## Half blind to acyclicity

The following definition comes from Spaltenstein though he attributes it to Bernstein.

**Definition**. A complex $P$ is called *K-projective* if \(\operatorname{Hom}_{K(\mathcal A)}(P,A) = 0\) for any acyclic complex $A$. Let K-Proj be the full subcategory of $K(\mathcal A)$ consisting of K-projective objects.

A complex $I$ is called *K-injective* if \(\operatorname{Hom}_{K(\mathcal A)}(A,I) = 0\) for any acyclic complex $A$.Let K-Inj be the full subcategory of $K(\mathcal A)$ consisting of K-injective objects.

A simple but very useful observation is the following.

**Proposition**. If $P$ is K-projective and $\phi: A \to B$ is a quasi-isomorphism, then \(\operatorname{Hom}_{K(\mathcal A)}(P,\phi) : \operatorname{Hom}_{K(\mathcal A)}(P,A) \to \operatorname{Hom}_{K(\mathcal A)}(P,B)\) is an isomorphism.

Similarly, if $I$ is K-injective, then $\operatorname{Hom}_{K(\mathcal A)}(\phi,I)$ is an isomorphism.

## **Proof**. (Expand to view)

We have an exact sequence $$ \cdots \to \operatorname{Hom}_{K(\mathcal A)}(P,CC(\phi)[-1]) \to \operatorname{Hom}_{K(\mathcal A)}(P,A) \to \operatorname{Hom}_{K(\mathcal A)}(P,B) \to \operatorname{Hom}_{K(\mathcal A)}(P,CC(\phi)[-1]) \to \cdots $$ with $$ \operatorname{Hom}_{K(\mathcal A)}(P,C(\phi)[-1]) = \operatorname{Hom}_{K(\mathcal A)}(P,CC(\phi)) = 0 $$ since $C(\phi)$ is acyclic. The case for $I$ is completely analogous. ■

**Definition**. A full subcategory $\mathcal S$ of a triangulated $\mathcal T$ is a *triangulated subcategory* if

- Given any object $S$ of $\mathcal S$, $S[1]$ and $S[-1]$ are also objects of $\mathcal S$.
- Given any map $f : S \to S^\prime$, then the cone over $f$ is also an object of $S$.

**Lemma**. Both K-Proj and K-Inj are triangulated subcategories.

## **Proof**. (Expand to view)

We have $$ \operatorname{Hom}(P[i],A) = \operatorname{Hom}(P,A[-i]) = 0 $$ since $A[-i]$ is also acyclic if $A$ is. Given a map $P \to Q$, we complete it to a triangle $$ P \to Q \to C \to P[1] $$ and apply $\operatorname{Hom}(-,A)$ to get a exact sequence $$ \cdots \leftarrow \operatorname{Hom}(Q,A) \leftarrow \operatorname{Hom}(C,A) \leftarrow \operatorname{Hom}(P[1],A) \leftarrow \cdots $$ with $$ \operatorname{Hom}(Q,A) = \operatorname{Hom}(P[1],A) = 0. $$ So $$ \operatorname{Hom}(C,A) = 0. $$ The argument for K-Inj is entirely analogous. We will handle closure under finite direct sums below. ■

Moreover, each is closed under larger operations.

**Lemma**. K-Proj is closed under coproducts. K-Inj is closed under products.

## **Proof**. (Expand to view)

For coproducts take in $K(\mathcal A)$ this clear: $$ \operatorname{Hom}(\bigoplus P_i, A) \cong \prod \opertorname{Hom}(P_i,A) $$ It is straightforward to check that coproducts in $Ch(\mathal A)$ are also coproducts in $K(\mathcal A)$. K-Inj works similarly. ■

**Lemma** Both K-Proj and K-Inj are closed under retracts.

## **Proof**. (Expand to view)

Given a retract $Q \to P \to Q$, we have a retract $$ \operatorname{Hom}(Q,A) \to \operatorname{Hom}(P,A) \to \operatorname{Hom}(Q,A) $$ of $0$ for acyclic $A$ and similarly for K-Inj. ■

We can construct some K-projective/K-injective objects.

**Lemma**. An object $P$ of $\mathcal A$ is projective if and only if $P[0]$ is K-projective.

An object $I$ of $\mathcal A$ is injective if and only if $I[0]$ is K-injective.

## **Proof**. (Expand to view)

We treat the case of $P$. Note that a chain map

We can now build out bounded complexes. But first we want to recall two different ways to truncate complexes.

**Definition**. The *stupid truncations* of $C$ are given by \((\sigma_{\leq i} C)^j = \begin{cases} C^j & j \leq i \\ 0 & j > i \end{cases}\) and \((\sigma_{\geq i} C)^j = \begin{cases} C^j & j \geq i \\ 0 & j < i \end{cases}\)

The other truncations (not really smart but less dumb) are \((\tau_{\leq i} C)^j = \begin{cases} C^j & j < i \\ \operatorname{ker} \delta_C^i & j = i \\ 0 & j > i \end{cases}\) and \((\tau_{\geq i} C)^j = \begin{cases} C^j & j > i \\ \operatorname{cok} \delta_C^{i-1} & j = i \\ 0 & j < i \end{cases}\)

**Corollary**. Any bounded complex of projectives is K-projective. Any bounded complex of injectives is K-injective.

## **Proof**. (Expand to view)

We proceed by induction on the length of the complex. The case of length is covered above. We have an isomorphism $$ \sigma_{\leq i} P = C(P^i[i+1] \to \sigma_{\leq i+1} P) $$ for the morphism

We can got a bit further in each direction separately.

**Lemma**. A bounded above complex of projectives is K-projective. A bounded below complex of injectives is K-injective.

## **Proof**. (Expand to view)

Given a bounded below complex of projectives $P$, the complex $P$ is the colimit over its truncations $$ \sigma{\geq i} P \to \sigma{\geq i-1} P $$ If acyclicity were closed under limits, we would be done quickly. However, this is not the case in general. We do have a quasi-isomorphism $$ C \left( \bigoplus \sigma_{\geq i} P \to \bigoplus \sigma_{\geq i} P \right) \to P $$ and as part of homework you will check directly that any quasi-isomorphism between bounded above complexes of projectives is invertible up to homotopy. Injectives are handled similarly. ■

You might hope that we can remove the one-sided boundedness conditions that remain. That is not the case.

**Example**. Let $R = k[x]/(x^2)$. Then the infinite complex \(P := \cdots \overset{x}{\to} R \overset{x}{\to} R \overset{x}{\to} \cdots\) is *not* K-projective.

If it were, since it is acyclic, it would have to be null-homotopic. In particular, for any other additive functor $F: \operatorname{Mod} R \to \operatorname{Ab}$, $F(P)$ would be acyclic.

But applying $- \otimes_R k$ yields \(\cdots \overset{0}{\to} k \overset{0}{\to} k \overset{0}{\to} \cdots\) which is not acyclic.

It is useful to note that any null-homotopic complex is both K-projective and K-injective so K-projectives do not need to have projective components.

Even with the difficulties, we can translate having enough projectives/injectives into K-projective/K-injective resolutions.

**Proposition**. If $\mathcal A$ has enough projectives, then for all $C \in Ch^-(\mathcal A)$ there exists a bounded below complex of projectives $P$ and a quasi-isomorphism \(\phi: P \to C\)

Similarly, $\mathcal A$ has enough injectives, then for all $C \in Ch^+(\mathcal A)$ there exists a bounded above complex of injectives $I$ and a quasi-isomorphism \(\phi: C \to I\)

## **Proof**. (Expand to view)

Via induction we can reduce to assuming that we have a commutative diagram

We say that K-projective $P$ with a quasi-isomorphism \(P \to C\) is a *resolution* of $C$ and similarly for K-injective.

Next we treat resolutions of unbounded complexes.

**Proposition**. Assume that $\mathcal A$ has enough projectives and countable coproducts. Then, any $C \in Ch(\mathcal A)$ has a K-projective resolution.

Assume that $\mathcal A$ has enough injectives and countable products. Then any $C$ has a K-injective resolution.

## **Proof**. (Expand to view)

The complex $C$ is the colimit of the (not dumb) truncations $$ \tau_{\leq i} C \to \tau_{\leq i + 1} C $$ We can choose resolutions $P_i \to \tau_{\geq i C}$. Since the cone over $P_{i+1} \to \tau_{\leq i+1} C$ is acyclic, we can lift $P_i \to \tau_{\leq i} C \to \tau_{\leq i+1} C$ to $P_i \to P_{i+1}$. Now take the cone $C\left(\bigoplus P_i \to \bigoplus P_i \right)$. This is quasi-isomorphic to the cone $C\left(\bigoplus \tau_{\geq i} C \to \bigoplus \tau_{\geq i}C \right)$ which is quasi-isomorphic to $C$ since $$ 0 \to \bigoplus \tau_{\geq i} C \to \bigoplus \tau_{\geq i} C \to C \to 0 $$ is exact. The case of K-injective resolutions is handled analogously. ■