# Enough injectives in the category of chain complexes

So I am a little confused about a question I thought would be obvious. Let $\mathcal{A}$ be an abelian category and let $\text{Ch}(\mathcal{A})$ be the category of chain complexes. Is it true that $\text{Ch}(\mathcal{A})^{\text{op}} \simeq \text{Ch}(\mathcal{A}^{\text{op}})$? From what I can understand, this is true. My concern is that it seems to lead to a corollary which I am doubtful is true.

Suppose $\mathcal{A}$ has enough projectives. Then the category of chain complexes $\text{Ch}(\mathcal{A})$ has enough projectives. You prove this by using the horseshoe lemma. In that case, the category $\mathcal{A}^{\text{op}}$ has enough injectives. If my supposedly obvious equivalence of categories above is true, then this would mean that $\text{Ch}(\mathcal{A}^{\text{op}})$ has enough injectives. Or in other words, the category of chain complexes of an abelian category with enough injectives itself has enough injectives. My concern with this is that there is no dual analogue for the horseshoe lemma, which makes me doubtful of this fact.

I know it is true that the category of chain complexes of a Grothendieck category is itself a Grothendieck category, and so in this situation having enough injectives is carried over to the category of chain complexes. But is it true for an abelian category with enough injectives that isn't Grothendieck?

As a side question, I am aware that in the category of chain complexes, the projective objects look like split exact sequences, each of whose objects are projective. Is this the case for injective chain complexes - namely that they are split exact sequences of chain complexes, each of whose objects are injective?

Another aside question, if this is true in the category of chain complexes, are both statements true in the category of bounded below (or above) complexes?

• What do you mean by a "dual" version of the horseshoe lemma? If we are thinking of the same thing, then there is such a "dual." – LPK Aug 21 '18 at 1:12

Chain complexes have an "index" category, which is a $\mathbf{Ab}$-category $\mathcal{I}$ whose objects are integers and whose hom-groups are given by

$$\hom(m,n) = \begin{cases} \mathbb{Z} & n \in \{m, m-1\} \\ 0 & \text{otherwise} \end{cases}$$

The generator of $\hom(m,n)$ is $1_m$. Let $\partial_m$ denote the generator of $\hom(m, m-1)$.

An alternative description of this $\mathbf{Ab}$-category is that the morphisms are freely generated by adjoining arrows $\partial_m : m \to (m-1)$, subject to the relations $\partial_{m-1} \partial_m = 0$.

Then, for any $\mathbf{Ab}$-category $\mathcal{A}$, you have an isomorphism of categories

$$\mathrm{Add}(\mathcal{I}, \mathcal{A}) \cong \mathrm{Ch}(\mathcal{A})$$

where $\mathrm{Add}$ is the category of additive functors. Now, we can compute

$$\mathrm{Ch}(\mathcal{A}^\mathrm{op}) \cong \mathrm{Add}(\mathcal{I}, \mathcal{A}^\mathrm{op})$$ $$\mathrm{Ch}(\mathcal{A})^\mathrm{op} \cong \mathrm{Add}(\mathcal{I}, \mathcal{A})^\mathrm{op} \cong \mathrm{Add}(\mathcal{I}^\mathrm{op}, \mathcal{A}^\mathrm{op})$$

$\mathcal{I}^\mathrm{op}$ is the index category for cochain complexes, so $\mathrm{Ch}(\mathcal{A})^\mathrm{op}$ is isomorphic to the category of cochain complexes on $\mathcal{A}^\mathrm{op}$.

However, by reindexing you can convert between chain and cochain complexes: $\mathcal{I}^\mathrm{op} \cong \mathcal{I}$, by the functor $n \mapsto -n$. So you do indeed have

$$\mathrm{Ch}(\mathcal{A}^\mathrm{op}) \cong \mathrm{Ch}(\mathcal{A})^\mathrm{op}$$

• @Luke: My conclusion was really bugging me, and I've found my mental block; there is a nice way to construct an isomorphism of these categories. – Hurkyl Aug 20 '18 at 9:19