## 13.18 Injective resolutions

In this section we prove some lemmas regarding the existence of injective resolutions in abelian categories having enough injectives.

Definition 13.18.1. Let $\mathcal{A}$ be an abelian category. Let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. An injective resolution of $A$ is a complex $I^\bullet$ together with a map $A \to I^0$ such that:

1. We have $I^ n = 0$ for $n < 0$.

2. Each $I^ n$ is an injective object of $\mathcal{A}$.

3. The map $A \to I^0$ is an isomorphism onto $\mathop{\mathrm{Ker}}(d^0)$.

4. We have $H^ i(I^\bullet ) = 0$ for $i > 0$.

Hence $A \to I^\bullet$ is a quasi-isomorphism. In other words the complex

$\ldots \to 0 \to A \to I^0 \to I^1 \to \ldots$

is acyclic. Let $K^\bullet$ be a complex in $\mathcal{A}$. An injective resolution of $K^\bullet$ is a complex $I^\bullet$ together with a map $\alpha : K^\bullet \to I^\bullet$ of complexes such that

1. We have $I^ n = 0$ for $n \ll 0$, i.e., $I^\bullet$ is bounded below.

2. Each $I^ n$ is an injective object of $\mathcal{A}$.

3. The map $\alpha : K^\bullet \to I^\bullet$ is a quasi-isomorphism.

In other words an injective resolution $K^\bullet \to I^\bullet$ gives rise to a diagram

$\xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d] \ar[r] & K^ n \ar[d] \ar[r] & K^{n + 1} \ar[d] \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^ n \ar[r] & I^{n + 1} \ar[r] & \ldots }$

which induces an isomorphism on cohomology objects in each degree. An injective resolution of an object $A$ of $\mathcal{A}$ is almost the same thing as an injective resolution of the complex $A$.

Lemma 13.18.2. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex of $\mathcal{A}$.

1. If $K^\bullet$ has an injective resolution then $H^ n(K^\bullet ) = 0$ for $n \ll 0$.

2. If $H^ n(K^\bullet ) = 0$ for all $n \ll 0$ then there exists a quasi-isomorphism $K^\bullet \to L^\bullet$ with $L^\bullet$ bounded below.

Proof. Omitted. For the second statement use $L^\bullet = \tau _{\geq n}K^\bullet$ for some $n \ll 0$. See Homology, Section 12.15 for the definition of the truncation $\tau _{\geq n}$. $\square$

Lemma 13.18.3. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives.

1. Any object of $\mathcal{A}$ has an injective resolution.

2. If $H^ n(K^\bullet ) = 0$ for all $n \ll 0$ then $K^\bullet$ has an injective resolution.

3. If $K^\bullet$ is a complex with $K^ n = 0$ for $n < a$, then there exists an injective resolution $\alpha : K^\bullet \to I^\bullet$ with $I^ n = 0$ for $n < a$ such that each $\alpha ^ n : K^ n \to I^ n$ is injective.

Proof. Proof of (1). First choose an injection $A \to I^0$ of $A$ into an injective object of $\mathcal{A}$. Next, choose an injection $I_0/A \to I^1$ into an injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose an injection $I^1/\mathop{\mathrm{Im}}(d^0) \to I^2$ into an injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. By Lemma 13.18.2 part (2) follows from part (3). Part (3) is a special case of Lemma 13.15.5. $\square$

Lemma 13.18.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be an acyclic complex. Let $I^\bullet$ be bounded below and consisting of injective objects. Any morphism $K^\bullet \to I^\bullet$ is homotopic to zero.

Proof. Let $\alpha : K^\bullet \to I^\bullet$ be a morphism of complexes. Assume that $\alpha ^ j = 0$ for $j < n$. We will show that there exists a morphism $h : K^{n + 1} \to I^ n$ such that $\alpha ^ n = h \circ d$. Thus $\alpha$ will be homotopic to the morphism of complexes $\beta$ defined by

$\beta ^ j = \left\{ \begin{matrix} 0 & \text{if} & j \leq n \\ \alpha ^{n + 1} - d \circ h & \text{if} & j = n + 1 \\ \alpha ^ j & \text{if} & j > n + 1 \end{matrix} \right.$

This will clearly prove the lemma (by induction). To prove the existence of $h$ note that $\alpha ^ n|_{d^{n - 1}(K^{n - 1})} = 0$ since $\alpha ^{n - 1} = 0$. Since $K^\bullet$ is acyclic we have $d^{n - 1}(K^{n - 1}) = \mathop{\mathrm{Ker}}(K^ n \to K^{n + 1})$. Hence we can think of $\alpha ^ n$ as a map into $I^ n$ defined on the subobject $\mathop{\mathrm{Im}}(K^ n \to K^{n + 1})$ of $K^{n + 1}$. By injectivity of the object $I^ n$ we can extend this to a map $h : K^{n + 1} \to I^ n$ as desired. $\square$

Remark 13.18.5. Let $\mathcal{A}$ be an abelian category. Using the fact that $K(\mathcal{A})$ is a triangulated category we may use Lemma 13.18.4 to obtain proofs of some of the lemmas below which are usually proved by chasing through diagrams. Namely, suppose that $\alpha : K^\bullet \to L^\bullet$ is a quasi-isomorphism of complexes. Then

$(K^\bullet , L^\bullet , C(\alpha )^\bullet , \alpha , i, -p)$

is a distinguished triangle in $K(\mathcal{A})$ (Lemma 13.9.14) and $C(\alpha )^\bullet$ is an acyclic complex (Lemma 13.11.2). Next, let $I^\bullet$ be a bounded below complex of injective objects. Then

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(C(\alpha )^\bullet , I^\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) \ar[lld] \\ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(C(\alpha )^\bullet [-1], I^\bullet ) }$

is an exact sequence of abelian groups, see Lemma 13.4.2. At this point Lemma 13.18.4 guarantees that the outer two groups are zero and hence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$.

Lemma 13.18.6. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

$\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^\beta \\ I^\bullet }$

where $I^\bullet$ is bounded below and consists of injective objects, and $\alpha$ is a quasi-isomorphism.

1. There exists a map of complexes $\beta$ making the diagram commute up to homotopy.

2. If $\alpha$ is injective in every degree then we can find a $\beta$ which makes the diagram commute.

Proof. The “correct” proof of part (1) is explained in Remark 13.18.5. We also give a direct proof here.

We first show that (2) implies (1). Namely, let $\tilde\alpha : K \to \tilde L^\bullet$, $\pi$, $s$ be as in Lemma 13.9.6. Since $\tilde\alpha$ is injective by (2) there exists a morphism $\tilde\beta : \tilde L^\bullet \to I^\bullet$ such that $\gamma = \tilde\beta \circ \tilde\alpha$. Set $\beta = \tilde\beta \circ s$. Then we have

$\beta \circ \alpha = \tilde\beta \circ s \circ \pi \circ \tilde\alpha \sim \tilde\beta \circ \tilde\alpha = \gamma$

as desired.

Assume that $\alpha : K^\bullet \to L^\bullet$ is injective. Suppose we have already defined $\beta$ in all degrees $\leq n - 1$ compatible with differentials and such that $\gamma ^ j = \beta ^ j \circ \alpha ^ j$ for all $j \leq n - 1$. Consider the commutative solid diagram

$\xymatrix{ K^{n - 1} \ar[r] \ar@/_2pc/[dd]_\gamma \ar[d]^\alpha & K^ n \ar@/^2pc/[dd]^\gamma \ar[d]^\alpha \\ L^{n - 1} \ar[r] \ar[d]^\beta & L^ n \ar@{-->}[d] \\ I^{n - 1} \ar[r] & I^ n }$

Thus we see that the dotted arrow is prescribed on the subobjects $\alpha (K^ n)$ and $d^{n - 1}(L^{n - 1})$. Moreover, these two arrows agree on $\alpha (d^{n - 1}(K^{n - 1}))$. Hence if

13.18.6.1
\begin{equation} \label{derived-equation-qis} \alpha (d^{n - 1}(K^{n - 1})) = \alpha (K^ n) \cap d^{n - 1}(L^{n - 1}) \end{equation}

then these morphisms glue to a morphism $\alpha (K^ n) + d^{n - 1}(L^{n - 1}) \to I^ n$ and, using the injectivity of $I^ n$, we can extend this to a morphism from all of $L^ n$ into $I^ n$. After this by induction we get the morphism $\beta$ for all $n$ simultaneously (note that we can set $\beta ^ n = 0$ for all $n \ll 0$ since $I^\bullet$ is bounded below – in this way starting the induction).

It remains to prove the equality (13.18.6.1). The reader is encouraged to argue this for themselves with a suitable diagram chase. Nonetheless here is our argument. Note that the inclusion $\alpha (d^{n - 1}(K^{n - 1})) \subset \alpha (K^ n) \cap d^{n - 1}(L^{n - 1})$ is obvious. Take an object $T$ of $\mathcal{A}$ and a morphism $x : T \to L^ n$ whose image is contained in the subobject $\alpha (K^ n) \cap d^{n - 1}(L^{n - 1})$. Since $\alpha$ is injective we see that $x = \alpha \circ x'$ for some $x' : T \to K^ n$. Moreover, since $x$ lies in $d^{n - 1}(L^{n - 1})$ we see that $d^ n \circ x = 0$. Hence using injectivity of $\alpha$ again we see that $d^ n \circ x' = 0$. Thus $x'$ gives a morphism $[x'] : T \to H^ n(K^\bullet )$. On the other hand the corresponding map $[x] : T \to H^ n(L^\bullet )$ induced by $x$ is zero by assumption. Since $\alpha$ is a quasi-isomorphism we conclude that $[x'] = 0$. This of course means exactly that the image of $x'$ is contained in $d^{n - 1}(K^{n - 1})$ and we win. $\square$

Lemma 13.18.7. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

$\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^{\beta _ i} \\ I^\bullet }$

where $I^\bullet$ is bounded below and consists of injective objects, and $\alpha$ is a quasi-isomorphism. Any two morphisms $\beta _1, \beta _2$ making the diagram commute up to homotopy are homotopic.

Proof. This follows from Remark 13.18.5. We also give a direct argument here.

Let $\tilde\alpha : K \to \tilde L^\bullet$, $\pi$, $s$ be as in Lemma 13.9.6. If we can show that $\beta _1 \circ \pi$ is homotopic to $\beta _2 \circ \pi$, then we deduce that $\beta _1 \sim \beta _2$ because $\pi \circ s$ is the identity. Hence we may assume $\alpha ^ n : K^ n \to L^ n$ is the inclusion of a direct summand for all $n$. Thus we get a short exact sequence of complexes

$0 \to K^\bullet \to L^\bullet \to M^\bullet \to 0$

which is termwise split and such that $M^\bullet$ is acyclic. We choose splittings $L^ n = K^ n \oplus M^ n$, so we have $\beta _ i^ n : K^ n \oplus M^ n \to I^ n$ and $\gamma ^ n : K^ n \to I^ n$. In this case the condition on $\beta _ i$ is that there are morphisms $h_ i^ n : K^ n \to I^{n - 1}$ such that

$\gamma ^ n - \beta _ i^ n|_{K^ n} = d \circ h_ i^ n + h_ i^{n + 1} \circ d$

Thus we see that

$\beta _1^ n|_{K^ n} - \beta _2^ n|_{K^ n} = d \circ (h_1^ n - h_2^ n) + (h_1^{n + 1} - h_2^{n + 1}) \circ d$

Consider the map $h^ n : K^ n \oplus M^ n \to I^{n - 1}$ which equals $h_1^ n - h_2^ n$ on the first summand and zero on the second. Then we see that

$\beta _1^ n - \beta _2^ n - (d \circ h^ n + h^{n + 1}) \circ d$

is a morphism of complexes $L^\bullet \to I^\bullet$ which is identically zero on the subcomplex $K^\bullet$. Hence it factors as $L^\bullet \to M^\bullet \to I^\bullet$. Thus the result of the lemma follows from Lemma 13.18.4. $\square$

Lemma 13.18.8. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet$ be bounded below complex consisting of injective objects. Let $L^\bullet \in K(\mathcal{A})$. Then

$\mathop{\mathrm{Mor}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Mor}}\nolimits _{D(\mathcal{A})}(L^\bullet , I^\bullet ).$

Proof. Let $a$ be an element of the right hand side. We may represent $a = \gamma \alpha ^{-1}$ where $\alpha : K^\bullet \to L^\bullet$ is a quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet$ is a map of complexes. By Lemma 13.18.6 we can find a morphism $\beta : L^\bullet \to I^\bullet$ such that $\beta \circ \alpha$ is homotopic to $\gamma$. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet$ such that there exists a quasi-isomorphism $\alpha : K^\bullet \to L^\bullet$ with $\beta \circ \alpha$ homotopic to zero. Then Lemma 13.18.7 shows that $\beta$ is homotopic to zero also, i.e., $b = 0$. $\square$

Lemma 13.18.9. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. For any short exact sequence $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ of $\text{Comp}^{+}(\mathcal{A})$ there exists a commutative diagram in $\text{Comp}^{+}(\mathcal{A})$

$\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_1^\bullet \ar[r] & I_2^\bullet \ar[r] & I_3^\bullet \ar[r] & 0 }$

where the vertical arrows are injective resolutions and the rows are short exact sequences of complexes. In fact, given any injective resolution $A^\bullet \to I^\bullet$ we may assume $I_1^\bullet = I^\bullet$.

Proof. Step 1. Choose an injective resolution $A^\bullet \to I^\bullet$ (see Lemma 13.18.3) or use the given one. Recall that $\text{Comp}^{+}(\mathcal{A})$ is an abelian category, see Homology, Lemma 12.13.9. Hence we may form the pushout along the injective map $A^\bullet \to I^\bullet$ to get

$\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I^\bullet \ar[r] & E^\bullet \ar[r] & C^\bullet \ar[r] & 0 }$

Note that the lower short exact sequence is termwise split, see Homology, Lemma 12.27.2. Hence it suffices to prove the lemma when $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ is termwise split.

Step 2. Choose splittings. In other words, write $B^ n = A^ n \oplus C^ n$. Denote $\delta : C^\bullet \to A^\bullet $ the morphism as in Homology, Lemma 12.14.10. Choose injective resolutions $f_1 : A^\bullet \to I_1^\bullet$ and $f_3 : C^\bullet \to I_3^\bullet$. (If $A^\bullet$ is a complex of injectives, then use $I_1^\bullet = A^\bullet$.) We may assume $f_3$ is injective in every degree. By Lemma 13.18.6 we may find a morphism $\delta ' : I_3^\bullet \to I_1^\bullet $ such that $\delta ' \circ f_3 = f_1 \circ \delta$ (equality of morphisms of complexes). Set $I_2^ n = I_1^ n \oplus I_3^ n$. Define

$d_{I_2}^ n = \left( \begin{matrix} d_{I_1}^ n & (\delta ')^ n \\ 0 & d_{I_3}^ n \end{matrix} \right)$

and define the maps $B^ n \to I_2^ n$ to be given as the sum of the maps $A^ n \to I_1^ n$ and $C^ n \to I_3^ n$. Everything is clear. $\square$

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