Abelian category with enough injectives but not functorially

Question

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.

Answer 1

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. $\square$