
Definition 4.24.1. Let $\mathcal{C}$, $\mathcal{D}$ be categories. Let $u : \mathcal{C} \to \mathcal{D}$ and $v : \mathcal{D} \to \mathcal{C}$ be functors. We say that $u$ is a left adjoint of $v$, or that $v$ is a right adjoint to $u$ if there are bijections

$\mathop{Mor}\nolimits _\mathcal {D}(u(X), Y) \longrightarrow \mathop{Mor}\nolimits _\mathcal {C}(X, v(Y))$

functorial in $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$.

In other words, this means that there is a given isomorphism of functors $\mathcal{C}^{opp} \times \mathcal{D} \to \textit{Sets}$ from $\mathop{Mor}\nolimits _\mathcal {D}(u(-), -)$ to $\mathop{Mor}\nolimits _\mathcal {C}(-, v(-))$. For any object $X$ of $\mathcal{C}$ we obtain a morphism $X \to v(u(X))$ corresponding to $\text{id}_{u(X)}$. Similarly, for any object $Y$ of $\mathcal{D}$ we obtain a morphism $u(v(Y)) \to Y$ corresponding to $\text{id}_{v(Y)}$. These maps are called the adjunction maps. The adjunction maps are functorial in $X$ and $Y$, hence we obtain morphisms of functors

$\text{id}_\mathcal {C} \to v \circ u\quad (\text{unit}) \quad \text{and}\quad u \circ v \to \text{id}_\mathcal {D}\quad (\text{counit}).$

Moreover, if $\alpha : u(X) \to Y$ and $\beta : X \to v(Y)$ are morphisms, then the following are equivalent

1. $\alpha$ and $\beta$ correspond to each other via the bijection of the definition,

2. $\beta$ is the composition $X \to v(u(X)) \xrightarrow {v(\alpha )} v(Y)$, and

3. $\alpha$ is the composition $u(X) \xrightarrow {u(\beta )} u(v(Y)) \to Y$.

In this way one can reformulate the notion of adjoint functors in terms of adjunction maps.

Lemma 4.24.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. If for each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the functor $x \mapsto \mathop{Mor}\nolimits _\mathcal {D}(u(x), y)$ is representable, then $u$ has a right adjoint.

Proof. For each $y$ choose an object $v(y)$ and an isomorphism $\mathop{Mor}\nolimits _\mathcal {C}(-, v(y)) \to \mathop{Mor}\nolimits _\mathcal {D}(u(-), y)$ of functors. By Yoneda's lemma (Lemma 4.3.5) for any morphism $g : y \to y'$ the transformation of functors

$\mathop{Mor}\nolimits _\mathcal {C}(-, v(y)) \to \mathop{Mor}\nolimits _\mathcal {D}(u(-), y) \to \mathop{Mor}\nolimits _\mathcal {D}(u(-), y') \to \mathop{Mor}\nolimits _\mathcal {C}(-, v(y'))$

corresponds to a unique morphism $v(g) : v(y) \to v(y')$. We omit the verification that $v$ is a functor and that it is right adjoint to $u$. $\square$

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

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

2. $v$ 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). $\square$

Remark 4.24.4. The proof of Lemma 4.24.3 shows that, in the situation of Definition 4.24.1, the composition $v \circ u$ is isomorphic to the identity via some natural isomorphism if and only if the unit is a natural isomorphism. Dually, the composition $u \circ v$ is isomorphic to the identity via some natural isomorphism if and only if the counit is a natural isomorphism.

Lemma 4.24.5. Let $u$ be a left adjoint to $v$ as in Definition 4.24.1.

1. Suppose that $M : \mathcal{I} \to \mathcal{C}$ is a diagram, and suppose that $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} M$ exists in $\mathcal{C}$. Then $u(\mathop{\mathrm{colim}}\nolimits _\mathcal {I} M) = \mathop{\mathrm{colim}}\nolimits _\mathcal {I} u \circ M$. In other words, $u$ commutes with (representable) colimits.

2. Suppose that $M : \mathcal{I} \to \mathcal{D}$ is a diagram, and suppose that $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} M$ exists in $\mathcal{D}$. Then $v(\mathop{\mathrm{lim}}\nolimits _\mathcal {I} M) = \mathop{\mathrm{lim}}\nolimits _\mathcal {I} v \circ M$. In other words $v$ commutes with representable limits.

Proof. A morphism from a colimit into an object is the same as a compatible system of morphisms from the constituents of the limit into the object, see Remark 4.14.4. So

$\begin{matrix} \mathop{Mor}\nolimits _\mathcal {D}(u(\mathop{\mathrm{colim}}\nolimits _{i \in \mathcal{I}} M_ i), Y) & = & \mathop{Mor}\nolimits _\mathcal {C}(\mathop{\mathrm{colim}}\nolimits _{i \in \mathcal{I}} M_ i, v(Y)) \\ & = & \mathop{\mathrm{lim}}\nolimits _{i \in \mathcal{I}^{opp}} \mathop{Mor}\nolimits _\mathcal {C}(M_ i, v(Y)) \\ & = & \mathop{\mathrm{lim}}\nolimits _{i \in \mathcal{I}^{opp}} \mathop{Mor}\nolimits _\mathcal {D}(u(M_ i), Y) \end{matrix}$

proves that $u(\mathop{\mathrm{colim}}\nolimits _{i \in \mathcal{I}} M_ i)$ is the colimit we are looking for. A similar argument works for the other statement. $\square$

Lemma 4.24.6. Let $u$ be a left adjoint of $v$ as in Definition 4.24.1.

1. If $\mathcal{C}$ has finite colimits, then $u$ is right exact.

2. If $\mathcal{D}$ has finite limits, then $v$ is left exact.

Proof. Obvious from the definitions and Lemma 4.24.5. $\square$

Lemma 4.24.7. Let $u_1, u_2 : \mathcal{C} \to \mathcal{D}$ be functors with right adjoints $v_1, v_2 : \mathcal{D} \to \mathcal{C}$. Let $\beta : u_2 \to u_1$ be a transformation of functors. Let $\beta ^\vee : v_1 \to v_2$ be the corresponding transformation of adjoint functors. Then

$\xymatrix{ u_2 \circ v_1 \ar[r]_\beta \ar[d]_{\beta ^\vee } & u_1 \circ v_1 \ar[d] \\ u_2 \circ v_2 \ar[r] & \text{id} }$

is commutative where the unlabeled arrows are the counit transformations.

Proof. This is true because $\beta ^\vee _ D : v_1D \to v_2D$ is the unique morphism such that the induced maps $\mathop{Mor}\nolimits (C, v_1D) \to \mathop{Mor}\nolimits (C, v_2D)$ is the map $\mathop{Mor}\nolimits (u_1C, D) \to \mathop{Mor}\nolimits (u_2C, D)$ induced by $\beta _ C : u_2C \to u_1C$. Namely, this means the map

$\mathop{Mor}\nolimits (u_1 v_1 D, D') \to \mathop{Mor}\nolimits (u_2 v_1 D, D')$

induced by $\beta _{v_1 D}$ is the same as the map

$\mathop{Mor}\nolimits (v_1 D, v_1 D') \to \mathop{Mor}\nolimits (v_1 D, v_2 D')$

induced by $\beta ^\vee _{D'}$. Taking $D' = D$ we find that the counit $u_1 v_1 D \to D$ precomposed by $\beta _{v_1D}$ corresponds to $\beta ^\vee _ D$ under adjunction. This exactly means that the diagram commutes when evaluated on $D$. $\square$

Lemma 4.24.8. Let $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ be categories. Let $v : \mathcal{A} \to \mathcal{B}$ and $v' : \mathcal{B} \to \mathcal{C}$ be functors with left adjoints $u$ and $u'$ respectively. Then

1. The functor $v'' = v' \circ v$ has a left adjoint equal to $u'' = u \circ u'$.

2. Given $X$ in $\mathcal{A}$ we have

4.24.8.1
$$\label{categories-equation-compose-counits} \epsilon _ X^ v \circ u(\epsilon ^{v'}_{v(X)}) = \epsilon ^{v''}_ X : u''(v''(X)) \to X$$

Where $\epsilon$ is the counit of the adjunctions.

Proof. Let us unwind the formula in (2) because this will also immediately prove (1). First, the counit of the adjunctions for the pairs $(u, v)$ and $(u', v')$ are maps $\epsilon _ X^ v : u(v(X)) \to X$ and $\epsilon _ Y^{v'} : u'(v'(Y)) \to Y$, see discussion following Definition 4.24.1. With $u''$ and $v''$ as in (1) we unwind everything

$u''(v''(X)) = u(u'(v'(v(X)))) \xrightarrow {u(\epsilon _{v(X)}^{v'})} u(v(X)) \xrightarrow {\epsilon _ X^ v} X$

to get the map on the left hand side of (4.24.8.1). Let us denote this by $\epsilon _ X^{v''}$ for now. To see that this is the counit of an adjoint pair $(u'', v'')$ we have to show that given $Z$ in $\mathcal{C}$ the rule that sends a morphism $\beta : Z \to v''(X)$ to $\alpha = \epsilon _ X^{v''} \circ u''(\beta ) : u''(Z) \to X$ is a bijection on sets of morphisms. This is true because, this is the composition of the rule sending $\beta$ to $\epsilon _{v(X)}^{v'} \circ u'(\beta )$ which is a bijection by assumption on $(u', v')$ and then sending this to $\epsilon _ X^ v \circ u(\epsilon _{v(X)}^{v'} \circ u'(\beta ))$ which is a bijection by assumption on $(u, v)$. $\square$

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