**Proof.**
Assume (1) holds. Then we have $H^ i(M) = \mathop{\mathrm{colim}}\nolimits H^ i(E_ n)$ by Derived Categories, Lemma 13.33.8. Thus it suffices to prove that $H^ i(E_ n)$ is countable for each $n$. By Proposition 22.36.4 we see that $E_ n$ is isomorphic in $D(A, \text{d})$ to a direct summand of a differential graded module $P$ which has a finite filtration $F_\bullet $ by differential graded submodules such that $F_ jP/F_{j - 1}P$ are finite direct sums of shifts of $A$. By assumption the groups $H^ i(F_ jP/F_{j - 1}P)$ are countable. Arguing by induction on the length of the filtration and using the long exact cohomology sequence we conclude that (2) is true. The interesting implication is the other one.

We claim there is a countable differential graded subalgebra $A' \subset A$ such that the inclusion map $A' \to A$ defines an isomorphism on cohomology. To construct $A'$ we choose countable differential graded subalgebras

\[ A_1 \subset A_2 \subset A_3 \subset \ldots \]

such that (a) $H^ i(A_1) \to H^ i(A)$ is surjective, and (b) for $n > 1$ the kernel of the map $H^ i(A_{n - 1}) \to H^ i(A_ n)$ is the same as the kernel of the map $H^ i(A_{n - 1}) \to H^ i(A)$. To construct $A_1$ take any countable collection of cochains $S \subset A$ generating the cohomology of $A$ (as a ring or as a graded abelian group) and let $A_1$ be the differential graded subalgebra of $A$ generated by $S$. To construct $A_ n$ given $A_{n - 1}$ for each cochain $a \in A_{n - 1}^ i$ which maps to zero in $H^ i(A)$ choose $s_ a \in A^{i - 1}$ with $\text{d}(s_ a) = a$ and let $A_ n$ be the differential graded subalgebra of $A$ generated by $A_{n - 1}$ and the elements $s_ a$. Finally, take $A' = \bigcup A_ n$.

By Lemma 22.37.1 the restriction map $D(A, \text{d}) \to D(A', \text{d})$, $M \mapsto M_{A'}$ is an equivalence. Since the cohomology groups of $M$ and $M_{A'}$ are the same, we see that it suffices to prove the implication (2) $\Rightarrow $ (1) for $(A', \text{d})$.

Assume $A$ is countable. By the exact same type of argument as given above we see that for $M$ in $D(A, \text{d})$ the following are equivalent: $H^ i(M)$ is countable for each $i$ and $M$ can be represented by a countable differential graded module. Hence in order to prove the implication (2) $\Rightarrow $ (1) we reduce to the situation described in the next paragraph.

Assume $A$ is countable and that $M$ is a countable differential graded module over $A$. We claim there exists a homomorphism $P \to M$ of differential graded $A$-modules such that

$P \to M$ is a quasi-isomorphism,

$P$ has property (P), and

$P$ is countable.

Looking at the proof of the construction of P-resolutions in Lemma 22.20.4 we see that it suffices to show that we can prove Lemma 22.20.3 in the setting of countable differential graded modules. This is immediate from the proof.

Assume that $A$ is countable and that $M$ is a countable differential graded module with property (P). Choose a filtration

\[ 0 = F_{-1}P \subset F_0P \subset F_1P \subset \ldots \subset P \]

by differential graded submodules such that we have

$P = \bigcup F_ pP$,

$F_ iP \to F_{i + 1}P$ is an admissible monomorphism,

isomorphisms of differential graded modules $F_ iP/F_{i - 1}P \to \bigoplus _{j \in J_ i} A[k_ j]$ for some sets $J_ i$ and integers $k_ j$.

Of course $J_ i$ is countable for each $i$. For each $i$ and $j \in J_ i$ choose $x_{i, j} \in F_ iP$ of degree $k_ j$ whose image in $F_ iP/F_{i - 1}P$ generates the summand corresponding to $j$.

Claim: Given $n$ and finite subsets $S_ i \subset J_ i$, $i = 1, \ldots , n$ there exist finite subsets $S_ i \subset T_ i \subset J_ i$, $i = 1, \ldots , n$ such that $P' = \bigoplus _{i \leq n} \bigoplus _{j \in T_ i} Ax_{i, j}$ is a differential graded submodule of $P$. This was shown in the proof of Lemma 22.36.3 but it is also easily shown directly: the elements $x_{i, j}$ freely generate $P$ as a right $A$-module. The structure of $P$ shows that

\[ \text{d}(x_{i, j}) = \sum \nolimits _{i' < i} x_{i', j'}a_{i', j'} \]

where of course the sum is finite. Thus given $S_0, \ldots , S_ n$ we can first choose $S_0 \subset S'_0, \ldots , S_{n - 1} \subset S'_{n - 1}$ with $\text{d}(x_{n, j}) \in \bigoplus _{i' < n, j' \in S'_{i'}} x_{i', j'}A$ for all $j \in S_ n$. Then by induction on $n$ we can choose $S'_0 \subset T_0, \ldots , S'_{n - 1} \subset T_{n - 1}$ to make sure that $\bigoplus _{i' < n, j' \in T_{i'}} x_{i', j'}A$ is a differential graded $A$-submodule. Setting $T_ n = S_ n$ we find that $P' = \bigoplus _{i \leq n, j \in T_ i} x_{i, j}A$ is as desired.

From the claim it is clear that $P = \bigcup P'_ n$ is a countable rising union of $P'_ n$ as above. By construction each $P'_ n$ is a differential graded module with property (P) such that the filtration is finite and the succesive quotients are finite direct sums of shifts of $A$. Hence $P'_ n$ defines a compact object of $D(A, \text{d})$, see for example Proposition 22.36.4. Since $P = \text{hocolim} P'_ n$ in $D(A, \text{d})$ by Lemma 22.23.2 the proof of the implication (2) $\Rightarrow $ (1) is complete.
$\square$

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