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Let $\text{Ch}⁺(R)$ be the category of non-negative chain complexes of $R$-modules where $R$ is a commutative ring. What is a cylinder object, in the sense of model categories, for a given complex $M^\bullet$? Thinking about algebraic topology (a circle and an annulus) I was thinking of tensoring $M^\bullet$ by $R$ (over $R$!) but this is quite trivial and surely I am making a mistake. Could somebody help me?

Thank you!

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@Zhen Lin: Yes, you are right, I was thinking with simplicial homology too but made a mistake, sorry. Therefore tensoring over $R$ with the chain complex $\dots\rightarrow R\rightarrow R\rightarrow 0$ should be a good candidate for the cylinder object. – bc87 Mar 10 '13 at 21:45
Oops, that's a circle... – Zhen Lin Mar 10 '13 at 22:36
up vote 5 down vote accepted

Let $M$ be a chain complex of $R$-modules, concentrated in degrees $\ge 0$. Define a chain complex $\textrm{Cyl}(M)$ as follows: $$\textrm{Cyl}(M)_n = M_n \oplus M_{n-1} \oplus M_n$$ $$\partial(a, b, c) = (\partial a + b, - \partial b, \partial c - b)$$ There are evident chain maps $i_0, i_1 : M \to \textrm{Cyl}(M)$ and $p : \textrm{Cyl}(M) \to M$: \begin{align} i_0 (m) & = (m, 0, 0) \\ i_1 (m) & = (0, 0, m) \\ p (a, b, c) & = a + c \end{align} Clearly, $p \circ i_0 = p \circ i_1 = \textrm{id}_M$. Moreover, there is a chain homotopy from $\textrm{id}_{\textrm{Cyl}(M)}$ to $i_0 \circ p$; indeed, define $h_n : \textrm{Cyl}(M)_n \to \textrm{Cyl}(M)_{n+1}$ by $$h_n(a, b, c) = (0, -c, 0)$$ and then $\textrm{id}_{\textrm{Cyl}(M)} - i_0 \circ p = \partial \circ h + h \circ \partial$. Hence, $i_0$ is a quasi-isomorphism and $p$ is an acyclic fibration.

We should show that $i : M \oplus M \to \textrm{Cyl}(M)$, defined by $i(a, c) = (a, 0, c)$, is a cofibration. Unfortunately, this happens if and only if $M$ is degreewise projective: indeed, $\operatorname{coker} i$ is manifestly (isomorphic to) $M[-1]$. In general, to find a cylinder object for $M$, one should look at the (cofibration, acyclic fibration) factorisation of the fold map $M \oplus M \to M$.

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Thank you very much! – bc87 Mar 11 '13 at 19:49

This is essentially the same answer as Zhen Lin, but I will offer a different point of view. We may define an interval object ,$\mathcal{I}$ in the category of chain complexes over $R$. We will define the zero degree R- module as $R[x,y]$ and the degree one R-moldule as $R[I]$, and all other degrees will be zero. The boundary map will be $$ \partial (I)=x-y$$. We now define $cyl(C_*)=C_*\otimes \mathcal{I}$, where the tensor product is that of chain complexes. Note that in general, we do not need all of the structure of a model category to define the notion of a cylinder functor. See . I will try to track down a source forg this description.

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Just in case anyone is wondering: The tensor product complex $\mathcal{I} \otimes C_*$ coincides with the complex given in Zhen Lin's answer (with the same signs). The tensor product $C_* \otimes \mathcal{I}$ coincides with his complex up to signs. – Ingo Blechschmidt Feb 7 at 15:57

You can see the cylinder $Cyl(f)$ defined as in Zhen's answer as the object in your category s.t. the following short sequence $0\rightarrow X \rightarrow Cyl(f) \rightarrow C(f) \rightarrow 0$ is exact, for any morphism $f: X\rightarrow Y$ and denoting by $C(f)$ the cone of $f$.

Alternatively you can see the above cylinder as the cone of the induced morphism $C(f)[-1]\rightarrow X$. The tricky part in dealing with cylinders and cones is that the 'innocent looking" short exact sequence above is not split, but rather semisplit in the sense of Gelfand Manin's "Methods of Homological Algebra' textbook.

In other words, the natural splittings $r: Cyl(f)\rightarrow X$ and $s:C(f)\rightarrow Cyl(f)$ are not morphisms in your category of complexes (they do not commute with the differentials, due to the twisted differentials on the cylinder and cone).

This fact is strongly used in defining the triangulated structure on homotopy categories and derived categories.

For a related discussion, please check Split-Lemma for chain complexes

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