hcju/stacksqa huggingface dataset:
This is clear for the case of open immersions since an open subspace of an open subspace is also an open subspace. \medskip\noindent Suppose $a : Z \to Y$ and $b : Y \to X$ are closed immersions of schemes. We will verify that $c = b \circ a$ is also a closed immersion. The assumption implies that $a$ and $b$ are homeomorphisms onto closed subsets, and hence also $c = b \circ a$ is a homeomorphism onto a closed subset. Moreover, the map $\mathcal{O}_X \to c_*\mathcal{O}_Z$ is surjective since it factors as the composition of the surjective maps $\mathcal{O}_X \to b_*\mathcal{O}_Y$ and $b_*\mathcal{O}_Y \to b_*a_*\mathcal{O}_Z$ (surjective as $b_*$ is exact, see Modules, Lemma \ref{modules-lemma-i-star-exact}). Hence by Lemma \ref{lemma-characterize-closed-immersions} above $c$ is a closed immersion. \medskip\noindent Finally, we come to the case of immersions. Suppose $a : Z \to Y$ and $b : Y \to X$ are immersions of schemes. This means there exist open subschemes $V \subset Y$ and $U \subset X$ such that $a(Z) \subset V$, $b(Y) \subset U$ and $a : Z \to V$ and $b : Y \to U$ are closed immersions. Since the topology on $Y$ is induced from the topology on $U$ we can find an open $U' \subset U$ such that $V = b^{-1}(U')$. Then we see that $Z \to V = b^{-1}(U') \to U'$ is a composition of closed immersions and hence a closed immersion. This proves that $Z \to X$ is an immersion and we win.
By Baer's criterion Lemma \ref{lemma-criterion-baer}, it suffices to show that if $\mathfrak{a} \subset R$ is an ideal, then any map $f : \mathfrak{a} \to \mathbf{M}_\alpha(N)$ extends to $R \to \mathbf{M}_\alpha(N)$. However, we know since $\alpha$ is a limit ordinal that $$ \mathbf{M}_{\alpha}(N) = \colim_{\beta < \alpha} \mathbf{M}_{\beta}(N), $$ so by Proposition \ref{proposition-modules-are-small}, we find that $$ \Hom_R(\mathfrak{a}, \mathbf{M}_{\alpha}(N)) = \colim_{\beta < \alpha} \Hom_R(\mathfrak a, \mathbf{M}_{\beta}(N)). $$ This means in particular that there is some $\beta' < \alpha$ such that $f$ factors through the submodule $\mathbf{M}_{\beta'}(N)$, as $$ f : \mathfrak{a} \to \mathbf{M}_{\beta'}(N) \to \mathbf{M}_{\alpha}(N). $$ However, by the fundamental property of the functor $\mathbf{M}$, see Lemma \ref{lemma-construction} part (3), we know that the map $\mathfrak{a} \to \mathbf{M}_{\beta'}(N)$ can be extended to $$ R \to \mathbf{M}( \mathbf{M}_{\beta'}(N)) = \mathbf{M}_{\beta' + 1}(N), $$ and the last object imbeds in $\mathbf{M}_{\alpha}(N)$ (as $\beta' + 1 < \alpha$ since $\alpha$ is a limit ordinal). In particular, $f$ can be extended to $\mathbf{M}_{\alpha}(N)$.
Let $\mathcal{G}_i$, $i \in I$ be a set of abelian sheaves such that every subsheaf of every $\mathbf{Z}_X^\#$ occurs as one of the $\mathcal{G}_i$. Apply Lemma \ref{lemma-map-into-smaller} to this collection to get an ordinal $\beta$. We claim that for any sheaf of abelian groups $\mathcal{F}$ the map $\mathcal{F} \to J_\beta(\mathcal{F})$ is an injection of $\mathcal{F}$ into an injective. Note that by construction the assignment $\mathcal{F} \mapsto \big(\mathcal{F} \to J_\beta(\mathcal{F})\big)$ is indeed functorial. \medskip\noindent The proof of the claim comes from the fact that by Lemma \ref{lemma-characterize-injectives} it suffices to extend any morphism $\gamma : \mathcal{G} \to J_\beta(\mathcal{F})$ from a subsheaf $\mathcal{G}$ of some $\mathbf{Z}_X^\#$ to all of $\mathbf{Z}_X^\#$. Then by Lemma \ref{lemma-map-into-smaller} the map $\gamma$ lifts into $J_\alpha(\mathcal{F})$ for some $\alpha < \beta$. Finally, we apply Lemma \ref{lemma-map-into-next-one} to get the desired extension of $\gamma$ to a morphism into $J_{\alpha + 1}(\mathcal{F}) \to J_\beta(\mathcal{F})$.
Please compare with the proof of Theorem \ref{theorem-baer-grothendieck}. Choose a generator $U$ of $\mathcal{A}$. For an object $M$ we define $\mathbf{M}(M)$ by the following pushout diagram $$ \xymatrix{ \bigoplus_{N \subset U} \bigoplus_{\varphi \in \Hom(N, M)} N \ar[r] \ar[d] & M \ar[d] \\ \bigoplus_{N \subset U} \bigoplus_{\varphi \in \Hom(N, M)} U \ar[r] &  \mathbf{M}(M). } $$ Note that $M \to \mathbf{M}(N)$ is a functor and that there exist functorial injective maps $M \to \mathbf{M}(M)$. By transfinite induction we define functors $\mathbf{M}_\alpha(M)$ for every ordinal $\alpha$. Namely, set $\mathbf{M}_0(M) = M$. Given $\mathbf{M}_\alpha(M)$ set $\mathbf{M}_{\alpha + 1}(M) = \mathbf{M}(\mathbf{M}_\alpha(M))$. For a limit ordinal $\beta$ set $$ \mathbf{M}_\beta(M) = \colim_{\alpha < \beta} \mathbf{M}_\alpha(M). $$ Finally, pick any ordinal $\alpha$ whose cofinality is greater than $|U|$. Such an ordinal exists by Sets, Proposition \ref{sets-proposition-exist-ordinals-large-cofinality}. We claim that $M \to \mathbf{M}_\alpha(M)$ is the desired functorial injective embedding. Namely, if $N \subset U$ is a subobject and $\varphi : N \to \mathbf{M}_\alpha(M)$ is a morphism, then we see that $\varphi$ factors through $\mathbf{M}_{\alpha'}(M)$ for some $\alpha' < \alpha$ by Proposition \ref{proposition-objects-are-small}. By construction of $\mathbf{M}(-)$ we see that $\varphi$ extends to a morphism from $U$ into $\mathbf{M}_{\alpha' + 1}(M)$ and hence into $\mathbf{M}_\alpha(M)$. By Lemma \ref{lemma-characterize-injective} we conclude that $\mathbf{M}_\alpha(M)$ is injective.
Please compare with the proof of Theorem \ref{theorem-baer-grothendieck} and Theorem \ref{theorem-injective-embedding-grothendieck}. Choose a cardinal $\kappa$ as in Lemmas \ref{lemma-acyclic-quotient-complexes-bounded-size} and \ref{lemma-characterize-K-injective}. Choose a set $(K_i^\bullet)_{i \in I}$ of bounded above, acyclic complexes such that every bounded above acyclic complex $K^\bullet$ such that $|K^n| \leq \kappa$ is isomorphic to $K_i^\bullet$ for some $i \in I$. This is possible by Lemma \ref{lemma-set-iso-classes-bounded-size}. Denote $\mathbf{M}^\bullet(-)$ the functor constructed in Lemma \ref{lemma-functorial-homotopies}. Denote $\mathbf{N}^\bullet(-)$ the functor constructed in Lemma \ref{lemma-functorial-injective}. Both of these functors come with injective transformations $\text{id} \to \mathbf{M}$ and $\text{id} \to \mathbf{N}$. \medskip\noindent By transfinite induction we define a sequence of functors $\mathbf{T}_\alpha(-)$ and corresponding transformations $\text{id} \to \mathbf{T}_\alpha$. Namely we set $\mathbf{T}_0(M^\bullet) = M^\bullet$. If $\mathbf{T}_\alpha$ is given then we set $$ \mathbf{T}_{\alpha + 1}(M^\bullet) = \mathbf{N}^\bullet(\mathbf{M}^\bullet(\mathbf{T}_\alpha(M^\bullet))) $$ If $\beta$ is a limit ordinal we set $$ \mathbf{T}_\beta(M^\bullet) = \colim_{\alpha < \beta} \mathbf{T}_\alpha(M^\bullet) $$ The transition maps of the system are injective quasi-isomorphisms. By AB5 we see that the colimit is still quasi-isomorphic to $M^\bullet$. We claim that $M^\bullet \to \mathbf{T}_\alpha(M^\bullet)$ does the job if the cofinality of $\alpha$ is larger than $\max(\kappa, |U|)$ where $U$ is a generator of $\mathcal{A}$. Namely, it suffices to check conditions (1) and (2) of Lemma \ref{lemma-characterize-K-injective}. \medskip\noindent For (1) we use the criterion of Lemma \ref{lemma-characterize-injective}. Suppose that $M \subset U$ and $\varphi : M \to \mathbf{T}^n_\alpha(M^\bullet)$ is a morphism for some $n \in \mathbf{Z}$. By Proposition \ref{proposition-objects-are-small} we see that $\varphi$ factor through $\mathbf{T}^n_{\alpha'}(M^\bullet)$ for some $\alpha' < \alpha$. In particular, by the construction of the functor $\mathbf{N}^\bullet(-)$ we see that $\varphi$ factors through an injective object of $\mathcal{A}$ which shows that $\varphi$ lifts to a morphism on $U$. \medskip\noindent For (2) let $w : K^\bullet  \to \mathbf{T}_\alpha(M^\bullet)$ be a morphism of complexes where $K^\bullet$ is a bounded above acyclic complex such that $|K^n| \leq \kappa$. Then $K^\bullet \cong K_i^\bullet$ for some $i \in I$. Moreover, by Proposition \ref{proposition-objects-are-small} once again we see that $w$ factor through $\mathbf{T}^n_{\alpha'}(M^\bullet)$ for some $\alpha' < \alpha$. In particular, by the construction of the functor $\mathbf{M}^\bullet(-)$ we see that $w$ is homotopic to zero. This finishes the proof.
We first prove $G$ is fully faithful, or equivalently that $F \circ G \to \text{id}$ is an isomorphism, see Categories, Lemma \ref{categories-lemma-adjoint-fully-faithful}. First, given an object $A$ the map $F(G(A)) \to A$ is surjective, because every map of $U \to A$ factors through $F(G(A))$ by construction. On the other hand, the map $F(G(A)) \to A$ is the adjoint of the map $\text{id} : G(A) \to G(A)$ and hence injective by Lemma \ref{lemma-F-G-monos}. \medskip\noindent The functor $F$ is right exact as it is a left adjoint. Since $\text{Mod}_R$ has enough projectives, to show that $F$ is exact, it is enough to show that the first left derived functor $L_1F$ is zero. To prove $L_1F(M) = 0$ for some $R$-module $M$ choose an exact sequence $0 \to K \to P \to M \to 0$ of $R$-modules with $P$ free. It suffices to show $F(K) \to F(P)$ is injective. Now we can write this sequence as a filtered colimit of sequences $0 \to K_i \to P_i \to M_i \to 0$ with $P_i$ a finite free $R$-module: just write $P$ in this manner and set $K_i = K \cap P_i$ and $M_i = \Im(P_i \to M)$. Because $F$ is a left adjoint it commutes with colimits and because $\mathcal{A}$ is a Grothendieck abelian category, we find that $F(K) \to F(P)$ is injective if each $F(K_i) \to F(P_i)$ is injective. Thus it suffices to check $F(K) \to F(P)$ is injective when $K \subset P = R^{\oplus n}$. Thus $F(K) \to U^{\oplus n}$ is injective by an application of Lemma \ref{lemma-F-G-monos}.
By the remark above the lemma the category $\text{Mod}_{R[G]}$ has functorial injective embeddings. Consider the forgetful functor $v : \text{Mod}_{R, G} \to \text{Mod}_{R[G]}$. This functor is fully faithful, transforms injective maps into injective maps and has a right adjoint, namely $$ u : M \mapsto u(M) = \{x \in M \mid \text{stabilizer of }x\text{ is open}\} $$ Since $v(M) = 0 \Rightarrow M = 0$ we conclude by Homology, Lemma \ref{homology-lemma-adjoint-functorial-injectives}.
For an abelian group $A$ we denote $j : A \to J(A)$ the functorial injective embedding constructed in More on Algebra, Section \ref{more-algebra-section-injectives-modules}. Let $\mathcal{F}$ be an abelian sheaf on $X$. By Sheaves, Example \ref{sheaves-example-sheaf-product-pointwise} the assignment $$ \mathcal{I} : U \mapsto \mathcal{I}(U) = \prod\nolimits_{x\in U} J(\mathcal{F}_x) $$ is an abelian sheaf. There is a canonical map $\mathcal{F} \to \mathcal{I}$ given by mapping $s \in \mathcal{F}(U)$ to $\prod_{x \in U} j(s_x)$ where $s_x \in \mathcal{F}_x$ denotes the germ of $s$ at $x$. This map is injective, see Sheaves, Lemma \ref{sheaves-lemma-sheaf-subset-stalks} for example. \medskip\noindent It remains to prove the following: Given a rule $x \mapsto I_x$ which assigns to each point $x \in X$ an injective abelian group the sheaf $\mathcal{I} : U \mapsto \prod_{x \in U} I_x$ is injective. Note that $$ \mathcal{I} = \prod\nolimits_{x \in X} i_{x, *}I_x $$ is the product of the skyscraper sheaves $i_{x, *}I_x$ (see Sheaves, Section \ref{sheaves-section-skyscraper-sheaves} for notation.) We have $$ \Mor_{\textit{Ab}}(\mathcal{F}_x, I_x) = \Mor_{\textit{Ab}(X)}(\mathcal{F}, i_{x, *}I_x). $$ see Sheaves, Lemma \ref{sheaves-lemma-stalk-skyscraper-adjoint}. Hence it is clear that each $i_{x, *}I_x$ is injective. Hence the injectivity of $\mathcal{I}$ follows from Homology, Lemma \ref{homology-lemma-product-injectives}.
For any ring $R$ and any $R$-module $M$ we denote $j : M \to J_R(M)$ the functorial injective embedding constructed in More on Algebra, Section \ref{more-algebra-section-injectives-modules}. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$-modules on $X$. By Sheaves, Examples \ref{sheaves-example-sheaf-product-pointwise} and \ref{sheaves-example-sheaf-product-pointwise-algebraic-structure} the assignment $$ \mathcal{I} : U \mapsto \mathcal{I}(U) = \prod\nolimits_{x\in U} J_{\mathcal{O}_{X, x}}(\mathcal{F}_x) $$ is an abelian sheaf. There is a canonical map $\mathcal{F} \to \mathcal{I}$ given by mapping $s \in \mathcal{F}(U)$ to $\prod_{x \in U} j(s_x)$ where $s_x \in \mathcal{F}_x$ denotes the germ of $s$ at $x$. This map is injective, see Sheaves, Lemma \ref{sheaves-lemma-sheaf-subset-stalks} for example. \medskip\noindent It remains to prove the following: Given a rule $x \mapsto I_x$ which assigns to each point $x \in X$ an injective $\mathcal{O}_{X, x}$-module the sheaf $\mathcal{I} : U \mapsto \prod_{x \in U} I_x$ is injective. Note that $$ \mathcal{I} = \prod\nolimits_{x \in X} i_{x, *}I_x $$ is the product of the skyscraper sheaves $i_{x, *}I_x$ (see Sheaves, Section \ref{sheaves-section-skyscraper-sheaves} for notation.) We have $$ \Hom_{\mathcal{O}_{X, x}}(\mathcal{F}_x, I_x) = \Hom_{\mathcal{O}_X}(\mathcal{F}, i_{x, *}I_x). $$ see Sheaves, Lemma \ref{sheaves-lemma-stalk-skyscraper-adjoint}. Hence it is clear that each $i_{x, *}I_x$ is an injective $\mathcal{O}_X$-module (see Homology, Lemma \ref{homology-lemma-adjoint-preserve-injectives} or argue directly). Hence the injectivity of $\mathcal{I}$ follows from Homology, Lemma \ref{homology-lemma-product-injectives}.
This reduces to the case of a single sheaf $\mathcal{G}$ by taking the direct sum of all the $\mathcal{G}_i$. \medskip\noindent Consider the sets $$ S = \coprod\nolimits_{U \in \Ob(\mathcal{C})} \mathcal{G}(U). $$ and $$ T_\beta = \coprod\nolimits_{U \in \Ob(\mathcal{C})} J_\beta(\mathcal{F})(U) $$ Then $T_\beta = \colim_{\alpha < \beta} T_\alpha$ with injective transition maps. A morphism $\mathcal{G} \to J_\beta(\mathcal{F})$ factors through $J_\alpha(\mathcal{F})$ if and only if the associated map $S \to T_\beta$ factors through $T_\alpha$. By Sets, Lemma \ref{sets-lemma-map-from-set-lifts} if the cofinality of $\beta$ is bigger than the cardinality of $S$, then the result of the lemma is true. Hence the lemma follows from the fact that there are ordinals with arbitrarily large cofinality, see Sets, Proposition \ref{sets-proposition-exist-ordinals-large-cofinality}.
Let $\mathcal{F} \to \mathcal{G}$ be an injective map of abelian sheaves. Suppose $\varphi : \mathcal{F} \to \mathcal{J}$ is a morphism. Arguing as in the proof of More on Algebra, Lemma \ref{more-algebra-lemma-injective-abelian} we see that it suffices to prove that if $\mathcal{F} \not = \mathcal{G}$, then we can find an abelian sheaf $\mathcal{F}'$, $\mathcal{F} \subset \mathcal{F}' \subset \mathcal{G}$ such that (a) the inclusion $\mathcal{F} \subset \mathcal{F}'$ is strict, and (b) $\varphi$ can be extended to $\mathcal{F}'$. To find $\mathcal{F}'$, let $X$ be an object of $\mathcal{C}$ such that the inclusion $\mathcal{F}(X) \subset \mathcal{G}(X)$ is strict. Pick $s \in \mathcal{G}(X)$, $s \not \in \mathcal{F}(X)$. Let $\psi : \mathbf{Z}_X^\# \to \mathcal{G}$ be the morphism corresponding to the section $s$ via (\ref{equation-free-sheaf-on}). Set $\mathcal{S} = \psi^{-1}(\mathcal{F})$. By assumption the morphism $$ \mathcal{S} \xrightarrow{\psi} \mathcal{F} \xrightarrow{\varphi} \mathcal{J} $$ can be extended to a morphism $\varphi' : \mathbf{Z}_X^\# \to \mathcal{J}$. Note that $\varphi'$ annihilates the kernel of $\psi$ (as this is true for $\varphi$). Thus $\varphi'$ gives rise to a morphism $\varphi'' : \Im(\psi) \to \mathcal{J}$ which agrees with $\varphi$ on the intersection $\mathcal{F} \cap \Im(\psi)$ by construction. Thus $\varphi$ and $\varphi''$ glue to give an extension of $\varphi$ to the strictly bigger subsheaf $\mathcal{F}' = \mathcal{F} + \Im(\psi)$.
Note that $\Ob(\mathcal{A})$ is a set by our conventions about categories. An isomorphism is a surjective morphism. The composition of surjective morphisms is surjective. And the base change of a surjective morphism in $\mathcal{A}$ is surjective, see Homology, Lemma \ref{homology-lemma-epimorphism-universal-abelian-category}.
Suppose that $f : V \to U$ is a surjective morphism of $\mathcal{A}$. Let $K = \Ker(f)$. Recall that $V \times_U V = \Ker((f, -f) : V \oplus V \to U)$, see Homology, Example \ref{homology-example-fibre-product-pushouts}. In particular there exists an injection $K \oplus K \to V \times_U V$. Let $p, q : V \times_U V \to V$ be the two projection morphisms. Note that $p - q : V \times_U V \to V$ is a morphism such that $f \circ (p  - q) = 0$. Hence $p - q$ factors through $K \to V$. Let us denote this morphism by $c : V \times_U V \to K$. And since the composition $K \oplus K \to V \times_U V \to K$ is surjective, we conclude that $c$ is surjective. It follows that $$ V \times_U V \xrightarrow{p - q} V \to U \to 0 $$ is an exact sequence of $\mathcal{A}$. Hence for an object $X$ of $\mathcal{A}$ the sequence $$ 0 \to \Hom_\mathcal{A}(U, X) \to \Hom_\mathcal{A}(V, X) \to \Hom_\mathcal{A}(V \times_U V, X) $$ is an exact sequence of abelian groups, see Homology, Lemma \ref{homology-lemma-check-exactness}. This means that $h_X$ satisfies the sheaf condition on $\mathcal{C}$. \medskip\noindent The functor is fully faithful by Categories, Lemma \ref{categories-lemma-yoneda}. The functor is a left exact functor between abelian categories by Homology, Lemma \ref{homology-lemma-check-exactness}. To show that it is right exact, let $X \to Y$ be a surjective morphism of $\mathcal{A}$. Let $U$ be an object of $\mathcal{A}$, and let $s \in h_Y(U) = \Mor_\mathcal{A}(U, Y)$ be a section of $h_Y$ over $U$. By Homology, Lemma \ref{homology-lemma-epimorphism-universal-abelian-category} the projection $U \times_Y X \to U$ is surjective. Hence $\{V = U \times_Y X \to U\}$ is a covering of $U$ such that $s|_V$ lifts to a section of $h_X$. This proves that $h_X \to h_Y$ is a surjection of abelian sheaves, see Sites, Lemma \ref{sites-lemma-mono-epi-sheaves}. \medskip\noindent The site $\mathcal{C}$ has enough points by Sites, Proposition \ref{sites-proposition-criterion-points}.
Please see Lemma \ref{lemma-criterion-baer} for the case of modules. Choose an injection $A \subset B$ and a morphism $\varphi : A \to I$. Consider the set $S$ of pairs $(A', \varphi')$ consisting of subobjects $A \subset A' \subset B$ and a morphism $\varphi' : A' \to I$ extending $\varphi$. Define a partial ordering on this set in the obvious manner. Choose a totally ordered subset $T \subset S$. Then $$ A' = \colim_{t \in T} A_t \xrightarrow{\colim_{t \in T} \varphi_t} I $$ is an upper bound. Hence by Zorn's lemma the set $S$ has a maximal element $(A', \varphi')$. We claim that $A' = B$. If not, then choose a morphism $\psi : U \to B$ which does not factor through $A'$. Set $N = A' \cap \psi(U)$. Set $M = \psi^{-1}(N)$. Then the map $$ M \to N \to A' \xrightarrow{\varphi'} I $$ can be extended to a morphism $\chi : U \to I$. Since $\chi|_{\Ker(\psi)} = 0$ we see that $\chi$ factors as $$ U \to \Im(\psi) \xrightarrow{\varphi''} I $$ Since $\varphi'$ and $\varphi''$ agree on $N = A' \cap \Im(\psi)$ we see that combined the define a morphism $A' + \Im(\psi) \to I$ contradicting the assumed maximality of $A'$.
Choose a generator $U$ of $\mathcal{A}$. Denote $c$ the function of Lemma \ref{lemma-surjection-bounded-size}. Set $\kappa = \sup \{c^n(|U|), n = 1, 2, 3, \ldots\}$. Let $n \in \mathbf{Z}$ and let $\psi : U \to M^n$ be a morphism. In order to prove (1) and (2) it suffices to prove there exists a subcomplex $N^\bullet \subset M^\bullet$ which is bounded above, acyclic, and $|N^m| \leq \kappa$, such that $\psi$ factors through $N^n$. To do this set $N^n = \Im(\psi)$, $N^{n + 1} = \Im(U \to M^n \to M^{n + 1})$, and $N^m = 0$ for $m \geq n + 2$. Suppose we have constructed $N^m \subset M^m$ for all $m \geq k$ such that \begin{enumerate} \item $\text{d}(N^m) \subset N^{m + 1}$, $m \geq k$, \item $\Im(N^{m - 1} \to N^m) = \Ker(N^m \to N^{m + 1})$ for all $m \geq k + 1$, and \item $|N^m| \leq c^{\max\{n - m, 0\}}(|U|)$. \end{enumerate} for some $k \leq n$. Because $M^\bullet$ is acyclic, we see that the subobject $\text{d}^{-1}(\Ker(N^k \to N^{k + 1})) \subset M^{k - 1}$ surjects onto $\Ker(N^k \to N^{k + 1})$. Thus we can choose $N^{k - 1} \subset M^{k - 1}$ surjecting onto $\Ker(N^k \to N^{k + 1})$ with $|N^{k - 1}| \leq c^{n - k + 1}(|U|)$ by Lemma \ref{lemma-surjection-bounded-size}. The proof is finished by induction on $k$.
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: \begin{enumerate} \item If $\alpha = \beta + 1$ and $K_\beta^\bullet = M^\bullet$ then we choose $K_\alpha^\bullet = K_\beta^\bullet$. \item 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 \ref{lemma-acyclic-quotient-complexes-bounded-size}. Finally, we let $K_\alpha^\bullet \subset M^\bullet$ be the inverse image of $N_\alpha^\bullet$. \item If $\alpha$ is a limit ordinal we set $K_\beta^\bullet = \colim K_\alpha^\bullet$. \end{enumerate} It is clear that $M^\bullet = K_\alpha^\bullet$ for a suitably large ordinal $\alpha$. We will prove that $$ \Hom_{K(\mathcal{A})}(K_\alpha^\bullet, I^\bullet) $$ 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 $$ \Hom_{K(\mathcal{A})}(K_\beta^\bullet, I^\bullet) \to \Hom_{K(\mathcal{A})}(K_\alpha^\bullet, I^\bullet) \to \Hom_{K(\mathcal{A})}(N_\alpha^\bullet, I^\bullet) $$ 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 $$ \Hom^\bullet(K_\alpha^\bullet, I^\bullet) = \lim_{\beta < \alpha} \Hom^\bullet(K_\beta^\bullet, I^\bullet) $$ with notation as in More on Algebra, Section \ref{more-algebra-section-hom-complexes}. These complexes compute morphisms in $K(\mathcal{A})$ by More on Algebra, Equation (\ref{more-algebra-equation-cohomology-hom-complex}). 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 \ref{homology-lemma-ML-over-ordinals} to conclude.
Choose a functorial injective embeddings $i_M : M \to I(M)$, see Theorem \ref{theorem-injective-embedding-grothendieck}. For every complex $M^\bullet$ denote $J^\bullet(M^\bullet)$ the complex with terms $J^n(M^\bullet) = I(M^n) \oplus I(M^{n + 1})$ and differential $$ d_{J^\bullet(M^\bullet)} = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$ There exists a canonical injective map of complexes $u_{M^\bullet} : M^\bullet \to J^\bullet(M^\bullet)$ by mapping $M^n$ to $I(M^n) \oplus I(M^{n + 1})$ via the maps $i_{M^n} : M^n \to I(M^n)$ and $i_{M^{n + 1}} \circ d : M^n \to M^{n + 1} \to I(M^{n + 1})$. Hence a short exact sequence of complexes $$ 0 \to M^\bullet \xrightarrow{u_{M^\bullet}} J^\bullet(M^\bullet) \xrightarrow{v_{M^\bullet}} Q^\bullet(M^\bullet) \to 0 $$ functorial in $M^\bullet$. Set $$ \mathbf{N}^\bullet(M^\bullet) = C(v_{M^\bullet})^\bullet[-1]. $$ Note that $$ \mathbf{N}^n(M^\bullet) = Q^{n - 1}(M^\bullet) \oplus J^n(M^\bullet) $$ with differential $$ \left( \begin{matrix} - d^{n - 1}_{Q^\bullet(M^\bullet)} & - v^n_{M^\bullet} \\ 0 & d^n_{J^\bullet(M)} \end{matrix} \right) $$ Hence we see that there is a map of complexes $j_{M^\bullet} : M^\bullet \to \mathbf{N}^\bullet(M^\bullet)$ induced by $u$. It is injective and factors through an injective subobject by construction. The map $j_{M^\bullet}$ is a quasi-isomorphism as one can prove by looking at the long exact sequence of cohomology associated to the short exact sequences of complexes above.