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*

*$A$ is projective in $\mathbf{Ab}^{\operatorname{f.t.}}$;*
*$A$ is projective in $\mathbf{Ab}$;*
*$A$ 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. $\square$

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$15more comments