The Stacks project

Lemma 46.8.6. Let $A$ be a ring.

  1. Any $A$-module has a pure projective resolution.

Let $M \to N$ be a map of $A$-modules. Let $P_\bullet \to M$ be a pure projective resolution and let $N_\bullet \to N$ be a universally exact resolution.

  1. There exists a map of complexes $P_\bullet \to N_\bullet $ inducing the given map

    \[ M = \mathop{\mathrm{Coker}}(P_1 \to P_0) \to \mathop{\mathrm{Coker}}(N_1 \to N_0) = N \]
  2. two maps $\alpha , \beta : P_\bullet \to N_\bullet $ inducing the same map $M \to N$ are homotopic.

Proof. Part (1) follows immediately from Lemma 46.8.2. Before we prove (2) and (3) note that by Lemma 46.8.4 we can split the universally exact complex $N_\bullet \to N \to 0$ into universally exact short exact sequences $0 \to K_0 \to N_0 \to N \to 0$ and $0 \to K_ i \to N_ i \to K_{i - 1} \to 0$.

Proof of (2). Because $P_0$ is pure projective we can find a map $P_0 \to N_0$ lifting the map $P_0 \to M \to N$. We obtain an induced map $P_1 \to F_0 \to N_0$ which ends up in $K_0$. Since $P_1$ is pure projective we may lift this to a map $P_1 \to N_1$. This in turn induces a map $P_2 \to P_1 \to N_1$ which maps to zero into $N_0$, i.e., into $K_1$. Hence we may lift to get a map $P_2 \to N_2$. Repeat.

Proof of (3). To show that $\alpha , \beta $ are homotopic it suffices to show the difference $\gamma = \alpha - \beta $ is homotopic to zero. Note that the image of $\gamma _0 : P_0 \to N_0$ is contained in $K_0$. Hence we may lift $\gamma _0$ to a map $h_0 : P_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{P, 1} : P_1 \to N_1$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in $K_1$. Since $P_1$ is pure projective may lift $\gamma _1'$ to a map $h_1 : P_1 \to N_2$. At this point we have $\gamma _1 = h_0 \circ d_{F, 1} + d_{N, 2} \circ h_1$. Repeat. $\square$

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Your email address will not be published.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 0703. Beware of the difference between the letter 'O' and the digit '0'.