Left adjoint to inclusion functor of torsion-free abelian groups in abelian groups Let $\text{Ab}$ be the category of abelian groups, $\text{TFAb}$ the full subcategory of torsion-free abelian groups, and $F \colon \text{TFAb} \to \text{Ab}$ the inclusion functor. I'm trying to show $F$ is monadic, and so first I need to show it has a left adjoint.
I don't really have any idea what the left adjoint should look like. If anyone could give a description that would be great and I can try to show it's an left adjoint myself.
 A: Yes, the left adjoint mods out the torsion subgroup. If $A$ is any abelian group and $B$ is torsion-free, then certainly any map $A\to B$ factors uniquely through this quotient, which is all that needs to be shown for a left adjoint to a fully faithful functor.
A: For every abelian group $A$ we can consider its set of torsion-elements,
$$
  T(A)
  :=
  \{
    a \in A
    \mid
    \text{there exists an nonzero $n \in \mathbb{Z}$ with $na = 0$}
  \} \,.
$$
This is a subgroup of $A$ because $\mathbb{Z}$ is an integral domain.
The quotient
$$
  F(A) := A / T(A)
$$
is torsion-free.
Indeed, if $[a]$ is an torsion element of $F(A)$ then there exists some nonzero scalar $n \in \mathbb{Z}$ with
$$
  0 = n [a] = [n a] \,.
$$
This means that $na \in T(A)$, i.e. that $na$ is a torsion element of $A$.
There hence exist some nonzero scalar $m \in \mathbb{Z}$ with $mna = 0$.
The scalar $mn$ is again nonzero because $\mathbb{Z}$ is an integral domain.
We thus find that $a$ is a torsion element of $A$, i.e. an element of $T(A)$.
This shows that $[a] = 0$.
For any two abelian groups $A$ and $B$ and every homomorphism of groups $f$ from $A$ to $B$ we have
$$
  f(T(A))
  \subseteq
  T(B) \,.
$$
It follows that the homomorphism $f$ descends to a homomorphism
$$
  F(f)
  \colon
  F(A)
  \to
  F(B) \,,
  \quad
  [a]
   \mapsto
  [f(a)] \,.
$$
This construction gives a functor
$$
  F
  \colon
  \mathbf{Ab}
  \to
  \mathbf{TFAb} \,.
$$
Suppose now that $A$ is any abelian group and that $B$ is a torsion free abelian group.
If $f$ is any homomorphism from $A$ to $B$ then
$$
  f(T(A)) \subseteq T(B) = 0 \,,
$$
so $f$ factors through a homomorphism $A / T(A) \to B$.
This shows that every homomorphism from $A$ to $B$ comes from a homomorphism from $F(A)$ to $B$.
In other words, the map
\begin{align*}
  \{ \text{homomorphisms $F(A) \to B$} \}
  &\longrightarrow
  \{ \text{homomorphisms $A \to B$} \} \,,
   \\
  f
  &\longmapsto
  f \circ \eta_A
\end{align*}
is a bijection, where $\eta_A \colon A \to F(A)$ is the canonical projection given by $a \mapsto [a]$.
If we denote the inclusion functor from $\mathbf{TFAb}$ to $\mathbf{Ab}$ by $R$ then we this gives us a bijection
\begin{align*}
  \varphi_{A,B}
  \colon
  \operatorname{Hom}_{\mathrm{TFAb}}(F(A), B)
  &\longrightarrow
  \operatorname{Hom}_{\mathrm{Ab}}(A, R(B)) \,,
   \\
  f
  &\longmapsto
  f \circ \eta_A \,.
\end{align*}
This bijection is natural in both $A$ and $B$ and does therefore show that $F$ is left adjoint to $R$.
