Lemma 4.24.3. Let $u$ be a left adjoint to $v$ as in Definition 4.24.1. Then

$u$ is fully faithful $\Leftrightarrow \text{id} \cong v \circ u$.

$v$ is fully faithful $\Leftrightarrow u \circ v \cong \text{id}$.

Lemma 4.24.3. Let $u$ be a left adjoint to $v$ as in Definition 4.24.1. Then

$u$ is fully faithful $\Leftrightarrow \text{id} \cong v \circ u$.

$v$ is fully faithful $\Leftrightarrow u \circ v \cong \text{id}$.

**Proof.**
Assume $u$ is fully faithful. We have to show the adjunction map $X \to v(u(X))$ is an isomorphism. Let $X' \to v(u(X))$ be any morphism. By adjointness this corresponds to a morphism $u(X') \to u(X)$. By fully faithfulness of $u$ this corresponds to a morphism $X' \to X$. Thus we see that $X \to v(u(X))$ defines a bijection $\mathop{Mor}\nolimits (X', X) \to \mathop{Mor}\nolimits (X', v(u(X)))$. Hence it is an isomorphism. Conversely, if $\varphi : \text{id} \to v \circ u$ is a natural isomorphism, then $u$ is faithful for the trivial reason that the composition $\mathop{Mor}\nolimits (X,X') \to \mathop{Mor}\nolimits (u(X),u(X')) \to \mathop{Mor}\nolimits (v(u(X)),v(u(X')))$ is a bijection. Furthermore, $u$ is full: A preimage for a morphism $\gamma : u(X) \to u(X')$ is $u(\varphi _{X'}^{-1} \circ v(\varepsilon _{u(X')} \circ u(\varphi _{X'}) \circ \gamma ) \circ \eta _ X)$, where $\eta : \text{id}_ C \to v \circ u$ is the unit and $\varepsilon : u \circ v \to \text{id}_ D$ is the counit of the adjunction.

Part (2) is dual to part (1). $\square$

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