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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.

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2 Answers

up vote 7 down vote accepted

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

$\qquad$http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible

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.

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Let me add that this complex is a counterexample to the claim in the question's second paragraph, improperly attributed to Dwyer and Spalinski. –  Fernando Muro Jul 31 '12 at 11:37
    
That's a good point. –  Ralph Jul 31 '12 at 21:43
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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 $Ch_R ^\geq0$ ? –  Shlomi A Aug 1 '12 at 10:39
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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. –  Ralph Aug 1 '12 at 13:00
    
Thanks a lot Ralph! A great answer! :-) –  Shlomi A Aug 1 '12 at 15:06
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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.

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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. –  Shlomi A Aug 1 '12 at 11:34
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Because $0$ is both initial and terminal, we have both $F \dashv U$ and $U \dashv F$. A highly unusual situation, to be sure! –  Zhen Lin Aug 1 '12 at 18:13
    
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? –  Shlomi A Aug 3 '12 at 8:57
    
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! –  Zhen Lin Aug 3 '12 at 10:20
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