# Injective resolution of a complex is unique up to homotopy type?

I was reading about how we can construct an injective resolution $$I^\bullet$$ of a bounded-below complex $$A^\bullet$$ in a category $$\mathcal{A}$$ with enough injective in this MSE post. But I'm wondering if the resolution $$I^\bullet$$ is unique up to homotopy type just like the injective resolution of a single object.

An injective resolution of $$A^\bullet$$ is a complex $$I^\bullet$$ consisting of injective objects such that there is a quasi-isomorphism $$f:A^\bullet \rightarrow I^\bullet$$.

When $$A$$ is a single object I can think of it as $$0\rightarrow A\rightarrow 0 \rightarrow 0 \rightarrow ...$$ and the injective resolution $$I^\bullet$$ is just the 'usual' injective resolution i.e. $$0\rightarrow A \rightarrow I^0\rightarrow I^1 \rightarrow ...$$ is exact. In this case I know (e.g. from P. J. Hilton's 'A Course in Homological Algebra', Proposition 4.3) that two injective resolutions $$I^\bullet, J^\bullet$$ are the same homotopy type (we can find $$f:I^\bullet\rightarrow J^\bullet$$ and $$g:J^\bullet\rightarrow I^\bullet$$ such that $$fg\simeq 1_I, gf\simeq 1_J$$).

My question is,

Is the injective resolution $$I^\bullet$$ of a bounded-below complex $$A^\bullet$$ in a category $$\mathcal{A}$$ with enough injective unique up to homotopy type? How do we show this?

The proof in know for single object case uses the fact that $$I^\bullet$$ is acyclic which is not true in the case of complex $$A^\bullet$$.

• If you know enough model categories and know the injective model structure, then since to injective resolutions are isomorphic in the homotopy category, and they are fibrant-cofibrant, it follows that there is a map in the complex category that is a quasi-isomorphism, and in fact there are maps in the two directions, and they are quasi-inverses. Then, again as they are fibrant-cofibrant, any good path-object has all the necessary homotopies, and choosing the correct path-object shows that you have actual homotopies. (cont.) – Max Jun 14 at 17:31
• So the complication here is really in proving that the inejctive model structure is a model structure; and then finding a suitable path-object that actually models the usual homotopies. The complication is in proving the lifting properties and the factorisation. The lifting properties should actually be of reasonable difficulty, it's mostly the factorisation property that will be complicated – Max Jun 14 at 17:34
• But if I remember correctly there is in fact a direct argument, I will try to look for it – Max Jun 14 at 17:35
• @Max Thank you for your reply. Unfortunately, I don't know much about model categories. Is it similar to saying that there is an equivalence $\mathbf{D}^b(\mathcal{A})\rightarrow \mathbf{K}^b(\mathcal{A})$ between derived and homotopic categories? I'm in fact asking the above question hoping to understand this statement better. So an elementary argument would be really appriciated. Thank you! – user113988 Jun 14 at 21:11

I'll use "complex" to mean "nonnegatively graded (cochain) complex". Of course if you have a bounded below cochain complex, up to a shift one may assume we are in this situation.

Here's a direct argument, that singles out the essential bits of the "model-category argument" that I sketched in the comments. It turns out that the important things you need are as follows (and they're not model-category-theoretic in nature, you actually need them to prove that the injective model-structure exists, or at least the first 2) :

1) Suppose $$f:A\to B$$ is a degree-wise monomorphism that is also a quasi-isomorphism, and $$g:C\to D$$ is a degree-wise epimorphism with a kernel that is a complex of injectives, then $$f$$ has the left lifting property against $$g$$, that is, every commutative square

$$\require{AMScd} \begin{CD} A @>{}>> C\\ @V{f}VV @VV{g}V\\ B @>>{}> D \end{CD}$$

has a diagonal map $$B\to C$$ that makes the whole thing commute (that is, the two triangles commute)

2) If $$\alpha\circ \beta$$ and $$\beta$$ are quasi-isomorphisms, then so is $$\alpha$$

3) Let $$I\to J$$ be a quasi-isomorphism of complexes of injectives. Then it is a chain homotopy equivalence.

With these 3 tools, you can conclude in the following way : Note that an injective resolution $$A\to I$$ is a degree-wise monomorphism that is also a quasi-isomorphism, and that if $$J$$ is a complex of injectives, $$J\to 0$$ is a degree-wise epimorphism with degree-wise injective kernel. Therefore, if $$A\to I,A\to J$$ are injective resolutions, we have the following square $$\require{AMScd} \begin{CD} A@>{}>> I\\ @V{}VV @VV{}V\\ J @>>{}> 0 \end{CD}$$

that satisfies the hypotheses of 1), so that there is a lift $$J\to I$$. The top triangle consists of 2 quasi-isomorphisms and the map $$J\to I$$ so by 2) $$J\to I$$ is a quasi-isomorphism. Thus by 3) it is a chain homotopy equivalence.

Proving 2) is fairly easy : prove that it holds for isomorphisms in any category, then note that a quasi-isomorphism is just some arrow $$\alpha$$ with $$H^*(\alpha)$$ being an isomorphism.

The idea for 1) is that each $$C^n\to D^n$$ has injective cokernel, so locally it is $$C^n\simeq D^n\oplus I^n$$ for some $$I^n$$, which you can then use to get a lift $$B\to C$$ by induction, because you already have $$B^n\to D^n$$, and you just need to pick a lift $$B^n\to I^n$$ of $$A^n\to I^n$$, but you can because $$I^n$$ is injective. Then you need to adjust the lift a bit to fit with the differentials that came before. For that you can use the cokernel of $$A\to B$$, which, by the long exact sequence in cohomology, is acyclic; and its acyclicity will allow you to do the adjustments. If you want I can give more details.

To prove 3), following Weibel, we first use a lemma : let $$C$$ be an exact cochain complex and $$I$$ a complex of injectives. Then any map $$C\to I$$ is nullhomotopic. This is a pretty standard/straightforward exercise in homological algebra that I'll leave to you.

Then let $$f:I\to J$$ denote our quasi-isomorphism, and consider $$cone(f)$$ : since $$f$$ is a quasi-isomorphism, it is exact. Moreover, we have a natural map $$cone(f) \to TI$$ where $$TI^n = I^{n+1}$$ which is a complex of injectives. It follows that the map $$cone(f) \to TI$$ is nullhomotopic.

Then one checks by some computation (see Weibel, proof of 10.4.6) that this provides a map $$J\to I$$ that, composed with $$I\to J$$ is homotopic to $$id_I$$.

In particular, the composition $$I\to J\to I$$ is a quasi-isomorphism, so the map $$J\to I$$ must be one too. Therefore, by what we just did, we get a map $$I\to J$$ such that the composite $$J\to I\to J$$ is homotopic to $$id_J$$.

Then in the homotopy category, $$J\to I$$ has a right-inverse and a left-inverse, it is therefore an isomorphism, and the two inverses are equal, so by going back to the original category, we see that the map $$J\to I$$ is a quasi-inverse to our original $$I\to J$$ (since its other inverse is homotopic to it, and homotopies compose)

We used that our complexes were nonnegatively graded in the proofs of 1) and 3) : in 1), it's to start off the induction and define the lift; in 3) it's to claim that a chain map from an exact complex to a complex of injectives is nullhomotopic, again it's to start off the induction (I left that as an exercise for you !)

ADDED (more words on 1) ) : I strongly recommend that you draw the diagrams yourself ! I will try to draw some here, but MSE is really annoying with commutative diagrams so I won't draw that many, and they won't be that good.

Let $$I$$ denote the kernel of $$C\to D$$: it is a complex of injectives, so locally we have a splitting $$C_n\simeq D_n\oplus I_n$$. Of course, a priori, this splitting is not a splitting of complexes, this is why we need to work to adjust the lifts along the way.

Let $$K$$ denote the cokernel of $$A\to B$$. By hypothesis and by the long exact sequence in cohomology, it is an exact complex. Note that the hypothesis of quasi-isomorphism together with the fact that our complexes are bounded and that $$A\to B$$ is a pointwise monomorphism in positive degree imply that $$A^0\to B^0$$ is a monomorphism too.

Let us build the lift by induction (we will call it $$h$$)

Let $$k^0 : B^0 \to I^0$$ be a lift of $$A^0\to C^0\to I^0$$ to $$B^0$$ : it exists as $$I^0$$ is injective and $$A^0\to B^0$$ is a monomorphism. Then $$h^0: B^0\to D^0\oplus I^0 \overset{\simeq}\to C^0$$ is clearly a lift on the degree $$0$$.

Now let's assume we have compatible lifts up to degree $$n$$ :

we have $$h^k:B^k\to C^k$$ that makes the square with diagonal commute in degree $$k$$ ,and $$d^k_C\circ h^k = h^{k+1}\circ d_B^k$$ for $$k+1\leq n$$.

Our goal is to build $$h^{n+1}$$. Note that $$C^{n+1}\simeq I^{n+1}\oplus D^{n+1}$$ and the map $$B^{n+1}\to D^{n+1}$$ is fully determined so we only need to get $$B^{n+1}\to I^{n+1}$$. Now $$A^{n+1}\to B^{n+1}$$ is a monomorphism and $$I^{n+1}$$ is injective, so we can extend it to $$B^{n+1}$$ : we get $$\tilde{h}^{n+1} : B^{n+1}\to C^{n+1}$$ that makes the square in degree $$k+1$$ commute, the problem being that it isn't necessarily compatible with the differentials: the discrepancy is "measured" by $$a:= d_C^n\circ h^n - \tilde{h}^{n+1}\circ d_B^n$$. (here I recommend drawing a sort of cubical diagram which consists in the commutative square in degrees $$n,n+1$$, the differentials between the two, and $$h^n,\tilde{h}^{n+1}$$ to see what's happening)

If we can write $$a=\epsilon\circ d_B^n$$ with $$\epsilon$$ with values in $$I^{n+1}$$, by putting $$h^{n+1} = \tilde{h}^{n+1}+\epsilon$$, we will be done and the induction may proceed (check this !). To do so we will use injectivity of $$I^{n+1}$$ and exactness of $$K$$.

Note that $$a$$ lands in $$I^{n+1}$$ and vanishes on $$A^n$$ (this is a simple diagram-chase by remembering how $$\tilde{h}^{n+1}$$ was defined and remembering why a few diagrams commute -use the diagram you drew above and compose with the correct things), therefore it factors as $$B^n\to K^n\overset{\alpha}\to I^{n+1}\to C^{n+1}$$

Then, note that by the induction hypothesis (compatibility of $$h^n$$ with lower degree differentials), $$a\circ d_B = 0$$ and this forces $$\alpha\circ d_K=0$$ (as the projection $$B\to K$$ is an epimorphism), so $$\alpha$$ factors as $$K^n\to K^n/\mathrm{im}(d) \to I^{n+1}$$. But $$K$$ is exact, therefore $$K^n/\mathrm{im}(d) = K^n/\ker(d) = \mathrm{im}(d)$$, and so we have a map $$\mathrm{im}(d) \overset{\beta}\to I^{n+1}$$

$$\mathrm{im}(d) \hookrightarrow K^{n+1}$$ and the injectivity of $$I^{n+1}$$ allow us to extend $$\beta$$ to $$b: K^{n+1}\to I^{n+1}$$, and thus by composing with the projection $$B^{n+1}\overset{\epsilon}\to I^{n+1}$$. It now remains to check that this $$\epsilon$$ satisfies what we wanted. First of all, it lands in $$I^{n+1}$$, as desired. And then the equation with $$\epsilon\circ d$$ is just a consequence of how everything was defined, and I'll leave that verification to you (it involves drawing a diagram containing $$B^n, B^{n+1}, K^n, K^{n+1}, I^{n+1}, \mathrm{im}(d),C^{n+1}$$)

Then proceed in the induction, and at the end you get your chain morphism $$h:B\to C$$ filling in the diagram. Here, we used that everything was bounded below (at $$0$$, but of course this is a notational simplification) to start off the induction.

• Thank you very much for your answer. It is quite long so I'm going to read it a bit later and possibly ask you some questions :) – user113988 Jun 20 at 13:20
• Absolutely no worries, take your time. I would adivse you to first make sure you understand the beginning up to "Proving 2) is fairly easy" : it's just the global argument, the nifty details appear after that (and aren't even completely written down, although I can if you ask me to) – Max Jun 20 at 13:29
• Hi @Max, sorry for the late response. I think I understood most things except the 1) point. Mainly I'm not sure how to adjust the lift so $B\rightarrow C$ becomes a morphism of complexes or how to use the fact that the cokernel of $A\rightarrow B$ is acyclic to help me. (I also thought the cokernel of $A\rightarrow B$ is exact because of the quasi-iso, not just acyclic?). If you don't mind, could you please help me more with this? Thank you. – user113988 Jun 26 at 18:29
• Yes of course, 1) isn't that easy and I left it with barely any indication. I'll add some words to that when I have the time. (As for your remark, acyclic = exact; having no cohomology means being exact) – Max Jun 26 at 19:14
• I added a few (a lot, actually) words on 1) – Max Jun 26 at 20:04