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

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

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 (15.71.0.1). Note that the transition maps in the system are surjective because $I^ j$ is injective 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. $\square$

Comment #8599 by nkym on

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

Comment #9511 by on

1. In the statement, to match the style used in Lemma 19.12.6 and in Derived Categories, Section 13.31, one could display “K-injective” instead of “$K$-injective.”

2. In the construction of the subcomplexes $K^\bullet_\alpha$, in case (3), $\alpha$ is a limit ordinal, we are inadvertently using that $\mathcal{A}$ has AB5 to get that the filtered colimit of subobjects is again a subobject (maybe it's worth mentioning AB5?).

3. When the proof says “note that the transition maps in the system are surjective” how is this fact being used? Lemma 12.31.8 does not require the transition maps being surjective as a hypothesis.

4. In “it is clear that $M^\bullet=K^\bullet_\alpha$ for a suitably large ordinal $\alpha$” one could add “(take an ordinal $\alpha$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$).”

5. For-posterity details for those who care: The reason we get exact is because $\operatorname{Hom}^\bullet(-,I^\bullet)$ is exact (this is equivalent to $I^j$ being injective for all $j$), so a s.e.s. of cochain complexes $S$ is sent to another s.e.s. of cochain complexes $\operatorname{Hom}^\bullet(S,I^\bullet)$. Application of Homology, Lemma 12.13.12 to $\operatorname{Hom}^\bullet(S,I^\bullet)$ gives exactness of \eqref{seq} (Homology, Equation 15.71.0.1).

Comment #9512 by on

Logic issues in the construction of the transfinite sequence of subcomplexes $K^\bullet_\alpha$. Cases (1) and (3) are independent of previous steps, but it seems (2) is not: it requires to make a choice since $N_\alpha^\bullet$ is not unique. Instead of transfinite recursion on $\alpha$, maybe what one actually needs is the generalized axiom of dependent choices? Let $\kappa$ be an aleph. This axiom says:

$\mathsf{DC}_\kappa$ [1, Sect. 8.1]. Let $S$ be a nonempty set and let $R$ be a binary relation such that for every $\alpha<\kappa$ and every $\alpha$-sequence $s=\left\langle x_{\xi}: \xi<\alpha\right\rangle$ of elements of $S$ there exists $y \in S$ such that $sRy$. Then there is a function $f: \kappa \rightarrow S$ such that for every $\alpha<\kappa,(f \mid \alpha) R f(\alpha)$.

(One has $\mathsf{AC}\Leftrightarrow\forall\kappa,\mathsf{DC}_\kappa$ [1, Theorem 8.1].)

The axioms $\mathsf{DC}_\kappa$ alone do not provide a transfinite sequence indexed on the class of all ordinals, but rather a transfinite sequence of length equal to an aleph. However, for our proof it suffices to consider some cardinal $\kappa$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$ (as it is hinted in #9511, 4).

A disclaimer: All set theory I know I have learnt it in the last month, from scratch. I hadn't ever been into any kind of foundations of math, logical or set-theoretic ones.

(Lest this comment not parse well in your device, you can consult the original code here.)

### References

1. T. Jech, The Axiom of Choice

Comment #9513 by on

Correction to #9512: it suffices to consider some cardinal $\kappa$ greater than $\prod_{i\in\mathbb{Z}}|M^i|$ and such that there is some other cardinal strictly in-between.

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).