Let R be an associative ring with a unit, and consider the standard projective model structure of non-negatively graded (left) R-module, $Ch_R$. A map $f:M\to N$ in $Ch_R$ is a weak equivalence if it induces isomorphisms $H_kM\to H_kN$ on all homologies, and a cofibration if each $f_k:M_k\to N_k$ is a monomorphism, with a projective R-module as its cokernel.

Let A be a projective R-module. The chain complex $(*)$ $0\to A\to A\to0$ is a projective object in the world of chain complexes, $Ch_R$. So is any direct sum of such chain complexes. It is a known result (see [DS95, 7.10] for instance) that any acyclic object in $Ch_R$ which is level-wise projective is isomorphic to a direct sum of chains complexes as in $(*)$, with $A$'s projective.

We have here two notions, of (i) a chain complex which is level-wise project, and of (ii) a projective object in $Ch_R$. My gut's feeling is that (i) doesn't imply (ii). In the other way around I'm not sure. I would be glad to have an answer in both directions. :-)

[DS95] = Dwyer & Spalinski, Homotopy theories and model categories.


Edit: I put my answer into a broader perspective.

In the following [We] refers to Weibel's "An Introduction to Homological Algebra".

Recall that a chain complex $C$ is split exact if it is acyclic and $Z_n$ (the cycles) is a direct summand of $C_n$ for each $n$ [We, Def. 1.4.1, Ex. 1.4.2]. Moreover, let $Ch$ be the category of (unbounded) chain complexes over an abelian category with enough projectives. Then, by [We, Ex. 2.2.1] and my answer in


we have:

For a chain complex $P$ the following are equivalent:

  1. $P$ is a projective object in $Ch$
  2. $P$ is a split exact complex of projectives
  3. $P$ is a contractible complex of projectives

Also from the link we obtain the following examples where $Ch_R$ denotes the category of unbounded chain complexes of modules over the ring $R$:

  • If $R$ is hereditary (e.g. PID's, Dedekind domains), then the projective objects of $Ch_R$ are exactly the acyclic complexes of projective $R$-modules.

  • If $R$ is any ring with unit and the acyclic complex $P$ of projective $R$-modules is bounded below, then $P$ is a projective object in $Ch_R$.

However, not all acyclic complexes of projective or free modules are projective objects in $Ch_R$. A counter-example (due to Dold) is given in [We, Example 1.4.2]:

  • Over $R=\mathbb{Z}/4$ the following complex is exact $$\cdots \to \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \to \cdots $$ But it's no projective object in $Ch_R$ since $Z_n = \mathbb{Z}/2$ can't be a direct summand of $C_n = \mathbb{Z}/4$.

Added: Let $Ch_b \subseteq Ch$ be the subcategory of chain complexes that are bounded below. In contrast to $Ch$ the following holds in $Ch_b$:

The projective objects in $Ch_b$ are exactly the acyclic chain complexes (bounded below) of projectives.

Proof: Let $P$ be an acyclic chain complex of projectives that is bounded below. We have already seen in example 2 above (compare also [We, Ex. 1.4.1 2.]) that $P$ is projective in $Ch$. Since $P \in Ch_b$ it's a projective object in $Ch_b$

Now let $P \in Ch_b$ be a projective object. The same proof as in $Ch$ shows that each $P_i$ is projective (consider objects of the abelian category as chain complexes concentrated in a single degree). Also the same proof as in $Ch$ can be used to see that $P$ is acyclic: The mapping cone of $id: P[1] \to P[1]$ yields a short exact sequence $$0 \to P[1] \to \operatorname{cone}(id_{P[1]}) \to P \to 0.$$ But $P, P[1] \in Ch_b$ and by definition $\operatorname{cone}(id)_i = P_i\oplus P_{i+1}$ whence it is also bunded below. So the short exact sequence is in $Ch_b$ and splits since $P$ is a projective object. Hence $id_{P[1]}$ is nullhomotopic. So in particular, $P[1]$ and thus $P$ is acyclic. q.e.d.

  • 3
    $\begingroup$ Let me add that this complex is a counterexample to the claim in the question's second paragraph, improperly attributed to Dwyer and Spalinski. $\endgroup$ – Fernando Muro Jul 31 '12 at 11:37
  • $\begingroup$ That's a good point. $\endgroup$ – Ralph Jul 31 '12 at 21:43
  • 1
    $\begingroup$ Ralph, the question is about non-negatively graded chain complexes. In view of Ex. 1.4.1(1) in [We], if we simply cut your example in dimension 0, we get a split-exact chain complex, thus projective in $Ch_R $, thus also in $Ch_R ^\geq0$. Can this be mended to give an example of a level-wise projective chain complex which is not a projective object in $Ch_R ^\geq0$ ? $\endgroup$ – Shlomi A Aug 1 '12 at 10:39
  • 1
    $\begingroup$ I added a characterization of the projective objects in the non-negatively graded case. Hence each complex of projective modules that is bounded below but not exact isn't a projective object. A simple example is $\cdots \to 0 \to R \xrightarrow{r} R \to 0 \to \cdots$ where $R$ is any ring and $r$ a non-unit. $\endgroup$ – Ralph Aug 1 '12 at 13:00

Here is a general nonsense fact: if $F \dashv U : \mathcal{A} \to \mathcal{B}$ is an adjunction and $U$ preserves epimorphisms, then $F$ preserves projective objects. Epimorphisms in $\textrm{Ch}(R)$ are precisely the levelwise epimorphisms in $\textrm{Mod}(R)$, so if we take $\mathcal{A} = \textrm{Mod}(R)$, $\mathcal{B} = \textrm{Ch}(R)$, and set $U(M)$ to be $M$ considered as a chain complex concentrated in degree $n$, and $F(K_\bullet) = K_n$, we have an adjunction $F \dashv U$ satisfying the above hypotheses, and therefore $F$ preserves projective objects. Thus,

Proposition. A projective chain complex is necessarily levelwise projective.  ◼

The converse is false, as you anticipated. Consider the following chain complexes: $$P_\bullet = L_\bullet = (R^3 \to R^3) \text{ with differential given by } \begin{pmatrix} 0 & -1 & 1 \newline 1 & 0 & -1 \newline -1 & 1 & 0 \end{pmatrix}$$ $$K_\bullet = (R^3 \to R^4) \text{ with differential given by } \begin{pmatrix} 0 & -1 & 0 \newline 1 & 0 & -1 \newline -1 & 1 & 0 \newline 0 & 0 & 1 \end{pmatrix}$$ Let $f_\bullet : K_\bullet \to L_\bullet$ be the map given in degree $1$ by the identity and in degree $0$ by the matrix $$f_0 = \begin{pmatrix} 1 & 0 & 0 & 1 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 0 \newline \end{pmatrix}$$ It is clear that $f_\bullet$ is an epimorphism. Geometrically, $P_\bullet$ is the simplicial chain complex of a triangle $S^1$, $K_\bullet$ is the simplicial chain complex of the interval $[0, 1]$ subdivided into 3 segments, and $f$ is the quotient map that identifies the endpoints of $[0, 1]$. Intuitively, we expect that there is no morphism $g : S^1 \to [0, 1]$ such that $f \circ g = \textrm{id}$, and this turns out to be true in the chain complex world as well... provided $R$ is not the trivial ring. Indeed, if $f_0 \circ g_0 = \textrm{id}$ and $f_1 \circ g_1 = \textrm{id}$, then by elementary algebra we must have $g_1 = \textrm{id}$ and $$g_0 = \begin{pmatrix} a & b & c \newline 0 & 1 & 0 \newline 0 & 0 & 1 \newline 1 - a & -b & -c \end{pmatrix}$$ for some $a, b, c$, but if $g_\bullet$ is to be a chain map, $a, b, c$ must satisfy $$\left\lbrace\begin{aligned} b & = c \newline a & = 1 + c \newline a & = b \newline b & = c \end{aligned}\right.$$ which implies $0 = 1$. So $P_\bullet$ cannot be projective in $\textrm{Ch}(R)$, even though it is degreewise projective.

  • 1
    $\begingroup$ Hi Zhen. After considering it, I agree that for an adjoint pair, if the right adjoint preserves epimorphisms, then the left adjoint preserves projectives. However, in the example you have given, $U$ is the left adjoint, and $F$ the right one (and not as written), so you must have meant $U\dashv F$. If I'm not mistaken, it follows from this argument that a projective $R$-module $M$ gives a projective object $U(M)$ in $Ch(R)$, and not in the other way around. $\endgroup$ – Shlomi A Aug 1 '12 at 11:34
  • 2
    $\begingroup$ Because $0$ is both initial and terminal, we have both $F \dashv U$ and $U \dashv F$. A highly unusual situation, to be sure! $\endgroup$ – Zhen Lin Aug 1 '12 at 18:13
  • $\begingroup$ If I understand you correctly, you claim that if $F$ is a functor between two categories which have a zero object, then a right adjoint of $F$ is also a left adjoint of it? Why is that so? Could you please formulate it precisely, and indicate how it could be proved? $\endgroup$ – Shlomi A Aug 3 '12 at 8:57
  • 1
    $\begingroup$ No, not at all. I'm remarking that for this functor $F$, the existence of a zero object is precisely what allows us to construct a simultaneous left and right adjoint. Try it and see! $\endgroup$ – Zhen Lin Aug 3 '12 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.