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The filtration in dual category is called the cofiltration. A subobject in the opposite category is a quotient object, so instead of having a sequence of inclusions, we have a sequence of projections. If is -filtered vector space, then the dual space is -cofiltered (or should one say -cofiltered).
I am these days interested in a special case of, so to speak, pro-finite-dimensional cofiltrations of vector spaces.
Let
be an exhaustive filtration of a vector space (i.e. ) by finite dimensional vector subspaces over a ground field . We also write . Such vector spaces equipped with exhaustive -filtrations by finite-dimensional vector spaces form a category which I will denote by which is symmetric monoidal with respect to the filtered algebraic tensor product: . Consider the full algebraic dual , ; it extends to a contravariant functor; the dual of a morphism is also called the transpose , and is defined by . Now the vector spaces comes with cofiltration by finite-dimensional vector spaces ; the connecting projections and the canonical projections are mutually compatible and simply the restriction maps along the inclusions and . The inverse limit is a vector subspace of the direct product whose elements are threads such that for all . Clearly, one can identify
Indeed the thread corresponds to the unique linear functional on such that .
Now consider the category of vector space cofiltered by finite-dimensional vector spaces and complete in the sense that they are equipped with an isomorphism with the inverse limit of its own cofiltration; for simplicity the morphisms will strictly respect the cofiltration (with a little more relaxed notion of a morphism we can get a subcategory of the category of pro-finite-dimensional vector spaces, but this would make the theory more cumbersome for my present purposes). I am claiming that is provided with a notion of a completed tensor product making into a symmetric monoidal category in such a way that the duality is equipped with a structure of a strong monoidal functor . In fact there is also the inverse functor, though is not isomorphic to for infinite-dimensional , the cofiltration is made out of finite-dimensional pieces which are rigid/dualizable objects with respect to the usual tensor product. This is likely true in much more general context than filtered vector spaces.
We proceed toward the study of the completed tensor product in We first see that there is an “ordinary” tensor product of cofiltered spaces (which is still cofiltered, but does not belong to in general as it is not complete) whose -th cofiltered piece is just
and the connecting projections are the unique epimorphisms between for varrying commuting with the projections from .
The completed tensor product in is given by the inverse limit
(Is there a universal property of this tensor product not referencing to the construction of and corresponding projections ?)
From now on we skip the field from the notation.
We now recall that for finite-dimensional vector spaces , .
We now claim that for , the following diagram commutes
where are the canonical embeddings. The right hand side of the diagram defines the cofiltration which together with its limit, by definition, forms the completed tensor product
The left hand side, is on the other hand, constructed by duality from the filtration on (indeed the connecting projections are dual to connecting injections) and it defines an isomorphic object in .
Heuristically we behave as if
The middle column is at this point just a heuristics (the sum is not direct has to be internal, but in which vector space…) and it should be in fact skipped. In the same vain, the linear maps need more explanation. First of all, by the definition and are the canonical projections.
Let us construct the isomorphism
So let be a representative on the lhs, , ; to start, we need to evaluate it on elements for some (by linearity on linear combination of such; if something belongs to tensor products for different we need to get the same). The formula is
Now this does not depend on a representative, as if we add to something in
hence the added term gives zero when evaluated on .
Suppose now for . Then hence again no difference. Thus the map above is well defined.
Now we have to show that it is an isomorphism.
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