A (free) functor $F\colon \mathcal C \to \mathcal D$ is left adjoint to a (forgetful) functor $G\colon \mathcal D \to \mathcal C$, (and $G$ is right adjoint to $F$) written $F \dashv G$ if,
- for every morphism $f: F(C) \to D$ there is a morphism $g: C \to G(D)$ and vice versa.
This "behaves well" (satisfies naturality) with functor-lifted morphisms.
The universal property is that this bijection must also satisfy two natural isomorphisms between the hom functors $\operatorname{Hom}(F(-), D)$ and $\operatorname{Hom}(-, G(D))$ and the hom functors $\operatorname{Hom}(F(C), -)$ and $\operatorname{Hom}(C, G(-))$
The naturality conditions look like this (these diagrams commute each in two directions):
For $f\colon X \to Y$ in $\mathcal C$
$$\begin{CD} \operatorname{Hom}(F(X), D) @= \operatorname{Hom}(X, G(D)) \\ @V\operatorname{Hom}(F(f), D)VV @VV\operatorname{Hom}(f, G(D))V \\ \operatorname{Hom}(F(Y), D) @= \operatorname{Hom}(Y, G(D)) \end{CD}$$
For $f\colon X \to Y$ in $\mathcal D$
$$\begin{CD} \operatorname{Hom}(F(C), X) @= \operatorname{Hom}(C, G(X)) \\ @V\operatorname{Hom}(F(C), f)VV @VV\operatorname{Hom}(C, G(f))V \\ \operatorname{Hom}(F(C), Y) @= \operatorname{Hom}(C, G(Y)) \end{CD}$$
Take the equal signs to mean isomorphism here.
- there are natural transformations $\eta_C: C \to G(F(C))$ (unit) and $\epsilon_D: F(G(D)) \to D$ (counit) that satisfy the triangle identities: