I am having trouble with the last part of this proof from Relative Homological Algebra by Jenda and Enoch. I was hoping someone could check my work, and give me hint for the last part of the proof. The complete question goes as follows:
Let $R$ be an integral domain and $\mathcal{F}$ be the class of torsion $R$-modules. Show that every $R$-module has an $\mathcal{F}$-cover that is an injection.
Here is what I have so far.
We begin by noting that $Tor(M)$ is a sub-module of any $R$-module $M$. We claim that the cover can be given by $$\iota : Tor(M) \to M, \quad x \mapsto x$$ The second condition is fairly straightforward to prove. Let $\phi: Tor(M)\mapsto Tor(M)$ be a homomorphism such that $\iota \circ\phi = \iota$. Then for $m \in Tor(M)$, if $\phi(m) = 0$, then $$(\iota \circ \phi)(m) =\iota(\phi(m)) = \iota(m) = m= 0$$ Shows that $\phi$ injects, and so it must be an automorphism.
The part I am having trouble with is showing that for any $T \in \mathcal{F}$, a commutative diagram with $\iota$ can be given.
Your proof of the second condition is not complete. You only show that $\phi$ is injective, but you need to show that $\phi$ is an isomorphism. But actually, $\phi = \mathrm{id}_{\mathrm{Tor}(M)}$ follows immediately from $\iota = \iota \circ \phi$, since $\iota$ is injective. And surely the identity is an automorphism.
The first condition is also easy to prove. If $f : N \to M$ is a homomorphism, where $N$ is torsion, we need to find a homomorphism $f' : N \to \mathrm{Tor}(M)$ with $\iota \circ f' = f$. In other words, we need to show $f(N) \subseteq \mathrm{Tor}(M)$. But if $n \in N$, then $rn = 0$ for some $0 \neq r \in R$ (since $N$ is torsion), hence $r f(n) = f(rn) = 0$, and we see that $f(n)$ is torsion.