On the difference between a projective chain complex and a level-wise projective chain complex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:08:39Z http://mathoverflow.net/feeds/question/103584 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103584/on-the-difference-between-a-projective-chain-complex-and-a-level-wise-projective On the difference between a projective chain complex and a level-wise projective chain complex Shlomi A 2012-07-31T06:02:39Z 2012-08-01T12:52:59Z <p>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 <em>weak equivalence</em> if it induces isomorphisms $H_kM\to H_kN$ on all homologies, and a <em>cofibration</em> if each $f_k:M_k\to N_k$ is a monomorphism, with a projective R-module as its cokernel.</p> <p>Let A be a projective R-<em>module</em>. 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.</p> <p>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) <em>doesn't</em> imply (ii). In the other way around I'm not sure. I would be glad to have an answer in both directions. :-)</p> <p>[DS95] = Dwyer &amp; Spalinski, <em>Homotopy theories and model categories</em>.</p> http://mathoverflow.net/questions/103584/on-the-difference-between-a-projective-chain-complex-and-a-level-wise-projective/103591#103591 Answer by Ralph for On the difference between a projective chain complex and a level-wise projective chain complex Ralph 2012-07-31T08:41:21Z 2012-08-01T12:52:59Z <p><strong>Edit:</strong> I put my answer into a broader perspective. </p> <p>In the following [We] refers to Weibel's "An Introduction to Homological Algebra". </p> <p>Recall that a chain complex $C$ is <em>split exact</em> 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 </p> <p>$\qquad$<a href="http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible/103071#103071" rel="nofollow">http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible</a> </p> <p>we have: </p> <blockquote> <p>For a chain complex $P$ the following are equivalent: </p> <ol> <li>$P$ is a projective object in $Ch$ </li> <li>$P$ is a split exact complex of projectives </li> <li>$P$ is a contractible complex of projectives</li> </ol> </blockquote> <p>Also from the link we obtain the following <strong>examples</strong> where $Ch_R$ denotes the category of unbounded chain complexes of modules over the ring $R$: </p> <ul> <li><p>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.</p></li> <li><p>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 $Ch_R$. </p></li> </ul> <p>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]: </p> <ul> <li>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 $C_n = \mathbb{Z}/4$. </li> </ul> <hr> <p><strong>Added:</strong> Let $Ch_b \subseteq Ch$ be the subcategory of chain complexes that are bounded below. In contrast to $Ch$ the following holds in $Ch_b$: </p> <blockquote> <p>The projective objects in $Ch_b$ are exactly the acyclic chain complexes (bounded below) of projectives. </p> </blockquote> <p>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 $Ch_b$ </p> <p>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. </p> http://mathoverflow.net/questions/103584/on-the-difference-between-a-projective-chain-complex-and-a-level-wise-projective/103593#103593 Answer by Zhen Lin for On the difference between a projective chain complex and a level-wise projective chain complex Zhen Lin 2012-07-31T09:05:16Z 2012-07-31T09:05:16Z <p>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,</p> <p><strong>Proposition.</strong> A projective chain complex is necessarily levelwise projective. 　◼</p> <p>The converse is false, as you anticipated. Consider the following chain complexes: $$P_\bullet = L_\bullet = (R^3 \to R^3) \text{ with differential given by } \begin{pmatrix} 0 &amp; -1 &amp; 1 \newline 1 &amp; 0 &amp; -1 \newline -1 &amp; 1 &amp; 0 \end{pmatrix}$$ $$K_\bullet = (R^3 \to R^4) \text{ with differential given by } \begin{pmatrix} 0 &amp; -1 &amp; 0 \newline 1 &amp; 0 &amp; -1 \newline -1 &amp; 1 &amp; 0 \newline 0 &amp; 0 &amp; 1 \end{pmatrix}$$ 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 $$f_0 = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 1 \newline 0 &amp; 1 &amp; 0 &amp; 0 \newline 0 &amp; 0 &amp; 1 &amp; 0 \newline \end{pmatrix}$$ 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 $$g_0 = \begin{pmatrix} a &amp; b &amp; c \newline 0 &amp; 1 &amp; 0 \newline 0 &amp; 0 &amp; 1 \newline 1 - a &amp; -b &amp; -c \end{pmatrix}$$ 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 &amp; = c \newline a &amp; = 1 + c \newline a &amp; = b \newline b &amp; = c \end{aligned}\right. which implies $0 = 1$. So $P_\bullet$ cannot be projective in $\textrm{Ch}(R)$, even though it is degreewise projective.</p>