The Stacks project

Proof. For every proper subobject $N' \subset N$ choose a morphism $\varphi _{N'} : U \to M$ such that $U \to M \to N$ does not factor through $N'$. Set

Then $M'$ works. $\square$

Proof. Choose a generator $U$ of $\mathcal{A}$. Denote $c$ the function of Lemma 19.12.1. 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 $N^ n = \mathop{\mathrm{Im}}(\psi )$, $N^{n + 1} = \mathop{\mathrm{Im}}(U \to M^ n \to M^{n + 1})$, 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

1. $\text{d}(N^ m) \subset N^{m + 1}$, $m \geq k$,

2. $\mathop{\mathrm{Im}}(N^{m - 1} \to N^ m) = \mathop{\mathrm{Ker}}(N^ m \to N^{m + 1})$ for all $m \geq k + 1$, and

3. $|N^ m| \leq c^{\max \{ n - m, 0\} }(|U|)$.

for some $k \leq n$. Because $M^\bullet $ is acyclic, we see that the subobject $\text{d}^{-1}(\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})) \subset M^{k - 1}$ surjects onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$. Thus we can choose $N^{k - 1} \subset M^{k - 1}$ surjecting onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$ with $|N^{k - 1}| \leq c^{n - k + 1}(|U|)$ by Lemma 19.12.1. The proof is finished by induction on $k$. $\square$

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

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

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

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

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. $\square$

Proof. For every $i \in I$ choose a (termwise) injective map of complexes $K_ i^\bullet \to L_ i^\bullet $ which is homotopic to zero with $L_ i^\bullet $ quasi-isomorphic to zero. For example, take $L_ i^\bullet $ to be the cone on the identity of $K_ i^\bullet $. We define $\mathbf{M}^\bullet (M^\bullet )$ by the following pushout diagram

Then $M^\bullet \to \mathbf{M}^\bullet (M^\bullet )$ is a functor. The right vertical arrow defines the functorial injective map $j_{M^\bullet }$. The cokernel of $j_{M^\bullet }$ is isomorphic to the direct sum of the cokernels of the maps $K_ i^\bullet \to L_ i^\bullet $ hence acyclic. Thus $j_{M^\bullet }$ is a quasi-isomorphism. Part (2) holds by construction. $\square$

Proof. Choose a functorial injective embeddings $i_ M : M \to I(M)$, see Theorem 19.11.7. 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

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

functorial in $M^\bullet $. Set

Note that

with differential

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. $\square$

Proof. Please compare with the proof of Theorem 19.2.8 and Theorem 19.11.7. Choose a cardinal $\kappa $ as in Lemmas 19.12.2 and 19.12.3. 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 19.11.4. Denote $\mathbf{M}^\bullet (-)$ the functor constructed in Lemma 19.12.4. Denote $\mathbf{N}^\bullet (-)$ the functor constructed in Lemma 19.12.5. Both of these functors come with injective transformations $\text{id} \to \mathbf{M}$ and $\text{id} \to \mathbf{N}$.

Using transfinite recursion 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

If $\beta $ is a limit ordinal we set

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 19.12.3.

For (1) we use the criterion of Lemma 19.11.6. 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 19.11.5 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 $U$.

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 19.11.5 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. $\square$