The Stacks project

Lemma 14.29.1. Let $\mathcal{A}$ be an abelian category. Let $A$ be a chain complex. Consider the covariant functor

\[ B \longmapsto \{ (a, b, h) \mid a, b : A \to B\text{ and }h\text{ a homotopy between }a, b \} \]

There exists a chain complex $\diamond A$ such that $\mathop{\mathrm{Mor}}\nolimits _{\text{Ch}(\mathcal{A})}(\diamond A, -)$ is isomorphic to the displayed functor. The construction $A \mapsto \diamond A$ is functorial.

Proof. We set $\diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$, and we define $d_{\diamond A, n}$ by the matrix

\[ d_{\diamond A, n} = \left( \begin{matrix} d_{A, n} & 0 & \text{id}_{A_{n - 1}} \\ 0 & d_{A, n} & -\text{id}_{A_{n - 1}} \\ 0 & 0 & -d_{A, n - 1} \end{matrix} \right) : A_ n \oplus A_ n \oplus A_{n - 1} \to A_{n - 1} \oplus A_{n - 1} \oplus A_{n - 2} \]

If $\mathcal{A}$ is the category of abelian groups, and $(x, y, z) \in A_ n \oplus A_ n \oplus A_{n - 1}$ then $d_{\diamond A, n}(x, y, z) = (d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z))$. It is easy to verify that $d^2 = 0$. Clearly, there are two maps $\diamond a, \diamond b : A \to \diamond A$ (first summand and second summand), and a map $\diamond A \to A[-1]$ which give a short exact sequence

\[ 0 \to A \oplus A \to \diamond A \to A[-1] \to 0 \]

which is termwise split. Moreover, there is a sequence of maps $\diamond h_ n : A_ n \to \diamond A_{n + 1}$, namely the identity from $A_ n$ to the summand $A_ n$ of $\diamond A_{n + 1}$, such that $\diamond h$ is a homotopy between $\diamond a$ and $\diamond b$.

We conclude that any morphism $f : \diamond A \to B$ gives rise to a triple $(a, b, h)$ by setting $a = f \circ \diamond a$, $b = f \circ \diamond b$ and $h_ n = f_{n + 1} \circ \diamond h_ n$. Conversely, given a triple $(a, b, h)$ we get a morphism $f : \diamond A \to B$ by taking

\[ f_ n = (a_ n, b_ n, h_{n - 1}). \]

To see that this is a morphism of chain complexes you have to do a calculation. We only do this in case $\mathcal{A}$ is the category of abelian groups: Say $(x, y, z) \in \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$. Then

\begin{eqnarray*} f_{n - 1}(d_ n(x, y, z)) & = & f_{n - 1}(d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z)) \\ & = & a_ n(d_ n(x)) + a_ n(z) + b_ n(d_ n(y)) - b_ n(z) - h_{n - 2}(d_{n - 1}(z)) \end{eqnarray*}

and

\begin{eqnarray*} d_ n(f_ n(x, y, z) & = & d_ n(a_ n(x) + b_ n(y) + h_{n - 1}(z)) \\ & = & d_ n(a_ n(x)) + d_ n(b_ n(y)) + d_ n(h_{n - 1}(z)) \end{eqnarray*}

which are the same by definition of a homotopy. $\square$

Comments (0)

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 01A2. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 14.29.1 as pdf