my question is related to the tensor product in $k$-linear additive categories. If you have such a category, an object $A$ and a finite dimensional $k$-vector space $V$, then one knows that the tensor product $V\otimes_kA$ exists.

One usually chooses a basis of $V$ and sets $A\otimes_kV$ simply as $A^n$ and verifies the universal property

$Hom(V\otimes_k A,B)\simeq Hom_k(V,Hom(A,B))$ directly.

Now it is not clear for me inasfar this construction is canonical, as you chose a basis of $V$.

And if $V$ is simply $Hom(A,B)$, then you have a morphism

$Hom(Hom(A,B)\otimes A,B)$ corresponding to the identity. How canonical is also this one?

Thank you very much!

  • 10
    $\begingroup$ If you construct an object by invoking the non-canonical Lord Of The Choices, and somehow you end up with an object which satisfies the universal property you want, then you know ---because of universality--- that you have found the unique-up-to-isomorphism-compatible-with-the-universal-property object. If your category is not skeletal, then you can always make other choices and construct another solution to theuniversal property. $\endgroup$ – Mariano Suárez-Álvarez Aug 16 '11 at 18:56
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    $\begingroup$ @Mariano: I think I shall phrase it that way from now on :) $\endgroup$ – Zev Chonoles Aug 16 '11 at 19:09
  • $\begingroup$ well, somehow the problem is that you verify also the universal property by choosing a basis of $V$. That's what's confusing me here. $\endgroup$ – Descartes Aug 16 '11 at 22:24
  • $\begingroup$ Once you verify the universal property, that implies the construction is independent of the choice of basis. And the morphism corresponding to the identity is completely canonical: it is the evaluation morphism. $\endgroup$ – David Roberts Sep 14 '11 at 5:33

You can make $A\otimes V$ canonical by defining it to be a family of objects indexed by the bases of $V$. But that's a big mess, so you probably don't want to do it. If all you want is to make $V\mapsto A\otimes V$ into a genuine functor, you can do that by choosing a suitable adjoint to the inclusion functor from vector spaces of the form $k^n$ to the category of all vector spaces.

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    $\begingroup$ Anafunctors (ncatlab.org/nlab/show/anafunctor) were introduced by Makkai to reduce the problem of defining a functor via universal constructions, such as this. There is a perfectly good bicategory of categories, anafunctors and transformations which if you allow Choice is equivalent to the usual 2-category of categories, functors and natural transformations. $\endgroup$ – David Roberts Aug 16 '11 at 23:40

If $C$ is a $k$-linear category for some ring $k$ (it does not have to be a field), then you can define $M \otimes_k X$ for $M \in \mathrm{Mod}(k), X \in C$ as $\mathrm{colim}_{m \in M} X$, if this colimit exists. Here the index category is $M$, where a morphism $m \to m'$ is an $r \in k$ such that $m'=rm$. Thus, $- \otimes_k X$ is the cocontinuous extension of $k \otimes_k X = X$. In particular, if $M$ is defined by some presentation $k^{(I)} \to k^{(J)} \to M \to 0$, then we get a presentation $X^{(I)} \to X^{(J)} \to M \otimes_k X \to 0$ (if these colimits exist). But of course, this is not the best way to define the tensor product. The best characterization is probably that $(-) \otimes_k X$ is left adjoint to $\mathrm{Hom}_C(X,-)$. As always, left adjoints just have to be defined pointwise, thus we may choose a presentation of $M$ in order to define $M \otimes_k X$, as soon as we ensure the universal property.

As for your last question, it is now clear how to define $\mathrm{Hom}_C(X,Y) \otimes X = \mathrm{colim}_{f : X \to Y} X \to Y$.

  • $\begingroup$ I would guess that you really need to the the colim over the category of finitely generated submodules, rather. $\endgroup$ – Mariano Suárez-Álvarez Nov 4 '11 at 13:28
  • $\begingroup$ It works as stated. $\endgroup$ – Martin Brandenburg Nov 22 '11 at 16:54

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