The Stacks project

Lemma 13.9.14. Let $\mathcal{A}$ be an additive category.

  1. Given a termwise split sequence of complexes $(\alpha : A^\bullet \to B^\bullet , \beta : B^\bullet \to C^\bullet , s^ n, \pi ^ n)$ there exists a homotopy equivalence $C(\alpha )^\bullet \to C^\bullet $ such that the diagram

    \[ \xymatrix{ A^\bullet \ar[r] \ar[d] & B^\bullet \ar[d] \ar[r] & C(\alpha )^\bullet \ar[r]_{-p} \ar[d] & A^\bullet [1] \ar[d] \\ A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r]^\delta & A^\bullet [1] } \]

    defines an isomorphism of triangles in $K(\mathcal{A})$.

  2. Given a morphism of complexes $f : K^\bullet \to L^\bullet $ there exists an isomorphism of triangles

    \[ \xymatrix{ K^\bullet \ar[r] \ar[d] & \tilde L^\bullet \ar[d] \ar[r] & M^\bullet \ar[r]_{\delta } \ar[d] & K^\bullet [1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C(f)^\bullet \ar[r]^{-p} & K^\bullet [1] } \]

    where the upper triangle is the triangle associated to a termwise split exact sequence $K^\bullet \to \tilde L^\bullet \to M^\bullet $.

Proof. Proof of (1). We have $C(\alpha )^ n = B^ n \oplus A^{n + 1}$ and we simply define $C(\alpha )^ n \to C^ n$ via the projection onto $B^ n$ followed by $\beta ^ n$. This defines a morphism of complexes because the compositions $A^{n + 1} \to B^{n + 1} \to C^{n + 1}$ are zero. To get a homotopy inverse we take $C^\bullet \to C(\alpha )^\bullet $ given by $(s^ n , -\delta ^ n)$ in degree $n$. This is a morphism of complexes because the morphism $\delta ^ n$ can be characterized as the unique morphism $C^ n \to A^{n + 1}$ such that $d \circ s^ n - s^{n + 1} \circ d = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.14.10. The composition $C^\bullet \to C(\alpha )^\bullet \to C^\bullet $ is the identity. The composition $C(\alpha )^\bullet \to C^\bullet \to C(\alpha )^\bullet $ is equal to the morphism

\[ \left( \begin{matrix} s^ n \circ \beta ^ n & 0 \\ -\delta ^ n \circ \beta ^ n & 0 \end{matrix} \right) \]

To see that this is homotopic to the identity map use the homotopy $h^ n : C(\alpha )^ n \to C(\alpha )^{n - 1}$ given by the matrix

\[ \left( \begin{matrix} 0 & 0 \\ \pi ^ n & 0 \end{matrix} \right) : C(\alpha )^ n = B^ n \oplus A^{n + 1} \to B^{n - 1} \oplus A^ n = C(\alpha )^{n - 1} \]

It is trivial to verify that

\[ \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) - \left( \begin{matrix} s^ n \\ -\delta ^ n \end{matrix} \right) \left( \begin{matrix} \beta ^ n & 0 \end{matrix} \right) = \left( \begin{matrix} d & \alpha ^ n \\ 0 & -d \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ \pi ^ n & 0 \end{matrix} \right) + \left( \begin{matrix} 0 & 0 \\ \pi ^{n + 1} & 0 \end{matrix} \right) \left( \begin{matrix} d & \alpha ^{n + 1} \\ 0 & -d \end{matrix} \right) \]

To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha )^\bullet \to A^\bullet [1]$ (see Definition 13.9.1) and $C(\alpha )^\bullet \to C^\bullet \to A^\bullet [1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : C^\bullet \to C(\alpha )^\bullet $ and check instead that the two maps $C^\bullet \to A^\bullet [1]$ agree. And note that $p \circ (s, -\delta ) = -\delta $ as desired.

Proof of (2). We let $\tilde f : K^\bullet \to \tilde L^\bullet $, $s : L^\bullet \to \tilde L^\bullet $ and $\pi : \tilde L^\bullet \to L^\bullet $ be as in Lemma 13.9.6. By Lemmas 13.9.2 and 13.9.13 the triangles $(K^\bullet , L^\bullet , C(f), i, p)$ and $(K^\bullet , \tilde L^\bullet , C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L^\bullet $ by $\tilde L^\bullet $ and $f$ by $\tilde f$. In other words we may assume that $f$ is a termwise split injection. In this case the result follows from part (1). $\square$

Comments (5)

Comment #296 by arp on


  1. In the second line of the proof, I think it should say "...because the compositions are zero."

  2. In "To see that this is homotopic to the identity map use the homotopy " there's a hanging parenthesis.

  3. In equation following "it is trivial to verify", I think the second term of the LHS should be And on the RHS, the first occurring should be not .

Comment #3049 by Dan Dore on

In the last line of the first paragraph of the proof, should be in a few places.

Comment #3553 by Nil on

In the first line of the last paragraph, should be .

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 014L. Beware of the difference between the letter 'O' and the digit '0'.