this might be a stupid question, but I'm not sure if this is true (at least in some class of cases).
Let $F : \mathcal{C} \rightarrow \mathcal{D}$ be left adjoint to an inclusion $\mathcal{D} \hookrightarrow \mathcal{C}$. Let $A, B \in \mathcal{D}$. Is it true that given a morphism $A \xrightarrow f B$ there exists a unique morphism $F(A) \xrightarrow g B$ such that $A \xrightarrow h F(A) \xrightarrow g B = A \xrightarrow f B$, where $h = \eta_A$ is the unit of the adjunction?
For $A = B$ it's just the triangle identity, but for $A \neq B$ I don't know. It's clear that this is true when the isomorphisms on the $\text{Hom}$'s are given by a composition too. Furthermore if the above assertion is not, in general, true does it hold that it's true iff the isomorphism on the $\text{Hom}$'s is given by composition with a morphism?
Let $A$ be in $\mathcal{C}$ and let $B$ in $\mathcal{D}$. Then yes, every morphism $f : A \to B$ factors through the unit $\eta_A : A \to F A$ in a unique way. Indeed, by the triangle identity and naturality, $$f = \epsilon_B \circ \eta_B \circ f = \epsilon_B \circ F f \circ \eta_A$$ and if $f = f' \circ \eta_A$, then $$\epsilon_B \circ F f = \epsilon_B \circ F f' \circ F \eta_A = f' \circ \epsilon_{F A} \circ F \eta_A = f'$$ so the factorisation is unique.
Thus, the bijection $$\mathcal{C} (A, B) \cong \mathcal{D} (F A, B)$$ is given by $f \mapsto \epsilon_B \circ F f$, which is not quite composition with a morphism, but its inverse is $f' \mapsto f' \circ \eta_A$.