Partial Serre duality and cocompact objects

Abstract

A successful theme in the development of triangulated categories has been the study of compact objects. A weak dual notion called 0-cocompact objects was introduced in Oppermann et al. (Adv Math 350:190–241, 2019), motivated by the fact that sets of such objects cogenerate co-t-structures, dual to the t-structures generated by sets of compact objects. In the present paper, we show that the notion of 0-cocompact objects also appears naturally in the presence of certain dualities. We introduce “partial Serre duality”, which is shown to link compact to 0-cocompact objects. We show that partial Serre duality gives rise to an Auslander–Reiten theory, which in turn implies a weaker notion of duality which we call “non-degenerate composition”, and throughout this entire hierarchy of dualities the objects involved are $0$-(co)compact. Furthermore, we produce explicit partial Serre functors for multiple flavors of homotopy categories, thus illustrating that this type of duality, as well as the resulting $0$-cocompact objects, are abundant in prevalent triangulated categories.

Appendices

Appendix A: Dual Brown representability

The aim of this appendix is to give a proof of a ‘constructive’ version of dual Brown representability for triangulated categories with enhancements, cogenerated by a set of $0$-cocompact objects. Note that it was already pointed out by Modoi in [30] that these categories do satisfy dual Brown representability, so our original contribution here is only the explicit description of the representing objects.

Throughout this appendix, ${{\,\mathrm{{\textsf{T}}}\,}}$ is a triangulated category at the base of a stable derivator. Moreover ${{\,\mathrm{{\textsf{T}}}\,}}$ is cogenerated by a set of $0$-cocompact objects. By Lemma 2.7 we know that products of $0$-cocompact objects are $0$-cocompact again, whence we may assume that ${{\,\mathrm{{\textsf{T}}}\,}}$ is cogenerated by a single $0$-cocompact object $S$, which we may moreover assume to be invariant under suspension.

With this setup, we will prove the following.

Theorem A.1

Let $F :{{\,\mathrm{{\textsf{T}}}\,}}\longrightarrow {{\,\mathrm{{{\textsf{A}}}{{\textsf{b}}}}\,}}$ be a homological functor commuting with products. For any commutative diagram

such that all the induced maps ${{\,\mathrm{{{\textsf{I}}}{{\textsf{m}}}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}(f_i, S) \longrightarrow F(S)$ are isomorphisms, we have

$$\begin{aligned} F \cong {{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i, -). \end{aligned}$$

In particular ${{\,\mathrm{{\textsf{T}}}\,}}$ satisfies dual Brown representability.

The proof of this theorem is less direct than one might imagine: We first show, in Proposition A.5, that if $F$ already is representable then we do get the desired isomorphism of functors. Then we show, in Proposition A.8, that in general $F$ is at least an epimorphic image of ${{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i, -)$. The argument for this part comes from [29]. Finally we complete the proof by employing a trick of Neeman’s [35].

While large parts of the argument are available in the literature, we found that the varying notation made it slightly challenging to read the entire proof. Therefore we believe it might be worthwhile to give a complete account here.

Let us start with two brief observations translating our assumptions on $F$.

Observation A.2

For any set of objects $T_i$, any natural transformation from $\coprod {{\,\mathrm{{\textsf{T}}}\,}}(T_i, -)$ to $F$ factors uniquely through $\coprod {{\,\mathrm{{\textsf{T}}}\,}}(T_i, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}( \prod T_i, -)$.

Indeed we have the following commutative square

where the lower horizontal map is an isomorphism by assumption.

Observation A.3

For any triangle $T_1 \longrightarrow T_2 \longrightarrow T_3 \longrightarrow T_1[1]$, and any natural transformation ${{\,\mathrm{{\textsf{T}}}\,}}(T_2,-) \longrightarrow F$ such that composition ${{\,\mathrm{{\textsf{T}}}\,}}(T_3, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(T_2, -) \longrightarrow F$ vanishes, there is a factorization as indicated by the following diagram.

This follows from the fact that $F$ is cohomological by employing the Yoneda lemma.

Of course the “in particular” part of Theorem A.1 above only follows if we can find at least one such diagram, given $F$. There is a straight-forward way of doing so:

Construction A.4

Pick an epimorphism of ${{\,\mathrm{\textsf{End}}\,}}(S)$-modules ${{\,\mathrm{\textsf{End}}\,}}(S)^{(I_0)} \longrightarrow F(S)$. Equivalently, we have a natural transformation ${{\,\mathrm{{\textsf{T}}}\,}}(S, -)^{(I_0)} \longrightarrow F$ which induces an epimorphism on $S$. By Observation A.2 this gives rise to a natural transformation ${{\,\mathrm{{\textsf{T}}}\,}}(S^{I_0}, -) \longrightarrow F$ which induces an epimorphism on $S$. We pick $T_0 = S^{I_0}$ and this natural transfomation.

Now assume a natural transformation $\psi _i :{{\,\mathrm{{\textsf{T}}}\,}}(T_i, - ) \longrightarrow F$ inducing an epimorphism $\psi _i^S :{{\,\mathrm{{\textsf{T}}}\,}}(T_i, S) \longrightarrow F(S)$ is already constructed. Pick an epimorphism of ${{\,\mathrm{\textsf{End}}\,}}(S)$-modules ${{\,\mathrm{\textsf{End}}\,}}(S)^{(I_{i+1})} \longrightarrow {{\,\mathrm{\textsf{Ker}}\,}}\psi _i^S$. Similarly to the first step, this gives rise to a natural transformation ${{\,\mathrm{{\textsf{T}}}\,}}( S^{I_{i+1}}, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(T_i, -)$ such that the sequence

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(T_i, -) \xrightarrow {\psi _i} F \end{aligned}$$

is exact on $S$. In particular the composition vanishes. Picking $T_{i+1}$ to be the object fitting in the triangle $T_{i+1} \longrightarrow T_i \longrightarrow S^{I_{i+1}}) \longrightarrow T_{i+1}[1]$ we can employ Observation A.3 and obtain a natural transformation ${{\,\mathrm{{\textsf{T}}}\,}}(T_{i+1}, -) \longrightarrow F$. Moreover, evaluating at $S$ we can use the exactness observed above to conclude that we have the desired image.

As an intermediate step towards Theorem A.1 we will prove the following.

Proposition A.5

Theorem A.1 holds under the additional assumption that $F$ is representable.

The proof is based on the following result due to Keller and Nicolás [23].

Theorem A.6

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be a triangulated category at the base of a stable derivator. Given a commutative diagram

there is a choice of cone morphisms

$$\begin{aligned} \cdots \longrightarrow {{\,\mathrm{\textsf{Cone}}\,}}(\varphi _2) \longrightarrow {{\,\mathrm{\textsf{Cone}}\,}}(\varphi _1) \longrightarrow {{\,\mathrm{\textsf{Cone}}\,}}(\varphi _1) \end{aligned}$$

such that there is a triangle

$$\begin{aligned} {{\,\mathrm{\textsf{holim}}\,}}X_i \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}Y_i \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}{{\,\mathrm{\textsf{Cone}}\,}}(\varphi _i) \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}X_i[1]. \end{aligned}$$

Moreover, we will utilize the following observation.

Lemma A.7

Let $\cdots \longrightarrow X_2 \longrightarrow X_1 \longrightarrow X_0$ be a sequence of contravariant $S$-ghosts. Then ${{\,\mathrm{\textsf{holim}}\,}}X_i = 0$.

Proof

The vanishing of all maps implies that the sequence

$$\begin{aligned} \cdots \longleftarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_2, S) \longleftarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_1, S) \longleftarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_0, S) \end{aligned}$$

has vanishing colimit and is dual ML. It follows, since $S$ is $0$-cocomapact and invariant under suspension, that ${{\,\mathrm{\textsf{holim}}\,}}X_i$ is an $S$-ghost. But since $S$ is a cogenerator, this means that ${{\,\mathrm{\textsf{holim}}\,}}X_i = 0$. $\square$

Proof of Proposition A.5

Let $F = {{\,\mathrm{{\textsf{T}}}\,}}(X, -)$. Thus, via the inverse of the Yoneda functor, we have the commutative diagram

Applying Theorem A.6, we obtain a triangle

$$\begin{aligned} {{\,\mathrm{\textsf{holim}}\,}}X \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}T_i \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}{{\,\mathrm{\textsf{Cone}}\,}}(\psi _i) \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}X[1] \end{aligned}$$

for suitable cone morphisms. Clearly ${{\,\mathrm{\textsf{holim}}\,}}X = X$, so it remains to show that ${{\,\mathrm{\textsf{holim}}\,}}{{\,\mathrm{\textsf{Cone}}\,}}(\psi _i) = 0$.

By definition, a cone morphism makes the following diagram commutative.

Applying ${{\,\mathrm{{\textsf{T}}}\,}}(-, S)$ this turns into

To see this, note that the epimorphisms follow from the fact that the image of the middle vertical map is ${{\,\mathrm{{\textsf{T}}}\,}}(X, S)$ by assumption. Since $S$ is assumed to be invariant under suspension, it follows that we also get the claimed monomorphisms.

Again invoking the fact that the image of the middle vertical map is ${{\,\mathrm{{\textsf{T}}}\,}}(X, S)$, we see that the left vertical map needs to vanish. In other words, all our cone morphisms are contravariant $S$-ghosts. Now the claim follows from Lemma A.7. $\square$

Now we return to the general situation, where $F$ is not assumed to be representable a priori.

Proposition A.8

In the situation of Theorem A.1, there is a natural epimorphism ${{\,\mathrm{{\textsf{T}}}\,}}({{\,\mathrm{\textsf{holim}}\,}}T_i, -) \longrightarrow F$.

This result, as well as the argument here are based on [29, Theorem 8].

For the proof we will need to show that, for any given $X \in {{\,\mathrm{{\textsf{T}}}\,}}$, the induced map ${{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i, X) \longrightarrow F(X)$ is surjective.

Given $X$, we consider the functor ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)$, and construct the diagram of functors

as described in Construction A.4. In particular $X_0 = S^{I_0}$, and ${{\,\mathrm{\textsf{Cone}}\,}}g_i = S^{I_i}$ for suitable sets $I_i$. By Proposition A.5 we know that $X = {{\,\mathrm{\textsf{holim}}\,}}X_i$.

With this setup, the key step in the proof of Proposition A.8 is the following.

Lemma A.9

Given a sequence of maps ${{\,\mathrm{{\textsf{T}}}\,}}(X_i, -) \longrightarrow F$ making the solid part of the following diagram commutative, we can find the dashed arrows making the entire diagram commutative.

Proof

Constructing from left to right, observe first that we can find $\varphi _0$ since $X_0 = S^{I_0}$ and ${{\,\mathrm{{\textsf{T}}}\,}}(T_0, S^{I_0}) \longrightarrow F(S^{I_0})$ is a surjection.

Next, assume we have already constructed $\varphi _i$. Note that we have a triangle $X_{i+1} \longrightarrow X_i \longrightarrow S^{I_{i+1}} \longrightarrow X_{i+1}$. Thus, in order to obtain a map $\varphi _{i+1}$ making the square between ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _i, -)$ and ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _{i+1}, -)$ commutative, it suffices to show that the composition

$$\begin{aligned} T_{i+1} \xrightarrow {f_{i+1}} T_i \xrightarrow {\varphi _i} X_i \longrightarrow S^{I_{i+1}} \end{aligned}$$

vanishes. Equivalently we may consider the sequence

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_i, -) \xrightarrow {{{\,\mathrm{{\textsf{T}}}\,}}(\varphi _i, -)} {{\,\mathrm{{\textsf{T}}}\,}}(T_i, -) \xrightarrow {{{\,\mathrm{{\textsf{T}}}\,}}(f_{i+1}, -)} {{\,\mathrm{{\textsf{T}}}\,}}(T_{i+1}, -), \end{aligned}$$

and moreover it suffices to consider the evaluation at $S$. Now note that, by assumption, ${{\,\mathrm{{{\textsf{I}}}{{\textsf{m}}}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}(f_{i+1}, S) = F(S)$, so the above vanishing is implied by the vanishing of the composition

$$\begin{aligned}&[ {{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}, S) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_i, S) \xrightarrow {{{\,\mathrm{{\textsf{T}}}\,}}(\varphi _i, S)} {{\,\mathrm{{\textsf{T}}}\,}}(T_i, S) \longrightarrow F(S) ] \\&\quad = [ {{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}, S) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(X_i, S) \xrightarrow {{{\,\mathrm{{\textsf{T}}}\,}}(g_{i+1}, S)} {{\,\mathrm{{\textsf{T}}}\,}}(X_{i+1}, S) \longrightarrow F(S) ], \end{aligned}$$

which holds since the first two maps come from consecutive maps in a triangle.

Note however that at this point we cannot be sure that the triangle involving ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _{i+1}, -)$ and $F$ commutes. Let

$$\begin{aligned} \delta = [ {{\,\mathrm{{\textsf{T}}}\,}}(X_{i+1}, -) \longrightarrow F] - [{{\,\mathrm{{\textsf{T}}}\,}}(T_{i+1}, -) \longrightarrow F] \circ {{\,\mathrm{{\textsf{T}}}\,}}( \varphi _{i+1}, -) \end{aligned}$$

be the obstruction to the triangle commuting. Of course $\delta \circ {{\,\mathrm{{\textsf{T}}}\,}}(g_{i+1}, -) = 0$. Since $F$ is homological, it follows that $\delta$ factors through the map

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}(X_{i+1}, -) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}[-1], -), \end{aligned}$$

say via $\delta '$. Since $S$ is assumed to be invariant under suspension we may disregard the shift. Now recall that ${{\,\mathrm{{\textsf{T}}}\,}}(T_{i+1}, S) \longrightarrow F(S)$ is a surjection. It follows that any map ${{\,\mathrm{{\textsf{T}}}\,}}(S^{I_{i+1}}[-1], - ) \longrightarrow F$, in particular $\delta '$, factors through ${{\,\mathrm{{\textsf{T}}}\,}}(T_{i+1}, -) \longrightarrow F$. Thus we find a map $\delta '' :T_{i+1} \longrightarrow S^{I_{i+1}}[-1]$ making the following diagram commutative.

It follows that we can replace $\varphi _{i+1}$ by $\varphi _{i+1} + [S^{I_{i+1}}[-1] \longrightarrow X_{i+1}] \circ \delta ''$, fixing the commutativity of the triangle involving ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _{i+1}, -)$ and $F$, while not affecting the previously commutative square involving ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _i, -)$ and ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _{i+1}, -)$. $\square$

Remark A.10

The iterative construction in the proof of Lemma A.9 does not actually require the entire diagram. In particular any map ${{\,\mathrm{{\textsf{T}}}\,}}(X_i, -) \longrightarrow F$ factors through ${{\,\mathrm{{\textsf{T}}}\,}}(\varphi _i, -)$.

Now we are ready to prove the proposition.

Proof of Proposition A.8

Let $X \in {{\,\mathrm{{\textsf{T}}}\,}}$. We have $X = {{\,\mathrm{\textsf{holim}}\,}}X_i$, with the sequence $X_i$ as discussed directly below the proposition. Recall that $F$ is a homological functor commuting with products. Thus the triangle

$$\begin{aligned} {{\,\mathrm{\textsf{holim}}\,}}X_i \longrightarrow \prod X_i \longrightarrow \prod X_i \longrightarrow {{\,\mathrm{\textsf{holim}}\,}}X_i [1] \end{aligned}$$

gives rise to the exact sequence

$$\begin{aligned} \prod F(X_i[-1]) \longrightarrow \prod F(X_i[-1]) \longrightarrow F({{\,\mathrm{\textsf{holim}}\,}}X_i) \longrightarrow \prod F(X_i) \longrightarrow \prod F(X_i). \end{aligned}$$

By definition, the kernel of the last map is ${{\,\mathrm{\textsf{lim}}\,}}F(X_i)$, while the cokernel of the first map is ${{\,\mathrm{\textsf{lim}}\,}}^1 F(X_i[-1])$. Thus we have the lower sequence in the following diagram. The upper sequence exists by the same argument applied to the functor ${{\,\mathrm{{\textsf{T}}}\,}}({{\,\mathrm{\textsf{holim}}\,}}T_i, -)$.

In order to show that the middle vertical map is surjective it suffices to show that the outer two are surjective.

We first consider the left hand side. Note that Remark A.10 holds analogously for $X_i[-1]$, so all the maps ${{\,\mathrm{{\textsf{T}}}\,}}(T_i, X_i) \longrightarrow F(X_i)$ are surjective. In particular the same holds for the maps ${{\,\mathrm{{\textsf{T}}}\,}}({{\,\mathrm{\textsf{holim}}\,}}T_i, X_i) \longrightarrow F(X_i)$. Now the left vertical map is onto by right exactness of ${{\,\mathrm{\textsf{lim}}\,}}^1$.

Next we look at the right hand side. Note that an element of ${{\,\mathrm{\textsf{lim}}\,}}F(X_i)$ is a sequence of elements $x_i \in F(X_i)$ such that $F(g_i)(x_i) = x_{i-1}$. Translating via the Yoneda lemma these are maps $x_i :{{\,\mathrm{{\textsf{T}}}\,}}(X_i, -) \longrightarrow F$ such that assumptions of Lemma A.9 are satisfied. By that lemma we obtain maps $\varphi _i :T_i \longrightarrow X_i$. Now $(\varphi _i \circ [{{\,\mathrm{\textsf{holim}}\,}}T_i \longrightarrow T_i] )$ is an element of ${{\,\mathrm{\textsf{lim}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i, X_i )$, and moreover a preimage of $(x_i)$. $\square$

Now we are ready to complete the proof of Theorem A.1. The missing piece is [35, Theorem 1.3].

Proof of Theorem A.1

By Proposition A.8 there is a natural epimorphism

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i, -) \longrightarrow F. \end{aligned}$$

Note that its kernel again satisfies the assumptions of Proposition A.8, and thus is an epimorphic image of some ${{\,\mathrm{{\textsf{T}}}\,}}( {{\,\mathrm{\textsf{holim}}\,}}T_i', -)$. In particular $F$ has a projective presentation as indicated in the left half of the following diagram.

Let $C$ denote the cocone of the map ${{\,\mathrm{\textsf{colim}}\,}}T_i \longrightarrow {{\,\mathrm{\textsf{colim}}\,}}T_i'$. We have the exact sequence as in the diagram above from this triangle. Finally, note that since $F$ is homological we get the dashed morphism making the triangle to its left commutative. It follows that $F$ is a direct summand of ${{\,\mathrm{{\textsf{T}}}\,}}(C, -)$, hence is representable.

Now the claim of Theorem A.1 follows from Proposition A.5. $\square$

Appendix B: Exactness of partial Serre functors

The aim of this appendix is to prove the following theorem, summing up functorial properties of partial Serre functors.

Theorem 3.3

Suppose ${{\,\mathrm{{\textsf{T}}}\,}}$ is a triangulated category, and let ${{\,\mathrm{{\textsf{X}}}\,}}$ be the full subcategory of all objects $X$ such that ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ is representable.

Then ${{\,\mathrm{{\textsf{X}}}\,}}$ is a triangulated subcategory of ${{\,\mathrm{{\textsf{T}}}\,}}$, and there is a partial Serre functor ${\mathbb {S}} :{{\,\mathrm{{\textsf{X}}}\,}}\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}$. Moreover, ${\mathbb {S}}$ is a triangle functor.

Most of this theorem is actually fairly easily seen. In fact, the existence of a partial Serre functor ${\mathbb {S}} :{{\,\mathrm{{\textsf{X}}}\,}}\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}$ follows easily from our assumption on representability — see Observation B.1 below. The main technical challenges in proving the theorem are showing that ${{\,\mathrm{{\textsf{X}}}\,}}$ is triangulated, which we will show in Corollary B.8 for the case that ${{\,\mathrm{{\textsf{T}}}\,}}$ is idempotent closed and in the very last subsection in general, and that ${\mathbb {S}}$ is a triangle functor—see Theorem B.10.

General observations on partial Serre functors

Observation B.1

Let ${{\,\mathrm{{\textsf{X}}}\,}}$ be a subcategory of ${{\,\mathrm{{\textsf{T}}}\,}}$ such that for any object $X \in {{\,\mathrm{{\textsf{X}}}\,}}$ the functor ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ is representable. Fix for all $X$ a representing object $\mathbb S X$ and a natural isomorphism

$$\begin{aligned} \eta _X :{{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(-, {\mathbb {S}} X ). \end{aligned}$$

Then ${\mathbb {S}}$ defines a functor ${{\,\mathrm{{\textsf{X}}}\,}}\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}$ by requiring the following square of functors to be commutative for any morphism $f :X_1 \longrightarrow X_2$ in ${{\,\mathrm{{\textsf{X}}}\,}}$.

Note that the lower natural transformation exists and is unique since $\eta _{X_1}$ is an isomorphism, and that it is uniquely representable as composition with some map by the Yoneda Lemma.

It follows directly from the commutative square defining $\mathbb S f$ that $\eta$ becomes a natural isomorphism of functors on ${{\,\mathrm{{\textsf{X}}}\,}}\times {{\,\mathrm{{\textsf{T}}}\,}}^{{{\,\mathrm{{{\textsf{o}}}{{\textsf{p}}}}\,}}}$, i.e. that ${\mathbb {S}}$ is a partial Serre functor.

Remark B.2

One may see that a Serre functor on ${{\,\mathrm{{\textsf{X}}}\,}}$ is unique up to unique natural isomorphism. Indeed, if $\widetilde{{\mathbb {S}}}$ is a different choice of a partial Serre functor on ${{\,\mathrm{{\textsf{X}}}\,}}$, with corresponding natural isomorphism ${\widetilde{\eta }} :{{\,\mathrm{{\textsf{T}}}\,}}(-, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(-, \widetilde{{\mathbb {S}}} -)$, then ${\widetilde{\eta }} \circ \eta ^{-1}$ is a natural isomorphism ${{\,\mathrm{{\textsf{T}}}\,}}(-, {\mathbb {S}} - ) \longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(-, \widetilde{{\mathbb {S}}} - )$, which, by the Yoneda lemma, comes from a natural isomorphism ${\mathbb {S}} \longrightarrow \widetilde{ {\mathbb {S}} }$.

Observation B.3

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be a $k$-category, and $[1]$ be an automorphism of ${{\,\mathrm{{\textsf{T}}}\,}}$. If ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ is representable for some object $X$, then so is ${{\,\mathrm{{\textsf{T}}}\,}}(X[1], -)^*$:

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}(X[1], -)^*\cong {{\,\mathrm{{\textsf{T}}}\,}}(X, - )^*\circ [-1] \cong {{\,\mathrm{{\textsf{T}}}\,}}(-, {\mathbb {S}} X) \circ [-1] \cong {{\,\mathrm{{\textsf{T}}}\,}}(-, ({\mathbb {S}} X) [1]). \end{aligned}$$

In particular, the subcategory of all objects $X$ such that ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ is representable, is invariant under all automorphisms of ${{\,\mathrm{{\textsf{T}}}\,}}$.

Note however that we may not be able to choose $\eta _{X[1]}$ to be induced by $\eta _X$ consistently over the entire subcategory ${{\,\mathrm{{\textsf{X}}}\,}}$. To account for this, we observe that at least there is a natural isomorphism controlling the difference between the two.

Observation B.4

Let ${\mathbb {S}}$ be a partial Serre functor on a subcategory ${{\,\mathrm{{\textsf{X}}}\,}}$ of ${{\,\mathrm{{\textsf{T}}}\,}}$. Assume ${{\,\mathrm{{\textsf{T}}}\,}}$ has an automorphism $[1]$, and that ${{\,\mathrm{{\textsf{X}}}\,}}$ is invariant under $[1]$. Then there is a unique natural isomorphism $\zeta :{\mathbb {S}} \circ [1] \longrightarrow [1] \circ {\mathbb {S}}$ such that the following diagram of isomorphisms commutes functorially in $X$ and in $T$.

$\varvec{\textsf{X}}$ is triangulated

Denote by ${{\,\mathrm{\textsf{Mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ the category of functors ${{\,\mathrm{{\textsf{T}}}\,}}^{{{\,\mathrm{{{\textsf{o}}}{{\textsf{p}}}}\,}}} \longrightarrow {{\,\mathrm{\textsf{Mod}}\,}}k$, and by ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ the subcategory of finitely presented functors. Recall that by [18, Theorem 3.1] the category ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ is Frobenius abelian, and its injectives are precisely the direct summands of representable functors. (Note that in [18] the objects of ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ are described as images of morphisms of representable functors, rather than as cokernels of such morphisms. However, since we can always complete triangles there is no difference between these categories.) Therefore we are particularly interested in injective functors.

Lemma B.5

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be an additive category, and let $T$ be an object in ${{\,\mathrm{{\textsf{T}}}\,}}$. Then ${{\,\mathrm{{\textsf{T}}}\,}}(T, -)^*$ is injective in ${{\,\mathrm{\textsf{Mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$.

Proof

We observe that for $F \in {{\,\mathrm{\textsf{Mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ we have the natural isomorphism

$$\begin{aligned} ({{\,\mathrm{\textsf{Mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}})(F, {{\,\mathrm{{\textsf{T}}}\,}}(T, -)^*) {\mathop {\longleftrightarrow }\limits ^{\cong }}F(T)^*. \end{aligned}$$

The map from left to right is given by sending a natural transformation $\eta$ to the composition $\left( F(T) {\mathop {\longrightarrow }\limits ^{\eta _T}}{{\,\mathrm{{\textsf{T}}}\,}}(T, T)^*\xrightarrow {{{\,\mathrm{{{\textsf{e}}}{{\textsf{v}}}}\,}}_{{{\,\mathrm{{{\textsf{i}}}{{\textsf{d}}}}\,}}}} I \right)$. The map from right to left sends a linear form $\phi$ to the natural transformation

$$\begin{aligned} F(X)&\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(T, X)^*\\ f&\longmapsto [ t \mapsto (\phi \circ F(t))(f)]. \end{aligned}$$

Since $F(T)^*$ is exact in $F$ it follows that ${{\,\mathrm{{\textsf{T}}}\,}}(T, -)^*$ is injective. $\square$

Proposition B.6

Suppose ${{\,\mathrm{{\textsf{T}}}\,}}$ is triangulated and idempotent complete, and let $T \in {{\,\mathrm{{\textsf{T}}}\,}}$. Then the functor ${{\,\mathrm{{\textsf{T}}}\,}}(T, -)^*$ is representable if and only if it is finitely presented.

Proof

Clearly any representable functor is finitely presented.

Assume conversely that ${{\,\mathrm{{\textsf{T}}}\,}}(T, -)^*$ is finitely presented. Since it is injective even in ${{\,\mathrm{\textsf{Mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$, it is also injective in ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$. Now, since ${{\,\mathrm{{\textsf{T}}}\,}}$ is assumed to be idempotent complete, the claim follows from [21, Proposition 15.1]. $\square$

Theorem B.7

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be triangulated and idempotent complete, and take a triangle $X \longrightarrow Y \longrightarrow Z \longrightarrow X[1]$ in ${{\,\mathrm{{\textsf{T}}}\,}}$.

If both ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ and ${{\,\mathrm{{\textsf{T}}}\,}}(Y, -)^*$ are representable, then ${{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*$ is representable.

Proof

By Proposition B.6, it suffices to show that ${{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*$ is finitely presented.

Consider the exact sequence

$$\begin{aligned} {{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(Y, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(X[1], -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(Y[1],-)^*. \end{aligned}$$

Since the leftmost two terms are finitely presented—in fact they are representable—it follows that the image of the map ${{\,\mathrm{{\textsf{T}}}\,}}(Y, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*$ is finitely presented. Since the rightmost two terms are finitely presented by Observation B.3, and ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$ is abelian, we also have that the image of the map ${{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}(X[1], -)^*$ is finitely presented. Now the claim follows, since extensions of finitely presented functors are finitely presented. $\square$

Corollary B.8

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be triangulated and idempotent complete. Then the collection of objects $X$ such that ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ is representable, is a triangulated subcategory of ${{\,\mathrm{{\textsf{T}}}\,}}$.

S is a triangle functor

Proposition B.9

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be a category with an automorphism $[1]$, and let ${{\,\mathrm{{\textsf{X}}}\,}}\subset {{\,\mathrm{{\textsf{T}}}\,}}$ be a $[1]$-invariant subcategory admitting a partial Serre functor ${\mathbb {S}}$. Let

$$\begin{aligned} X {\mathop {\longrightarrow }\limits ^{x}}Y {\mathop {\longrightarrow }\limits ^{y}}Z {\mathop {\longrightarrow }\limits ^{z}}X[1] \,\text {and} \, T {\mathop {\longrightarrow }\limits ^{t}}U {\mathop {\longrightarrow }\limits ^{u}}V {\mathop {\longrightarrow }\limits ^{v}}T[1] \end{aligned}$$

be sequences of objects and morphisms in ${{\,\mathrm{{\textsf{X}}}\,}}$ and ${{\,\mathrm{{\textsf{T}}}\,}}$, respectively.

Assume that for any $f$ and $g$ in the following diagram, there is a morphism $h$ making the diagram commutative.

Then also in the following diagram we have that for any $f$ and $g$ there is $h$ making it commutative.

Proof

The assumption may be reformulated into the statement that the morphism between the kernels in the following commutative diagram, is onto.

Here the vertical maps are projections to the respective summands, and the horizontal maps are given by

$$\begin{aligned} \left[ \begin{matrix} t \circ \star &{} - (\star \circ x) &{} 0 \\ 0 &{} u \circ \star &{} - (\star \circ y) \\ - (\star [1] \circ z) &{} 0 &{} v \circ \star \end{matrix} \right] \text { and \,} [ t \circ \star \; - (\star \circ x) ], \end{aligned}$$

respectively. Here $t \circ \star$ stands for $f \longmapsto t \circ f$, and similar.

We may consider the kernels of the vertical projections, and a cokernel morphism as indicated in the following diagram.

By the Snake Lemma we observe that the map ${{\,\mathrm{\textsf{Ker}}\,}}_1 \longrightarrow {{\,\mathrm{\textsf{Ker}}\,}}_2$ being epi is equivalent to the map ${{\,\mathrm{\textsf{Cok}}\,}}_0 \longrightarrow {{\,\mathrm{\textsf{Cok}}\,}}_1$ being mono.

Dualizing the upper two rows and the rightmost three columns of this diagram, and identifying via the natural isomorphism $\eta$ defining the partial Serre functor, we obtain the diagram

with the horizontal maps given by

$$\begin{aligned} \left[ \begin{matrix} \star \circ t &{} 0 &{} - (\zeta _X \circ {\mathbb {S}} z \circ \star )[-1] \\ - ({\mathbb {S}} x \circ \star ) &{} \star \circ u &{} 0 \\ 0 &{} -({\mathbb {S}} y \circ \star ) &{} \star \circ v \end{matrix} \right] \text { and } [- {\mathbb {S}} y \circ \star \qquad \; \star \circ v]. \end{aligned}$$

Most of these entries are immediate from the naturality of $\eta$, only the term in the right upper corner warrants further explanation. Note that the map in question is induced via $\eta$ by the composition along the upper row of the following diagram. Hence it is precisely the composition appearing in the lower row of that diagram.

Here the left square commutes by the naturality of $\eta$, and the pentagon on the right comes from Observation B.4.

Now we can translate the statement that the morphism between kernels is surjective, back to a commutative diagram: It means that for any $f \in {{\,\mathrm{{\textsf{T}}}\,}}(V, {\mathbb {S}} Y)$ and $g \in {{\,\mathrm{{\textsf{T}}}\,}}(T[1], {\mathbb {S}} Z)$ such that ${\mathbb {S}} y \circ f = g \circ v$—i.e. for each element of the lower kernel—there is $h \in {{\,\mathrm{{\textsf{T}}}\,}}(U, {\mathbb {S}} X)$ such that $h \circ t = (\zeta _X \circ {\mathbb {S}} z \circ g)[-1]$ and ${\mathbb {S}} x \circ h = f \circ u$ — i.e. a preimage in the upper kernel. In diagrammatic language we thus have

as desired. $\square$

Theorem B.10

Let ${{\,\mathrm{{\textsf{T}}}\,}}$ be a triangulated category, and let ${{\,\mathrm{{\textsf{X}}}\,}}$ be a triangulated subcategory admitting a partial Serre functor ${\mathbb {S}}$.

Then ${\mathbb {S}} :{{\,\mathrm{{\textsf{X}}}\,}}\longrightarrow {{\,\mathrm{{\textsf{T}}}\,}}$ is a triangle functor.

Proof

The first ingredient to a triangle functor is a natural isomorphism

$$\begin{aligned} {\mathbb {S}} \circ [1] \longrightarrow [1]\circ {\mathbb {S}}. \end{aligned}$$

We have already constructed such a natural isomorphism $\zeta$ in Observation B.4. However, here we will choose the natural isomorphism $- \zeta$.

It remains for us to show that for any triangle $X {\mathop {\longrightarrow }\limits ^{x}}Y {\mathop {\longrightarrow }\limits ^{y}}Z {\mathop {\longrightarrow }\limits ^{z}}X[1]$ in ${{\,\mathrm{{\textsf{X}}}\,}}$, the sequence

$$\begin{aligned} {\mathbb {S}} X {\mathop {\longrightarrow }\limits ^{{\mathbb {S}} x}}{\mathbb {S}} Y {\mathop {\longrightarrow }\limits ^{{\mathbb {S}} y}}{\mathbb {S}} Z \xrightarrow {(- \zeta _X) \circ {\mathbb {S}} z} ({\mathbb {S}} X)[1] \end{aligned}$$

is a triangle in ${{\,\mathrm{{\textsf{T}}}\,}}$.

Note that there exists a triangle $({\mathbb {S}} Z)[-1] \longrightarrow U \longrightarrow {\mathbb {S}} Y {\mathop {\longrightarrow }\limits ^{{\mathbb {S}} y}}{\mathbb {S}} Z$. By the axioms of triangulated categories, any two triangles satisfy the assumptions of Proposition B.9. We will apply that proposition to our original triangle and the one involving $U$. Choosing the two free vertical maps in the conclusion to be identities, we obtain the commutative diagram

Let $T \in {{\,\mathrm{{\textsf{T}}}\,}}$. Applying ${{\,\mathrm{{\textsf{T}}}\,}}(T, -)$ to the above diagram we obtain

The leftmost upper square commutes by naturality of $\zeta$. For the second left square from the top, note that ${\mathbb {S}}( z [-1] ) \circ \zeta _{Z[-1]}[-1] = \zeta _X[-1] \circ ({\mathbb {S}} z)[-1]$ by naturality of $\zeta$, and this in turn is equal to $h \circ t$.

The leftmost column of this diagram is exact, and so is the rightmost column, since it is the dual of an exact sequence. Thus the middle column is also exact, so the Five lemma applies, and tells us that $h \circ \star$ is an isomorphism. By the Yoneda lemma this implies that $h$ is an isomorphism.

Now

$$\begin{aligned} {\mathbb {S}} X {\mathop {\longrightarrow }\limits ^{ {\mathbb {S}} x }}{\mathbb {S}} Y {\mathop {\longrightarrow }\limits ^{{\mathbb {S}} y}}{\mathbb {S}} Z \xrightarrow {(- \zeta _X) \circ {\mathbb {S}} z} ({\mathbb {S}} X)[1] \end{aligned}$$

is isomorphic to the triangle $U {\mathop {\longrightarrow }\limits ^{u}}{\mathbb {S}} Y {\mathop {\longrightarrow }\limits ^{{\mathbb {S}} y}}{\mathbb {S}} Y \xrightarrow {-t[1]} U[1]$, hence it is a triangle itself. $\square$

The case that $\textsf{T}$ in not idempotent closed

By [4], any triangulated category ${{\,\mathrm{{\textsf{T}}}\,}}$ is canonically embedded into an idempotent closed triangulated category ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$. More precisely, ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$ is the category of injective objects in ${{\,\mathrm{\textsf{mod}}\,}}{{\,\mathrm{{\textsf{T}}}\,}}$.

We denote by ${{\,\mathrm{{\textsf{X}}}\,}}$ and ${{\widehat{{{\,\mathrm{{\textsf{X}}}\,}}}}}$ the subcategories of objects X and ${{\widehat{X}}}$ such that ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ and ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}} ({{\widehat{X}}}, -)^*$ are representable, respectively. One easily observes, using the fact that any object in ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$ is a direct summand of an object in ${{\,\mathrm{{\textsf{T}}}\,}}$, that ${{\,\mathrm{{\textsf{X}}}\,}}\subset {{\widehat{{{\,\mathrm{{\textsf{X}}}\,}}}}}$. Note that by Theorem B.7 we know that ${{\widehat{{{\,\mathrm{{\textsf{X}}}\,}}}}}$ is a triangulated subcategory of ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$.

Now let $X \longrightarrow Y \longrightarrow Z \longrightarrow X[1]$ be a triangle in ${{\,\mathrm{{\textsf{T}}}\,}}$, such that ${{\,\mathrm{{\textsf{T}}}\,}}(X, -)^*$ and ${{\,\mathrm{{\textsf{T}}}\,}}(Y, -)^*$ are representable, say by ${\mathbb {S}} X$ and ${\mathbb {S}} Y$. By the above comment we know that ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}(Z, -)^*$ is representable, say by $\widehat{{\mathbb {S}} Z} \in {{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$. Finally we apply Theorem B.10, which tells us that ${\mathbb {S}} X \longrightarrow {\mathbb {S}} Y \longrightarrow \widehat{ {\mathbb {S}} Z } \longrightarrow {\mathbb {S}} X [1]$ is a triangle in ${{\widehat{{{\,\mathrm{{\textsf{T}}}\,}}}}}$. But since ${{\,\mathrm{{\textsf{T}}}\,}}$ is a triangulated subcategory of $\widehat{{\,\mathrm{{\textsf{T}}}\,}}$, and two of the terms of the triangle lie in ${{\,\mathrm{{\textsf{T}}}\,}}$, so does $\widehat{ {\mathbb {S}} Z}$. Thus we have shown that ${{\,\mathrm{{\textsf{T}}}\,}}(Z, -)^*$ is representable.