explicit formula for contraction of conus of homotopy equivalence

Question

Let $f: K^\bullet \to L^\bullet$ be homotopy equivalence. From theory of triangulated categories it follows that $C(f)^\bullet$ is contractible. But how can i produce explicit formula for homotopy of $Id_{C(f)^\bullet}$ and $0_{C(f)^\bullet}$?

Answer 1

It is not that difficult! Let us write down carefully all the morphisms and identities that we have. We have morphisms of complexes $$f^n\colon K^n \to L^n, \quad f^{n+1} \circ d^n_K = d_L^n\circ f^n$$ and $$g^n\colon L^n \to K^n, \quad g^{n+1} \circ d^n_L = d_K^n\circ g^n,$$ and homotopies $h_K^n\colon K^n\to K^{n-1}$, $h_L^n\colon L^n\to L^{n-1}$ such that $$\mathrm{id}_{K^{n+1}} - g^{n+1}\circ f^{n+1} = d_K^n\circ h_K^{n+1} + h_K^{n+2}\circ d_K^{n+1}$$ and $$\mathrm{id}_{L^n} - f^n\circ g^n = d_L^{n-1}\circ h_L^n + h_L^{n+1}\circ d_L^n.$$ The cone $C (f)$ is formed by objects $C (f)^n \mathrel{\mathop:}= L^n\oplus K^{n+1}$ and differentials $$d^n \mathrel{\mathop:}= \begin{pmatrix} d^n_L & f^{n+1} \\ 0 & -d^{n+1}_K \end{pmatrix}.$$

We want to construct certain homotopy $h^n\colon C (f)^n \to C (f)^{n-1}$. In theory, it is possible to come up directly with a null-homotopy, i.e. find some $h^n$ such that $$d^{n-1} \circ h^n + h^{n+1}\circ d^n \stackrel{???}{=} \mathrm{id}_{C (f)^n} = \begin{pmatrix} \mathrm{id}_{L^n} & 0 \\ 0 & \mathrm{id}_{K^{n+1}} \end{pmatrix},$$ but instead, let us try a rather obvious candidate (which won't be a null-homotopy, but we'll see how to correct it) $$h^n \mathrel{\mathop:}= \begin{pmatrix} h^n_L & 0 \\ g^n & -h_K^{n+1} \end{pmatrix}.$$

Multiplying matrices, we get $$d^{n-1} \circ h^n + h^{n+1}\circ d^n =$$ $$\begin{pmatrix} d^{n-1}_L & f^n \\ 0 & -d^n_K \end{pmatrix} \circ \begin{pmatrix} h^n_L & 0 \\ g^n & -h_K^{n+1} \end{pmatrix} + \begin{pmatrix} h^{n+1}_L & 0 \\ g^{n+1} & -h_K^{n+2} \end{pmatrix} \circ \begin{pmatrix} d^n_L & f^{n+1} \\ 0 & -d^{n+1}_K \end{pmatrix} =$$

$$\begin{pmatrix} d_L^{n-1}\circ h_L^n + f^n\circ g^n & -f^n\circ h_K^{n+1} \\ -d_K^n\circ g^n & d^n_K\circ h_K^{n+1} \end{pmatrix} + \begin{pmatrix} h_L^{n+1}\circ d_L^n & h_L^{n+1}\circ f^{n+1} \\ g^{n+1}\circ d^n_L & g^{n+1}\circ f^{n+1} + h^{n+2}_K\circ d^{n+1}_K \end{pmatrix} =$$

$$\begin{pmatrix} \mathrm{id}_{L^n} & - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \\ 0 & \mathrm{id}_{K^{n+1}} \end{pmatrix}.$$

Sadly, $k^n \mathrel{\mathop:}= - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \ne 0$, so $h^n$ is not a null-homotopy but a homotopy between the zero morphism and some morphism $\phi^n \mathrel{\mathop:}= \begin{pmatrix} \mathrm{id}_{L^n} & k^n \\ 0 & \mathrm{id}_{K^{n+1}} \end{pmatrix}$. However, the latter is visibly an isomorphism of complexes, having $\psi^n \mathrel{\mathop:}= \begin{pmatrix} \mathrm{id}_{L^n} & -k^n \\ 0 & \mathrm{id}_{K^{n+1}} \end{pmatrix}$ as its inverse. Hence the required null-homotopy is given by $h^n\circ \psi^n$: $$d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ \psi^{n+1}\circ d^n = d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ d^n\circ \psi^n = \phi^n\circ \psi^n = \mathrm{id}_{C (f)^n}.$$

By the way, it is also true that if you have a chain contraction for the cone of $f^\bullet$, then $f^\bullet$ is a homotopy (proof: write the chain contraction $2\times 2$ matrix and look at the identity $d^{n-1} \circ h^n + h^{n+1}\circ d^n = \mathrm{id}_{C (f)^n}$).

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And to make you laugh, here is this construction as described in a real GTM textbook (Rosenberg, "Algebraic K-Theory and Its Applications", p. 45):