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Let $\mathcal{A}$ be an abelian category with enough injectives. Let $F:\mathcal{A}\to\mathcal{B}$ be a left-exact additive functor to $\mathcal{B}$ another abelian category

If $M$ has an $F$-acyclic resolution: $$0\longrightarrow M\longrightarrow X^0 \longrightarrow X^1 \longrightarrow X^2\longrightarrow\cdots$$

It is claimed (In Lang (2002) on page 797 line 11) that I can find an injective resolution: $$0\longrightarrow M\longrightarrow I^0 \longrightarrow I^1 \longrightarrow I^2\longrightarrow\cdots$$

and a cochain map from the first resolution to the second, such that each $i\geq 0$ the morphism $X^i\to I^i$ is a monomorphism.

Question: How do I find such an injective resolution, with such a chain map?

My attempts are below!

1) Using that we have enough injectives, I can find a monomorphism $X^0\to I^0$ for some $I^0$ injective. Then by composition I can find a morphism $M\to X^0\to I^0$. This gives me the monomorphism $M\hookrightarrow I^0$. Similarly using enough injectives, we can find a monomorphism $X^1\hookrightarrow I^1$ for $I^1$ injective. Then using the composite $X^0\to X^1\to I^1$ and that $I^1$ is injective, we induce a morphism from $I^0\to I^1$. We can do this inductively to obtain a cochain map between $0\to M\to X^\bullet$ and $0\to M\to I^\bullet$, with these monomorphisms for all $k\geq 0$ $X^k\hookrightarrow I^k$ for $I^k$ injective. But I can't show that the bottom row is exact, and hence I cannot show that this $I^\bullet$ an injective resolution. In fact part of me doubts that it even is (even trying to chase elements fails, since I have no epimorphisms appearing).

1b) I tried taking cokernels in each degree so that I could try to do something with the epimorphisms it gives me. Mainly I want to apply $F$ and use the long exact cohomology sequence, to make use of the $F$-acyclicity. But unless I can show exactness on the second row, I can't ensure $F$ will preserve the short exact sequence of complexes.

2) I tried using that we have enough injectives to induces a monomorphism from $M\to I_M^0$ and a monomorphism $X^0\to I_{X^0}^0$ and taking the direct sum, which I can show is also an injective object, and then repeating this for $I_{X^0}^0\oplus I_M^0$ and $X^1$, so on. But this seems highly non-exact (in fact all the morphisms in this bottom row are probably injective...)

3) I think the category of chain complexes has enough injectives, but I don't think an injective object in the chain category looks like what I want.

4) I couldn't think of any way to take the injective resolution first, and then choose compatible morphisms such that the result follows.

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  • $\begingroup$ Just a thing I noticed: In 1.) you did not even use, that the $X^i$ are acyclic. $\endgroup$ Aug 25, 2018 at 15:52
  • $\begingroup$ @red_trumpet it's a good point, see my 1b that I forgot to mention $\endgroup$
    – user587121
    Aug 25, 2018 at 15:53
  • $\begingroup$ The assumption of $F$-acyclicity is totally irrelevant (and so you should ignore it). Indeed, you can see that it must be irrelevant because if $F$ is an exact functor, then every resolution is $F$-acyclic. $\endgroup$ Aug 26, 2018 at 15:27
  • $\begingroup$ @EricWofsey How would you suggest approaching the problem? I am fine with inducing a morphism between the complexes which is unique up to chain homotopy, but without the maps at each degree being monomorphisms. Whatever I try such that they would be monomorphisms, it doesn't form an exact sequence in the second row. $\endgroup$
    – user587121
    Aug 26, 2018 at 22:57
  • $\begingroup$ I would suggest that to construct $I^1$, you first try and construct an object that has all the required properties of $I^1$ except for injectivity. Then you can just choose $I^1$ to be an injective object with a monomorphism from that object. $\endgroup$ Aug 26, 2018 at 23:26

1 Answer 1

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$\require{AMScd}$

Let $0 \to M \to X^0 \to X^1 \to \cdots$ be an exact sequence. We form a Cartan-Eilenberg resolution of this: \begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VVV \\ 0 @>>> M @>>> X^0 @>>> X^1 @>>> \cdots \\ @. @VVV @VVV @VVV \\ 0 @>>> J^0 @>>> K^{0,0} @>>> K^{1,0} @>>> \cdots \\ @. @VVV @VVV @VVV \\ 0 @>>> J^1 @>>> K^{0,1} @>>> K^{1,1} @>>> \cdots \\ @. @VVV @VVV @VVV \\ @. \vdots @. \vdots @. \vdots \\ \end{CD} The rows are exact and the columns are injective resolutions. If we define $I^n := \oplus_{p+q=n} K^{p,q} \ (n \ge 0)$, $0 \to M \to I^0 \to I^1 \to \cdots$ is a injective resolution. $M \to I^0 = K^{0,0}$ is the natural map and the differential after this is that of the total complex. The chain map from $0 \to M \to X^0 \to X^1 \to \cdots$ to $0 \to M \to I^0 \to I^1 \to \cdots$ is induced from the injection $X^p \to K^{p,0}$.

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  • $\begingroup$ I'm not seeing how you get the 0 morphism by applying the differential of the total complex twice. I am assuming you are defining the differential of the total complex by the vertical differential plus the horizontal differential. $\endgroup$
    – Charuvinda
    Jul 25, 2020 at 3:21
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    $\begingroup$ Yes. If we denote $K^{p,q} \to K^{p+1,q}$ by $\delta^{p,q}$ and $K^{p,q} \to K^{p,q+1}$ by $d^{p,q}$, $I^n \to I^{n+1} (n \ge 0)$ is defined by $I^n := \bigoplus_{p=0}^n K^{p,n-p} \ni \sum a_p \mapsto \sum (\delta^{p,n-p} + (-1)^p d^{p,n-p})a_p \in \bigoplus_{p=0}^{n+1} K^{p,n+1-p} =: I^{n+1}$. I used module-like notation. $\endgroup$
    – hipokaba
    Aug 16, 2020 at 20:26

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