# Does every complex have a quasi-isomorphic projective complex?

Let $C^{\textbf{.}}$ be a complex in some abelian category (edit: assuming it has enough projectives). I would like to know if there exist a complex $X^{\textbf{.}}$ consisting of projective objects and a quasi-isomorphism $f:C^{\textbf{.}}\to X^{\textbf{.}}$.

This question comes from "Cohomology of Number Fields", page 110 in the second edition. In the book it is assumed that the category is of modules over a dedekind domain, so mabye it is only true with this hypothesis. Another hypothesis about the special case in which the authors use this statement, is where $C^{\textbf{.}}$ is bounded above.

• Some abelian category? No chance, if the abelian category does not have enough projectives. Otherwise, if we are talking about chain complexes of modules over a ring, then this is a classical result of Cartan and Eilenberg. However, the superscript bullet makes me think you want something about cochain complexes... Apr 5, 2013 at 18:58
• @Zhen Lin: This is an answer, not a comment. Apr 5, 2013 at 18:59
• I would have posted it as an answer if the OP used subscript bullets! (Is it true that every cochain complex (in non-negative cochain degrees) has a projective resolution?) Apr 5, 2013 at 19:04
• @Zhen Lin: It is true, for modules anyway. One way of stating this is that the category of (possibly unbounded) complexes has a model structure in which quasi-isomorphisms are weak equivalences and cofibrant objects are certain kinds of complexes of projectives. Check out chapter 2 in Mark Hovey's book on model categories for a more precise statement. Apr 5, 2013 at 21:05
• What is the difference between cochain complexes and chain complexes? isnt it just a matter of notation? Apr 6, 2013 at 11:49

The statement you've asked for is true provided some conditions on the abelian category: there are enough projectives, all colimits exist, and all filtered colimits are exact.

Suppose first that a cochain complex $A^\bullet$ is bounded above. In this case, we can find a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives as long as there are enough projectives in the abelian category. This is a classical result: you can do this by constructing a Cartan-Eilenberg resolution of $A^\bullet$ and taking its total complex, but you can also give an elementary proof by induction. Inductively, suppose that for all $i > n$, we have constructed projective objects $P^i$, differentials $d_P^i : P^i \to P^{i+1}$, and morphisms $f^i : P^i \to A^i$ which commute with the differentials and which induce isomorphisms $\ker(d_P^{i+1})/\mathrm{im}(d_P^i) \to H^{i+1}(A^\bullet)$. Since $A^\bullet$ is bounded above, the base case of the induction is satisfied since we can just take $P^i = 0$ for all big enough $i$. The fact that the map $f^{n+1} : P^{n+1} \to A^n$ commutes with the differentials tells us that there is an induced map $\ker(d_P^{n+1}) \to Z^{n+1}(A^\bullet)$. We know that $B^{n+1}(A^\bullet) \subset Z^{n+1}(A^\bullet)$. If we form the pullback $\ker(d_P^{n+1}) \times_{Z^{n+1}(A^\bullet)} B^{n+1}(A^\bullet)$, since there are enough projectives we can find an epimorphism $$P^n \twoheadrightarrow \ker(d_P^{n+1}) \times_{Z^{n+1}(A^\bullet)} B^{n+1}(A^\bullet).$$ Now there is obviously a map $d_P^n : P^n \to P^{n+1}$ which is factors through $\ker(d_P^{n+1})$ so $d_P^{n+1} \circ d_P^n = 0$. There is also a map $P^n \to B^{n+1}(A^\bullet)$, and we have an epimorphism $A^n \twoheadrightarrow B^{n+1}(A^\bullet)$, so we can construct a map $f^n : P^n \to A^n$ using the fact that $P^n$ is projective. Now if you do a little diagram chasing, you can show that the inductive hypotheses continue to be satisfied, which completes the induction and produces us a quasi-isomorphism $f^\bullet : P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives.

The difference between cochain complexes and chain complexes is notational, but sometimes people like their cochain complexes to be concentrated in non-negative degrees and their chain complexes to be concentrated in non-negative degrees. The above proof proves that we can find quasi-isomorphisms $P^\bullet \to A^\bullet$ when $A^\bullet$ is, for instance, concentrated in non-positive degrees, which, after a change of notation, proves that we can find quasi-isomorphisms $P_\bullet \to A_\bullet$ of chain complexes when $A_\bullet$ is concentrated in non-negative degrees.


This gives a proof of the statement you asked about, but I should probably point out that asking for a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a complex of projectives is not a very useful thing to do unless $A^\bullet$ is bounded above. More specifically, I mean the following. A complex $P^\bullet$ of projective objects is called dg-projective if it satisfies the following property: whenever $C^\bullet$ is acyclic, any map $P^\bullet \to C^\bullet$ is null-homotopic. For bounded above complexes, being dg-projective is equivalent to being projective in all degrees, but this is not true for unbounded complexes. When $A^\bullet$ is bounded above, the proof we gave above shows that we can always find a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a dg-projective complex, but for unbounded $A^\bullet$, proving that there is always a quasi-isomorphism $P^\bullet \to A^\bullet$ with $P^\bullet$ a dg-projective complex is more involved.

• Unfortunately this doesn't work. Unless I'm missing something, the procedure as described gives a zero complex. It is true that if one manages to circumvent the zero step and produce a suitable isomorphism in degree 0 this will induce monomorphisms in homology, however.
– Pedro
Jul 23, 2015 at 18:39
• I think you're right. I'll fix this soon. Jul 23, 2015 at 21:05

$$\require{AMScd}$$ As noted above, the first argument provides an injection in homology, but not an isomorphism. I will set things up inductively so that we get a diagram $$\begin{CD} \\ &{}&{}& C_n @>>> C_{n-1} \\ {}&{}& {} @Af_n AA @A A f_{n-1} A \\ {}&{}&{}&P_{n} @>>d_n> P_{n-1} \\ \end{CD}$$

so that $$f_{n-1}$$ induces an isomorphism in homology and $$f_n$$ induces an epimorphism from $$\ker d_n$$ to $$H_n(C_\cdot)$$. When $$n=0$$ this is almost trivial: take $$f_{n-1}=0$$, $$d_{-1}=0$$ and $$f_0$$ induced from covering the zeroeth homology $$\rho:P_0\to H_0$$ by a projective and obtaining $$f_0$$ induced from the projection $$\pi_0:C_0\to H_0$$. This gives a commutative diagram

$$\begin{CD} \\ &{}&{}& C_0 @>\partial_{0}>> 0 \\ {}&{}& {} @Af_0 AA @AAA \\ {}&{}&{}&P_0 @>>d_{0}> 0\\ \end{CD}$$

Assume inductively that we have obtained a commutative diagram $$\begin{CD} \\ &{}&{}& C_n @>>> C_{n-1} \\ {}&{}& {} @Af_n AA @AA f_{n-1} A \\ {}&{}&{}&P_{n} @>>d_n> P_{n-1} \\ \end{CD}$$ such that $$f_{n-1}$$ induces an isomorphism in homology and $$f_n$$ induces an epimorphism $$\ker d_n\to H_n(C_\cdot )$$. Take a cover $$\rho_{n+1}:P_{n+1}' \to H_{n+1}$$ with $$P_{n+1}'$$ projective and $$f_{n+1}'$$ induced from the projection $$\ker{\partial_{n+1}}\to H_{n+1}$$. Now consider the pullback diagram (at $$Q_{n+1}$$)

$$\begin{CD} \\ &{}&{}& {\rm im}\; \partial_{n+1} @>\iota>> \ker \partial_{n+1} @>\widetilde{j}>> C_n \\ {}&{}& {} @A\psi_n AA @A\widetilde{f_n} AA @Af_nAA \\ P_{n+1}'' @>\Lambda_{n+1}>> Q_{n+1} @>\varphi_n>> \ker d_n @>j>> P_n \\ \end{CD}$$

where we have taken $$P_{n+1}''$$ projective and $$\Lambda_{n+1}$$ an epimorphism. Define $$d_{n+1}''= j\varphi_n\Lambda_{n+1}$$. This also gives an arrow $$\psi_n\Lambda_{n+1}:P_{n+1}''\longrightarrow {\rm im}\; \partial_{n+1}$$ and we obtain $$f_{n+1}'':P_{n+1}''\longrightarrow C_{n+1}$$ from the epimorphism $$\tilde\partial_{n+1}:C_{n+1} \longrightarrow {\rm im}\; \partial_{n+1}$$. Now let $$d_{n+1}:P_{n+1}\longrightarrow C_{n+1}$$ where $$P_{n+1}=P_{n+1}'\oplus P_{n+1}''$$ and $$(p',p'')\mapsto d_{n+1}''(p)$$. Set $$f_{n+1}=f_{n+1}'+f_{n+1}''$$. Then

$$(1)$$ Since the image of $$f_{n+1}'$$ lies in the kernel of $$\partial_{n+1}$$, we have

\begin{align} \partial_{n+1}f_{n+1}&= \partial_{n+1}f_{n+1}''\\&= \tilde j\iota \tilde \partial_{n+1} f_{n+1}''\\&=\tilde j\iota \psi_n\Lambda_{n+1}\\&=\tilde j\tilde{f_n}\varphi_n \Lambda_{n+1}\\&=f_n j\varphi_n \Lambda_{n+1}\\&=f_n d_{n+1}''=f_n d_{n+1}\end{align}

$$(2)$$ Evidently $$d_nd_{n+1}=0$$ since $$d_{n+1}''$$ factors through the kernel of $$d_n$$. Assume that $$z\in\ker d_n''$$ is such that $$f_n(z)=\partial_{n+1}z ''$$. Then $$(\partial_{n+1}z'',z)$$ lies in $$Q_{n+1}$$ so there is some $$p\in P_{n+1}''$$ so that $$\Lambda_{n+1}(p)$$ equals this. It then follows $$d_{n+1}''(p)=z$$ and since the images of $$d$$ and $$d''$$ coincide we obtain the desired isomorphism.

$$(3)$$ We have that $$\ker d_{n+1}=P_{n+1}'\oplus \ker d_{n+1}''$$ maps by $$f_{n+1}$$ onto $$H_{n+1}(C_\cdot)$$, since the first summand is already onto.

• For this you need the complex to start somewhere. Jul 25, 2015 at 1:50
• @Mariano Yes, I am assuming the complex is zero in negative degrees (i.e. the first part of the argument in the accepted answer)
– Pedro
Jul 25, 2015 at 2:05
• @user1987 Yes, $p$ should be $p''$. What I mean is that $f_n$ sends $(p',p'')$ to $f_n'(p')+f_n''(p'')$. Note that $d_n''$ comes already equipped with the inclusion into $P_n$ as one of the coordinates, so the last equation should make sense. At the moment I cannot look into the details.
– Pedro
Jan 16, 2016 at 2:01