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Let $\mathcal{A}$ be a $k$-linear abelian symmetric tensor category with unit $\mathcal{O}_A$; here $k$ is a comm. ring. By that I assume implicitly that $\otimes$ is finitely cocontinuous in each variable.

Assume that $\mathcal{L}$ is a line object on $\mathcal{A}$, i.e. it is invertible with respect to $\otimes$ and the symmetry $\mathcal{L} \otimes \mathcal{L} \cong \mathcal{L} \otimes \mathcal{L}$ is the identity.

Question: Is every epimorphism $\mathcal{O}_A \to \mathcal{L}$ an isomorphism?

Remark that this is true in the following two cases:

a) $\mathcal{A}$ is a category of modules over a ring, or a category which "locally" looks like this (for example $\text{Qcoh}(X)$ for some scheme $X$).

b) $2$ is invertible in $k$ (which I learned from t3suji).

Is it true in general, or is there a nasty counterexample over $k=\mathbb{F}_2$? What about the special case $\mathcal{L} = \mathcal{O}_A$?

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1 Answer 1

up vote 2 down vote accepted

Assuming that $\otimes$ preserves finite colimits in each separate argument, I think it's true in general.

Since it's very easy to get twisted around here, let me abstract the situation a bit. Recall that if $\alpha: F' \to F$ is a map between left adjoints in a bicategory (with right adjoints $G'$ and $G$ respectively), then the mate $\alpha^\ast: G \to G'$ is defined by the composite

$$G \stackrel{\eta' G}{\to} (G'F')G \stackrel{(G'\alpha) G}{\to} (G'F)G \cong G'(FG) \stackrel{G'\varepsilon}{\to} G'$$

where $\eta'$ and $\varepsilon$ are the appropriate unit and counit of adjunctions. Here of course we are thinking of our monoidal category as a one-object bicategory.

If $\pi: \mathcal{O} \to L$ is an epi, then I claim its mate $\pi^\ast: L^{-1} \to \mathcal{O}$ is a mono. For, if $f, g: A \to L^{-1}$ are distinct arrows, then the two composites

$$A \stackrel{\overset{f}{\to}}{\underset{g}{\to}} L^{-1} \stackrel{\pi^\ast}{\to} \mathcal{O}$$

are equal to the composites

$$A \cong \mathcal{O} \otimes A \stackrel{\pi \otimes A}{\to} L \otimes A \stackrel{\overset{L \otimes f}{\to}}{\underset{L \otimes g}{\to}} L \otimes L^{-1} \stackrel{\varepsilon}{\cong} \mathcal{O}$$

by the usual yoga of mates and adjunctions; note that $L \otimes f$ and $L \otimes g$ are distinct since tensoring with $L$ is an equivalence, and note that $\pi \otimes A$ is an epi because tensoring with $A$ is right exact. So these two composites are distinct. Hence $\pi^\ast$ is monic.

On the other hand, the mate $\pi^\ast$ is also epic, because it is given by the composite

$$L^{-1} \cong \mathcal{O} \otimes L^{-1} \stackrel{\pi \otimes L^{-1}}{\to} L \otimes L^{-1} \stackrel{\varepsilon}{\cong} \mathcal{O}$$

where the middle arrow is epi again because tensoring with an object is right exact.

Since $\pi^\ast$ is mono and epi in an abelian category, it is an isomorphism, and therefore its mate $\pi$ is also an isomorphism.

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