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 (15.71.0.1). 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. $\square$

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