{op} \rightarrow \mathbb{R}\mathbf{Mod}$$

    $$X \mapsto \mathbb{R}\mathbf{Mod}(X,\mathbb{R}).$$

    It seems likely that we can, since I tend to think of dual vector spaces as "completing" a vector space to a gadget in which certain infinite sums exist, and then forgetting this extra structure.

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