By an abelian $\otimes$-category I mean a symmetric monoidal category $(\mathcal{A},\otimes,\mathcal{O})$, such that $\mathcal{A}$ also is an abelian category and for every $M \in \mathcal{A}$ the functor $M \otimes - $ is cocontinuous (i.e. right exact and preserves coproducts; in particular additive). A line bundle is defined as an object $\mathcal{L}$ of $\mathcal{A}$ such that there is some object $\mathcal{L}'$ such that $\mathcal{L} \otimes \mathcal{L}' \cong \mathcal{O}$. An example is the category of (quasi-coherent) modules on a locally ringed space. The line bundles then coincide with the modules which are locally free of rank $1$ (see here).

Now I want to show that in general these line bundles have similar properties as in the case of the module category. For example it's not hard to show that if $\mathcal{L}$ is a line bundle, then it is flat in the sense that $\mathcal{L} \otimes -$ is exact (it is even an automorphism of $\mathcal{A}$ with inverse $\mathcal{L}^{-1} \otimes -$). The isomorphism classes of line bundles yield a group, which may be denoted as $\text{Pic}(\mathcal{A})$.

Question 1. Is there any literature about these abelian $\otimes$-categories which treats them systematically? Perhaps the "usual" definition differs a little from mine, this does not matter.

Question 2. Let $\mathcal{L}$ be a line bundle and $\phi : \mathcal{L} \to \mathcal{L}$ an epimorphism. Does it follow that $\phi$ is an isomorphism?

Question 3. Let $\mathcal{L}$ be a line bundle and assume $\phi : \mathcal{L} \to \mathcal{L}$ is an epimorphism. Does it follow that there is an epimorphism $\psi : \mathcal{L}^{-1} \to \mathcal{L}^{-1}$ such that $\phi \otimes \psi$ corresponds to the identity of $\mathcal{O}$ under the isomorphism $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{O}$?

Question 4. Assume we also have a $\lambda$-structure on $\mathcal{A}$ which is compatible with the given data. Is it possible to give a reasonable definition of a locally free object of rank $n$? See also this question.

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    $\begingroup$ For questions 1, 2, and 3, it does not seem necessary to require that the monoidal product be symmetric. $\endgroup$ – André Henriques Nov 23 '10 at 19:41
  • $\begingroup$ What about doing algebraic geometry over $\mathcal{A}$ in the sense of Toën and Vaquié arXiv:math/0509684 (published in J. K-theory 3 (2009), 437-500)? This way, you will be able to speak of locally free objects as usual (this might answer questions 1 and 5). Also, the comment of Theo Johnson-Freyd below suggests that the answer to Question 2 is yes: apply his argument to the smallest tensor abelian category generated by $\mathcal{L}$ (which is small). $\endgroup$ – Denis-Charles Cisinski Nov 27 '10 at 23:36
  • $\begingroup$ Could you elaborate what the paper has to do with my question? I don't want to enrich my data with topologies, it should stay homological. Otherwise it's just a trivial translation. Also I'm not sure if the Theorem cited by Theo applies here, does it really give a full faithful exact tensor functor? $\endgroup$ – Martin Brandenburg Nov 28 '10 at 15:14
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    $\begingroup$ The point of Toën and Vaquié is precisely that the monoidal structure itself defines canonically a Zariski topology, as well as a faithully flat topology (so, without any extra data, if you think of your $\mathcal{A}$ as an abstract category of modules over a commutative ring, you will have a lot of the tools you are used to with usual commutative algebra). This idea of doing algebraic geometry over an abelian tensor category was also used by Deligne to study tannakian categories; see the Grothendieck Festschrift vol II, p. 111-195. With Toën and Vaquié, you don't need the abelian structure. $\endgroup$ – Denis-Charles Cisinski Nov 28 '10 at 22:38
  • $\begingroup$ Q2: No, take Z[x]-modules on which x acts injective. Q3: No, since this is Q2 by Steve's answer. Q4: Yes, this will appear in my thesis ... $\endgroup$ – Martin Brandenburg Feb 1 '13 at 13:19

Question 2: this does not really depend on line bundles. If $\phi:\mathcal {L\to L}$ is a non-invertible epimorphism, then $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is epi, since $-\otimes\mathcal L'$ is exact, and non-invertible, since otherwise $\phi\otimes 1\otimes 1:\mathcal{L\otimes L'\otimes L\to L\otimes L'\otimes L}$ would be invertible. Thus there is a non-invertible epimorphism $\mathcal {O\to O}$.

Question 3: this has the same answer as Question 2. If the answer to Q2 is yes, then we can take $\psi$ to be the identity and get a positive answer to Q3. If the answer to Q2 is no, then (as above) if $\phi:\mathcal{ L\to L}$ is a non-invertible epimorphism, also $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is a non-invertible epimorphism. But now if $\psi:\mathcal {L'\to L'}$ is any map, then $\phi\otimes\psi=(1\otimes\psi)\circ(\phi\otimes 1)$ and if this is invertible then $\phi\otimes 1$ is split monic and so invertible (since it is already known to be epi).

  • $\begingroup$ Q2: Ok this reduces it to the case of maps $\mathcal{O} \to \mathcal{O}$. Are there examples? Q3: If $\phi$ is an isomorphism, then $\phi \otimes L'$ does not have to be the identity! $\endgroup$ – Martin Brandenburg Nov 23 '10 at 23:21
  • $\begingroup$ Q3: you're right, I should have used $\mathcal{(\theta^{-1}\otimes 1_{L'})(1\otimes\phi^{-1}\otimes 1)(\theta\otimes_{L'}}$ for $\psi$, where $\theta:\mathcal{\O\cong L\otimes L'}$. $\endgroup$ – Steve Lack Nov 24 '10 at 0:05
  • $\begingroup$ sorry, try again: $(\theta^{-1}\otimes 1_{L'})(1_{L'}\otimes \phi^{-1}\otimes 1_{L'})(\theta\otimes 1_{L'})$ where $\theta:O\cong L'\otimes L$. (Hope this works, it's tricky without preview facility) $\endgroup$ – Steve Lack Nov 24 '10 at 0:08
  • $\begingroup$ If you are going to find a counterexample to Q2 (i.e. a $\otimes$ category when 1 has a non-invertible epi), you'll have to look pretty far. Most such categories (say, the small ones) embed in R-mod-R for some ring R, and the endomorphisms of 1 in R-mod-R are the center of R, and the only epis are invertible. $\endgroup$ – Theo Johnson-Freyd Nov 25 '10 at 19:53
  • $\begingroup$ @Theo: You quote arxiv.org/abs/math/0004160, right? But there we only have a right exact embedding. $\endgroup$ – Martin Brandenburg Dec 3 '10 at 14:48

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