A category with enough injectives but not enough projectives

Kernels, images, and cokernels of homomorphisms between torsion abelian groups are torsion again, so $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ is an abelian full subcategory of the abelian category from [L1]. If $A$ is torsion, [L3] embeds it into a divisible group $D$; the torsion subgroup $t(D)$ is still divisible and still contains $A$. Hence [L2] makes $t(D)$ injective, so $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ has enough injectives.

Suppose $P$ were a nonzero projective torsion group. Let $P^{\times }=P\setminus \{0\}$, and for each $x\in P^{\times }$ let $n_x$ be the order of $x$. Form $G:={⨁}_{x\in P^{\times }}\mathbb{Z}\mathrm{/}{n}_x\mathbb{Z}$ with generators $e_x$, and define the surjection $\pi :G\to P$ by $\pi (e_x)=x$. Projectivity gives a section $s:P\to G$. Since $s(P)\mathrm{\ne }0$, some coordinate projection restricts to a nonzero map $P\to \mathbb{Z}\mathrm{/}{n}_x\mathbb{Z}$. Let $H$ be its nonzero cyclic image, write $\mathrm{\mid }H\mathrm{\mid }=d>1$, and regard the resulting map $u:P\twoheadrightarrow H\cong \mathbb{Z}\mathrm{/}d\mathbb{Z}$ as surjective.

Choose $x\in P$ with $u(x)=1\in \mathbb{Z}\mathrm{/}d\mathbb{Z}$. For each $n\ge 1$, projectivity lifts $u$ through the reduction $\mathbb{Z}\mathrm{/}{d}^n\mathbb{Z}\twoheadrightarrow \mathbb{Z}\mathrm{/}d\mathbb{Z}$ to a map $u_n:P\to \mathbb{Z}\mathrm{/}{d}^n\mathbb{Z}$. The element $u_n(x)$ is congruent to $1$ modulo $d$, hence is relatively prime to $d$ and has order $d^n$. But the order of $u_n(x)$ must divide the fixed finite order of $x$, impossible for arbitrarily large $n$.

Therefore $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ has enough injectives but no nonzero projective objects, so it does not have enough projectives. This refutes the statement.