Let $\mathcal{A}$ be an abelian category, $\mathcal{T}, \mathcal{F}$ two strictly full additive subcategories of $\mathcal{A}$. Then according to nLab and other sources including Constructing Torsion Pairs by Assem and Kerner, $(\mathcal{T}, \mathcal{F})$ is a torsion pair/theory if the following hold:

  1. $\operatorname{Hom}_{\mathcal{A}}(\mathcal{T}, \mathcal{F}) = 0$.
  2. For every $X \in \mathcal{A}$, there are objects $T \in \mathcal{T}$ and $F \in \mathcal{F}$ and a short exact sequence $$0 \rightarrow T \rightarrow X \rightarrow F \rightarrow 0$$

or equivalently if

  1. $\operatorname{Hom}_{\mathcal{A}}(X,\mathcal{F}) = 0 $ if and only if $X \in \mathcal{T}$.
  2. $\operatorname{Hom}_{\mathcal{A}}(\mathcal{T}, X) = 0$ if and only if $X \in \mathcal{F}$.

Now, I can see how $(1) \& (2) \Rightarrow (3)\&(4)$ and $(3)\&(4) \Rightarrow (1)$.

The problem I'm having is $(3)\&(4) \Rightarrow (2)$.

Say you have some $X \in \mathcal{A}$ and $X \notin \mathcal{T}$. Then there is an object $F \in \mathcal{F}$ and a nonzero map $f : X \rightarrow F$. Let $\pi : X \rightarrow \operatorname{Im}(f)$ and $i : \operatorname{Im}(f) \rightarrow F$ the canonical projection and inclusion maps. For any $T \in \mathcal{T}$ and any nonzero map $t : T \rightarrow X$, we have $ft = i \pi t = 0$, so $\pi t \subseteq \operatorname{Ker}(i) = 0$. For any nonzero map $t' : T \rightarrow \operatorname{Im}(f)$, the composition $it' = 0$, so $\operatorname{Im}(f) \in \mathcal{F}$.

So $\mathcal{F}$ is closed under subobjects, similarly we show that $\mathcal{T}$ is closed under factors. How then, starting with any object $X$, can I build a sequence as above?

  • $\begingroup$ Does $\operatorname{Hom}_\mathcal{A}(X,\mathcal{F})=0$ mean that $\operatorname{Hom}_\mathcal{A}(X,Y)=0$ for all $Y\in \mathcal{F}$ ? $\endgroup$ – Captain Lama May 15 at 22:23
  • $\begingroup$ @CaptainLama Yes, exactly. $\endgroup$ – Auclair May 15 at 22:30

A nonzero map $T\to im(f)$ composes with the inclusion to give a nonzero map $T\to F$ (the inclusion is monic), so $im(f)\in\mathcal{F}$.

More generally, $\mathcal{F}$ is closed under subobjects and $\mathcal{T}$ under quotients if we have 3 and 4

I don't see where in the nLab they state this equivalence, they seem to always assumes condition 2 (which is called "the last condition" in the nLab page here).

Here's a proof in nice cases : we assume $\mathcal{A}$ has direct sums, and is well-powered, that is, for every object there is a set of subobjects such that any subobject is isomorphic to one in this set (by duality, we could assume that $\mathcal{A}$ has products and is well-co-powered). Then it is clear from conditions 3 and 4 that $\mathcal{T}$ is closed under direct sums.

Now let $X$ be any object, and let $I$ be a set together with subobject $t_i : T_i \to X$ with $T_i\in\mathcal{T}$ such that any subobject of $X$ that is in $\mathcal{T}$ is isomorphic to one of them (as a subobject ! - this is possible because $\mathcal{A}$ is well-powered). Then as $\mathcal{T}$ is closed under quotients and direct sums, the image of $\bigoplus_i T_i \to X$, which we denote by $T=\sum_i T_i$, is in $\mathcal{T}$. The idea is to prove that $X/T$ is in $\mathcal{F}$.

We use condition 4. for this : let $L\in\mathcal{T}$ and $f: L\to X/T$ be a map. Then its image is of the form $H/T$ for some subobject $H$ of $X$ bigger than $T$ (this is obvious for module categories, and you can prove it by hand, but doing so depends on your level of comfort with abelian categories, and can be very tedious). As $\mathcal{T}$ is closed under quotients and $L$ is in it, $H/T\in\mathcal{T}$.

Moreover we have the short exact sequence $0\to T\to H\to H/T\to 0$ from which it follows that $H\in\mathcal{T}$. But then $H$ is a subobject of $X$ that is in $\mathcal{T}$ so it's isomorphic to some $T_i$ as a subobject, from which it follows that $H$ is a subobject of $T$; but it was also bigger than $T$ so by some standard nonsense about subobjects, we get that they're isomorphic (as subobjects) and so $H/T = 0$, so that $im(f) = 0$ and so $f=0$. Therefore by condition 4., $X/T\in\mathcal{F}$ and so $0\to T\to X\to X/T\to 0$ is the desired sequence.

I don't have a counterexample for the general case, but here's an explanation for why the conditions seem natural : if you fiddle around with the conditions, you'll see that any sequence $T\to^g X\to^f F$ is a chain complex, that is, the composition is $0$, so $im(g)\subset \ker(f)$. So if you have the desired short exact sequence, since you have equality, the image of $g$ will be bigger than all images of all morphisms $T\to X$, and $\ker (f)$ will be smaller than all kernels of all morphisms $X\to F$.

So you need to somehow make sure that you can take a supremum of all images, or an infimum of all kernels; and make sure that they are attained. Now these conditions are closely related to completeness assumptions on the category; and well-poweredness (if you risk having a proper class of non-isomorphic subobjects, then the usual assumptions of completeness/cocompleteness will not give you anything).

Then if you think in analogy with torsion groups/torsion-free groups, it becomes clear that you'll need to take the sum of all sub-torsion groups; or take the intersection of all kernels into torsion-free groups by taking their product for instance, hence the assumptions we made here, or their dual

Moreover, note that these assumptions are met in usual abelian categories, so if it doesn't hold in general, finding a counterexample will not be easy. Essentially if you look at the proof, you have to find a category with an object $X$ such that the collection of subobjects of $X$ that are in $\mathcal{T}$ has no supremum, or such that its supremum isn't in $\mathcal{T}$ (but that's probably harder to force: it would imply that the maps $T_i\to T$ aren't jointly epic, which I can't prove is absurd right now, but seems more complicated to achieve)

  • $\begingroup$ Ah, right. I made an edit to my post correcting that. Do you know how to find a short exact sequence as in (2) starting with (3) and (4)? $\endgroup$ – Auclair May 16 at 9:45
  • $\begingroup$ Where is the equivalence stated ? $\endgroup$ – Max May 16 at 12:17
  • $\begingroup$ Well, not anywhere that I can find. But sources give these two definitions for torsion pairs, so I just assumed they are equivalent definitions. $\endgroup$ – Auclair May 16 at 12:27
  • $\begingroup$ But the nLab doesn't give the second definition it seems $\endgroup$ – Max May 16 at 12:44
  • $\begingroup$ Here it is: core.ac.uk/download/pdf/82276884.pdf $\endgroup$ – Auclair May 16 at 13:48

$\def\cV{\mathcal{V}}\def\cA{\mathcal{A}}\def\cF{\mathcal{F}}\def\cT{\mathcal{T}}$In general, (3)+(4) don't imply (2). Let $k$ be a field. Let $\cV$ be the category whose objects are triples $(V,W,a)$ where $V$ and $W$ are $k$-vector spaces and $a: V \to W$ is a $k$-linear map. A morphism $(V,W,a) \to (V', W', a')$ is a pair of $k$-linear maps $b:V \to V'$ and $c:W \to W'$ forming a commutative square $$\begin{matrix} V & \overset{a}{\longrightarrow} & W \\ b \downarrow & & c \downarrow \\ V' & \overset{a'}{\longrightarrow} & W' \\ \end{matrix} \qquad (\ast)$$ (In other words, representations of the quiver $\bullet \longrightarrow \bullet$.) This is easily seen to be abelian.

Let $\cA$ be the full subcategory of $\cV$ on objects $(V,W,a)$ where $\mathrm{Ker}(a)$ and $\mathrm{CoKer}(a)$ are finite dimensional.

Lemma $\cA$ is closed within $\cV$ under taking kernels and cokernels.

Proof I attributed this to the snake lemma before. It's true, but not that easy. Here is a proof. Consider $(\ast)$ as a double complex and form the two spectral sequences. If we do the horizontal maps first, then all the spaces on the next page are finite dimensional by hypothesis, so the spectral sequence must converge to something finite dimensional.

If we take the vertical maps first, we get $$\begin{matrix} \mathrm{Ker}(b) & \longrightarrow & \mathrm{Ker}(c) \\ & & \\ \mathrm{CoKer}(b) & \longrightarrow & \mathrm{CoKer}(c) \\ \end{matrix} \quad (\dagger)$$ on the next page. On the pages after that one, there are no more maps, so the maps in $(\dagger)$ must have finite dimensional kernel and cokernel; this is what we want to show. $\square$

I'm sure this can be done without spectral sequences, but this is the first proof I found.

A full subcategory of an abelian category, closed under kernel and cokernel, is itself an abelian category, so $\cA$ is abelian.

Now, let $\cT$ be the subcategory of $\cA$ on objects of the form $0 \to k^N$ and let $\cF$ be the subcategory on objects of the form $k^M \to 0$. Verification of (3) and (4): We certainly have $\mathrm{Hom}(0 \to k^N, k^M \to 0)=0$. For any object $X = (V,W,a)$, if $\mathrm{Hom}(0 \to k, (V,W,a))=0$ then $W=0$ so $(V,W,a) \in \cF$ and, if $\mathrm{Hom}((V,W,a), k \to 0)$ is $0$ then $V=0$, so $(V,W,a) \in \cT$.

Now, let $X$ be $k^{\infty} \to k^{\infty}$, where the map is an isomorphism. There does not exist any $F \in \cF$ with a map $X \to F$ whose kernel is in $\cT$, so there is no extension as in (2).

  • $\begingroup$ +1. So what fails with respect to my answer is that $\mathcal{A}$ doesn't have all direct sums, or all direct products $\endgroup$ – Max 2 days ago

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