The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

4.24 Adjoint functors

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

    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 ( 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 $(u, v)$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0036. Beware of the difference between the letter 'O' and the digit '0'.