The Stacks project

Lemma 12.29.6. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ be a functor. If there exists a subset $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ such that

  1. every object of $\mathcal{B}$ is a quotient of an element of $\mathcal{P}$, and

  2. for every $P \in \mathcal{P}$ there exists an object $Q$ of $\mathcal{A}$ such that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(Q, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A))$ functorially in $A$,

then there exists a left adjoint $v$ of $u$.

Proof. By the Yoneda lemma (Categories, Lemma 4.3.5) the object $Q$ of $\mathcal{A}$ corresponding to $P$ is defined up to unique isomorphism by the formula $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(Q, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A))$. Let us write $Q = v(P)$. Denote $i_ P : P \to u(v(P))$ the map corresponding to $\text{id}_{v(P)}$ in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P), v(P))$. Functoriality in (2) implies that the bijection is given by

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P), A) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A)),\quad \varphi \mapsto u(\varphi ) \circ i_ P \]

For any pair of elements $P_1, P_2 \in \mathcal{P}$ there is a canonical map

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, P_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_2), v(P_1)),\quad \varphi \mapsto v(\varphi ) \]

which is characterized by the rule $u(v(\varphi )) \circ i_{P_2} = i_{P_1} \circ \varphi $ in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, u(v(P_1)))$. Note that $\varphi \mapsto v(\varphi )$ is compatible with composition; this can be seen directly from the characterization. Hence $P \mapsto v(P)$ is a functor from the full subcategory of $\mathcal{B}$ whose objects are the elements of $\mathcal{P}$.

Given an arbitrary object $B$ of $\mathcal{B}$ choose an exact sequence

\[ P_2 \to P_1 \to B \to 0 \]

which is possible by assumption (1). Define $v(B)$ to be the object of $\mathcal{A}$ fitting into the exact sequence

\[ v(P_2) \to v(P_1) \to v(B) \to 0 \]

Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(B), A) & = \mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_1), A) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_2), A)) \\ & = \mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_1, u(A)) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, u(A))) \\ & = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(B, u(A)) \end{align*}

Hence we see that we may take $\mathcal{P} = \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$, i.e., we see that $v$ is everywhere defined. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 12.29: Injectives and adjoint functors

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 0793. Beware of the difference between the letter 'O' and the digit '0'.