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I am thinking about some problems in noncommutative projective geometry, to do with graded module categories and their quotients by torsion modules. The following question arose in this more specific setting but I'm phrasing it in generality as I don't believe the answer is likely to be specific to graded modules.

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories and let $\mathcal{C}$ be a thick subcategory of $\mathcal{A}$. If $F\colon \mathcal{A} \to \mathcal{B}$ is an equivalence of categories, does it descend to an equivalence $\bar{F}\colon \mathcal{A}/\mathcal{C} \to \mathcal{B}/F(\mathcal{C})$?

Using the universal property of the Serre quotient (Stacks Project, 02MS), I can show that if $F$ is an isomorphism of categories, then this descends to an isomorphism $\bar{F}$. ($\bar{F}$ exists by using the functors $\pi_{\mathcal{A}}$ and $\pi_{\mathcal{B}}F$, where $\pi_{A}\colon \mathcal{A}\to \mathcal{A}/\mathcal{C}$ is the canonical functor and similarly $\pi_{B}\colon \mathcal{B}\to \mathcal{B}/F(\mathcal{C})$. It's clear that $\mathcal{C}$ is in the kernel of $\pi_{\mathcal{B}}F$, of course. Similarly one may run the argument using $F^{-1}$ and then comparing $\bar{F}\overline{F^{-1}}$ with the identity functor on $\mathcal{A}/\mathcal{C}$, and using the uniqueness in the universal property, we're basically done.)

For obvious philosophical reasons, one would expect the same to be true if $F$ is only an equivalence, and indeed much of the argument runs through. One can construct an $\bar{F}$ and a $\bar{G}$ (where $G$ is the functor such that $GF\simeq \mathbb{1}_{\mathcal{A}}$ and vice versa) but the tricky bit seems to be deducing that $\bar{G}\bar{F} \simeq \mathbb{1}_{\mathcal{A}/\mathcal{C}}$ and its counterpart.

I would not be surprised if this is well-known, or if there is a simple standard way to complete the argument, but I'm not quite able to see it myself. Either a reference or a proof (or indeed both) would be gratefully received.

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You get an isomorphism of functors $\bar{G}\bar{F}\simeq1_\mathcal{A/C}$ by applying $\pi_\mathcal{A}$ to an isomorphism $GF\simeq1_\mathcal{A}$.

Just to check we're thinking of the quotient category in the same way: $\mathcal{A/C}$ has the same objects as $\mathcal{A}$, and the maps are constructed by adjoining inverses to every $s\in S$, where $S$ is the collection of maps in $\mathcal{A}$ whose kernel and cokernel are in $\mathcal{C}$, and the natural functor $\pi_\mathcal{A}:A\to\mathcal{A/C}$ is the identity on objects (i.e., $\pi_\mathcal{A}(X)=X$).

Also, on objects $\bar{F}(X)=F(X)$.

Suppose we have isomorphisms $\theta_X:GF(X)\to X$ in $\mathcal{A}$ giving a natural isomorphism from $GF$ to $1_\mathcal{A}$. Then in $\mathcal{A/C}$ we have isomorphisms $$\bar{\theta}_X=\pi_\mathcal{A}(\theta_X): GF(X)\to X.$$ To prove naturality, it suffices to check that $f\bar{\theta}_X=\bar{\theta}_Y\bar{G}\bar{F}(f)$ on maps $f:X\to Y$ of the form $\pi_\mathcal{A}(g)$ for $g:X\to Y$ a map in $\mathcal{A}$, and of the form $\pi_\mathcal{A}(s)^{-1}$ for $s:Y\to X$ in $S$, since every map in $\mathcal{A/C}$ is a composition of such maps. But this is straightforward: in the first case apply $\pi_\mathcal{A}$ to $g\theta_X=\theta_YGF(g)$, and in the second case apply $\pi_\mathcal{A}$ to $\theta_XGF(s)=s\theta_Y$.

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  • $\begingroup$ Yes, this is exactly what I wanted - thanks Jeremy. $\endgroup$ – Jan Grabowski Jan 13 '17 at 11:42

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