The Stacks project

Lemma 19.12.3. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $\kappa $ be a cardinal as in Lemma 19.12.2. Suppose that $I^\bullet $ is a complex such that

  1. each $I^ j$ is injective, and

  2. for every bounded above acyclic complex $M^\bullet $ such that $|M^ n| \leq \kappa $ we have $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$.

Then $I^\bullet $ is an $K$-injective complex.

Proof. Let $M^\bullet $ be an acyclic complex. We are going to construct by induction on the ordinal $\alpha $ an acyclic subcomplex $K_\alpha ^\bullet \subset M^\bullet $ as follows. For $\alpha = 0$ we set $K_0^\bullet = 0$. For $\alpha > 0$ we proceed as follows:

  1. If $\alpha = \beta + 1$ and $K_\beta ^\bullet = M^\bullet $ then we choose $K_\alpha ^\bullet = K_\beta ^\bullet $.

  2. If $\alpha = \beta + 1$ and $K_\beta ^\bullet \not= M^\bullet $ then $M^\bullet /K_\beta ^\bullet $ is a nonzero acyclic complex. We choose a subcomplex $N_\alpha ^\bullet \subset M^\bullet /K_\beta ^\bullet $ as in Lemma 19.12.2. Finally, we let $K_\alpha ^\bullet \subset M^\bullet $ be the inverse image of $N_\alpha ^\bullet $.

  3. If $\alpha $ is a limit ordinal we set $K_\beta ^\bullet = \mathop{\mathrm{colim}}\nolimits K_\alpha ^\bullet $.

It is clear that $M^\bullet = K_\alpha ^\bullet $ for a suitably large ordinal $\alpha $. We will prove that

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\alpha ^\bullet , I^\bullet ) \]

is zero by transfinite induction on $\alpha $. It holds for $\alpha = 0$ since $K_0^\bullet $ is zero. Suppose it holds for $\beta $ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence of complexes

\[ 0 \to K_\beta ^\bullet \to K_\alpha ^\bullet \to N_\alpha ^\bullet \to 0 \]

Since each component of $I^\bullet $ is injective we see that we obtain an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\beta ^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K_\alpha ^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N_\alpha ^\bullet , I^\bullet ) \]

By induction the term on the left is zero and by assumption on $I^\bullet $ the term on the right is zero. Thus the middle group is zero too. Finally, suppose that $\alpha $ is a limit ordinal. Then we see that

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (K_\alpha ^\bullet , I^\bullet ) = \mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } \mathop{\mathrm{Hom}}\nolimits ^\bullet (K_\beta ^\bullet , I^\bullet ) \]

with notation as in More on Algebra, Section 15.71. These complexes compute morphisms in $K(\mathcal{A})$ by More on Algebra, Equation ( Note that the transition maps in the system are surjective because $I^ j$ is surjective for each $j$. Moreover, for a limit ordinal $\alpha $ we have equality of limit and value (see displayed formula above). Thus we may apply Homology, Lemma 12.31.8 to conclude. $\square$

Comments (1)

Comment #8599 by nkym on

In the thrid last line the proof says surjective but they are injective. Also, in the last exact sequence the arrows should point backward.

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 079L. Beware of the difference between the letter 'O' and the digit '0'.