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Let $\mathcal{A}$ be an Abelian category with enough injectives. Is it always possible to make the injective embedding functorial? By this I mean that there should exist a functor $I \colon \mathcal{A} \to \mathcal{A}$ and a natural transformation $\operatorname{id} \to I$ such that for all objects $A$, the mapping $A \to I(A)$ is a monomorphism. This should be the same as this definition from the Stacks project. (edit: in a first version, I was requiring the functor to be additive, which is not what I had in mind even in the case of modules, as pointed out by Jeremy Rickard)

The Stacks project distinguishes categories with enough injectives from categories with functorial injective embeddings, so the two notions should be different. But I realized that I cannot think of an example of a category with enough injectives that does not admit functorial injective embeddings.

edit removed a motivation comment that was sparking more discussion than necessary and distracting from the main question.

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    $\begingroup$ I don't have an answer to your question, but I agree it probably won't be anything obvious. I do have a comment about the motivation. Even without functorial injective embeddings, one can choose (assuming a strong form of the axiom of choice) an injective resolution for each object of $\mathcal{A}$. The standard arguments, guarantee the existence of a morphism between chosen injective resolutions lifting a given one, and this is unique up to homotopy. This should give functoriality of derived functors. Aside from a philosophical objection to the first step, I'm not sure I see a problem. $\endgroup$ – Donu Arapura Apr 5 at 14:56
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    $\begingroup$ Well, it would be $\mathbb{Q}/\mathbb{Z}$, right? I don't understand why it would have to be annihilated by $2$. In fact Theorem 19.2.8 in the stacks project constructs a functorial injective embedding for $\operatorname{Mod}_R$. $\endgroup$ – Andrea Ferretti Apr 6 at 13:37
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    $\begingroup$ But the functor constructed there is not additive, which was one of the conditions you listed. $\endgroup$ – Jeremy Rickard Apr 6 at 13:51
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    $\begingroup$ Welcome back Andrea =] $\endgroup$ – Harry Gindi Apr 7 at 10:43
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    $\begingroup$ @JeremyRickard: I think the following is a functorial injective resolution on countable abelian groups. Let $I(A)$ be the quotient of $\mathbf Q^{(A)}$ by the subgroup (not $\mathbf Q$-subspace) generated by $e_{a+b} - e_a - e_b$ for $a,b \in A$. The natural presentation of $A$ as $\mathbf Z^{(A)}$ modulo the same relations gives an injection $A \hookrightarrow I(A)$. Clearly $I(A)$ is divisible and countable, because the same holds for $\mathbf Q^{(A)}$. For example, $I(0) = (\mathbf Q/\mathbf Z)e_0$, and $I(\mathbf Z/2\mathbf Z) = (\mathbf Q/\mathbf Z)e_0 \oplus (\mathbf Q/2\mathbf Z)e_1$. $\endgroup$ – R. van Dobben de Bruyn Apr 8 at 18:58
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Since the dual of an abelian category is also an abelian category, the question is equivalent to the same question for projective resolutions.

I will show that the category $\mathbf{Ab}^{\operatorname{f.t.}}$ of finitely generated abelian groups has enough projectives, but no functorial projective cover. The idea is that multiplication by any $n \in \mathbf Z$ is central in $\mathbf{Ab}$, and we do some representation theory to show that $F(\mathbf Z)$ has to have infinite rank by considering the action of multiplication by $n$ on $\mathbf Z/m$ for all $m$.

First some trivial lemmas:

Lemma 1. The category $\mathbf{Ab}^{\operatorname{f.t.}}$ has enough projectives. For $A \in \mathbf{Ab}^{\operatorname{f.t.}}$, the following are equivalent

  1. $A$ is projective in $\mathbf{Ab}^{\operatorname{f.t.}}$;
  2. $A$ is projective in $\mathbf{Ab}$;
  3. $A$ is finite free.

Proof. Implications (2) $\Rightarrow$ (1) and (2) $\Leftrightarrow$ (3) are clear. This immediately gives the first statement. For (1) $\Rightarrow$ (3), choose a surjection $F \twoheadrightarrow A$ with $F$ finite free. By assumption (1) it splits, so $A$ is a summand of a finite free module, hence finite free. $\square$

Lemma 2. Let $F \twoheadrightarrow G$ be an epimorphism of functors $F, G \colon \mathscr C \to \mathscr D$. If $G$ is faithful, then so is $F$.

Proof. Two maps $f, g \colon A \rightrightarrows B$ give a commutative diagram $$\begin{array}{ccc}F(A) & \twoheadrightarrow & G(A)\\\downdownarrows & & \downdownarrows\\F(B) & \twoheadrightarrow & G(B).\!\end{array}$$ Since the top map is an epimorphism, we see $F(f) = F(g) \Rightarrow G(f) = G(g)$, which by assumption implies $f = g$. $\square$

We are now ready for the main result.

Proposition. Let $F \colon \mathbf{Ab}^{\operatorname{f.t.}} \to \mathbf{Ab}$ be a functor taking every object to a projective object, together with a natural surjection $F \twoheadrightarrow \iota$ onto the inclusion $\iota \colon \mathbf{Ab}^{\operatorname{f.t.}} \to \mathbf{Ab}$. Then $F(\mathbf Z)$ has infinite rank. In particular, there is no such functor landing in $\mathbf{Ab}^{\operatorname{f.t.}}$.

By Lemma 1, this shows that there is no functorial projective hull on $\mathbf{Ab}^{\operatorname{f.t.}}$.

Proof. First note that Lemma 2 implies that $F$ is faithful, i.e. for all $A, B \in \mathbf{Ab}^{\operatorname{f.t.}}$, the map \begin{align*} \operatorname{Hom}(A,B) &\to \operatorname{Hom}(F(A),F(B))\\ f &\mapsto f_* \end{align*} is injective. For any $n > 1$, we can equip every $F(A)_{\mathbf Q} = F(A) \otimes_{\mathbf Z} \mathbf Q$ for $A \in \mathbf{Ab}^{\operatorname{f.t.}}$ with the structure of a $\mathbf Q[x]$-module by letting $x$ act by $n_*$, where $n \colon A \to A$ is multiplication by $n$ (so $x^k$ acts by $(n_*)^k = (n^k)_*$ for $k \geq 0$). For any $m$, the natural surjection $\pi \colon \mathbf Z \to \mathbf Z/m$ gives a commutative diagram $$\begin{array}{ccc}\mathbf Z & \stackrel{n^k}\to & \mathbf Z \\ \!\!\!\!\!\!{\scriptsize \pi_*}\downarrow & & \downarrow{\scriptsize \pi_*}\!\!\!\!\!\! \\ \mathbf Z/m & \underset{n^k}\to & \mathbf Z/m,\!\end{array}$$ which by functoriality gives a commutative diagram $$\begin{array}{ccc}F(\mathbf Z) & \stackrel{n^k_*}\to & F(\mathbf Z) \\ \!\!\!\!\!\!{\scriptsize \pi_*}\downarrow & & \downarrow{\scriptsize \pi_*}\!\!\!\!\!\! \\ F(\mathbf Z/m) & \underset{n^k_*}\to & F(\mathbf Z/m).\!\end{array}$$ Thus the image of the map $\operatorname{Hom}(\mathbf Z,\mathbf Z/m) \to \operatorname{Hom}_{\mathbf Q}(F(\mathbf Z)_{\mathbf Q},F(\mathbf Z/m)_{\mathbf Q})$ is contained in $\operatorname{Hom}_{\mathbf Q[x]}(F(\mathbf Z)_{\mathbf Q},F(\mathbf Z/m)_{\mathbf Q})$.

If $F(\mathbf Z)$ has finite rank, then $F(\mathbf Z)_\mathbf Q$ has finite length over $\mathbf Q[x]$, hence is supported at finitely many maximal ideals $\mathfrak m \subseteq \mathbf Q[x]$. Since $n$ acts invertibly of order $m$ on $\mathbf Z/(n^m-1)$, the $\mathbf Q[x]$-module $F(\mathbf Z/(n^m-1))_\mathbf Q$ is supported at $$\mathbf Q[x]\big/\big(x^m-1\big) \cong \prod_{d \mid m} \mathbf Q\big(\zeta_d\big),\tag{1}\label{1}$$ where $\mathbf Q(\zeta_d)$ is the $d$-th cyclotomic field. Choose $m = p \gg 0$ prime so that $F(\mathbf Z)_\mathbf Q$ is not supported at $\mathbf Q(\zeta_p)$. Then $\operatorname{Hom}_{\mathbf Q[x]}(F(\mathbf Z)_\mathbf Q, F(\mathbf Z/(n^p-1))_\mathbf Q)$ is only supported at $\mathbf Q(\zeta_1) = \mathbf Q[x]/(x-1)$ by (\ref{1}), i.e. the action of $x$ is trivial. But this contradicts faithfulness of $F$: the maps $n^k\pi \colon \mathbf Z \to \mathbf Z/(n^p-1)$ for $k \in \{0,\ldots,p-1\}$ are pairwise distinct, hence the same goes for the $n^k_*\pi_*$. We conclude that $F(\mathbf Z)_\mathbf Q$ cannot have finite length as $\mathbf Q[x]$-module, so $F(\mathbf Z)$ has infinite rank. $\square$

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    $\begingroup$ When considering this question the other day, I had convinced myself there was an obstacle to approaches like this one, and I don’t see how this answer gets around the obstacle (though I don’t see any flaw in your argument either). The obstacle is this: It seems that after the use of Lemma 2, the proof only uses the value of $F$ on cyclic groups. But if I’m not mistaken, there is a functor defined on $\mathbb{Z}$ and all finite groups giving a faithful f.g. projective cover (details in next comment). So any answer must use values of $F$ on non-cyclic infinite groups. $\endgroup$ – Peter LeFanu Lumsdaine Apr 9 at 9:53
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    $\begingroup$ The functor is given as follows: For finite $A$, take $F(A)$ to be free on the set $A \setminus \{0\}$; this is easily seen to be a functorial f.g. projective cover. Take $F(\mathbb{Z})$ to be $\mathbb{Z}$ with the identity cover. The action on maps to/from $\mathbb{Z}$ is “the first thing you guess” in each case; checking functoriality has to work by cases depending whether the source/target of maps are $\mathbb{Z}$ or finite, but all cases work out, essentially since there are no nonzero maps from finite groups to $\mathbb{Z}$. Am I mistaken, or does your answer get around this somehow? $\endgroup$ – Peter LeFanu Lumsdaine Apr 9 at 10:01
  • $\begingroup$ @PeterLeFanuLumsdaine I don't see how you get a functor, but maybe "the first thing I guess" is not what you thought would be!. If you take the map $\mathbb{Z}\to\mathbb{Z}$ given by multiplication by $2n$ and the nonzero map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$, the composition is zero, independently of $n$. How do you define the functor on these maps? $\endgroup$ – Jeremy Rickard Apr 9 at 14:50
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    $\begingroup$ @AndreaFerretti I thought we didn't want $F$ to be additive? $\endgroup$ – Jeremy Rickard Apr 9 at 16:06
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    $\begingroup$ @JeremyRickard: You’re right, that functoriality case (with the maps from $\mathbb{Z}$) does fail — I’d been sloppy in thinking it through! So the “obstacle” is not a problem; sorry for the noise. (The action on finite groups does work fine, though, that I did check carefully: it’s perhaps clearer if you present $F(A)$ as $\mathbb{Z}^{\oplus A}/(e_0)$ rather than $\mathbb{Z}^{\oplus (A \setminus \{0\})}$. The reason for killing $e_0$, not just using $\mathbb{Z}^{\oplus A}$, was because I thought that would allow the functor to be extended to a fin. gen. value on $\mathbb{Z}$.) $\endgroup$ – Peter LeFanu Lumsdaine Apr 9 at 16:39

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