**Proof.**
It is clear that (2) implies (1). The implication (1) $\Rightarrow $ (2) follows from Lemma 52.6.1. A free resolution of the $\mathcal{O}$-module $\mathcal{O}_ f$ is given by

\[ 0 \to \bigoplus \nolimits _{n \in \mathbf{N}} \mathcal{O} \to \bigoplus \nolimits _{n \in \mathbf{N}} \mathcal{O} \to \mathcal{O}_ f \to 0 \]

where the first map sends a local section $(x_0, x_1, \ldots )$ to $(x_0, x_1 - fx_0, x_2 - fx_1, \ldots )$ and the second map sends $(x_0, x_1, \ldots )$ to $x_0 + x_1/f + x_2/f^2 + \ldots $. Applying $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(-, \mathcal{I}^\bullet )$ where $\mathcal{I}^\bullet $ is a K-injective complex of $\mathcal{O}$-modules representing $K$ we get a short exact sequence of complexes

\[ 0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, \mathcal{I}^\bullet ) \to \prod \mathcal{I}^\bullet \to \prod \mathcal{I}^\bullet \to 0 \]

because $\mathcal{I}^ n$ is an injective $\mathcal{O}$-module. The products are products in $D(\mathcal{O})$, see Injectives, Lemma 19.13.4. This means that the object $T(K, f)$ is a representative of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K)$ in $D(\mathcal{O})$. Thus the equivalence of (1) and (3).
$\square$

## Comments (2)

Comment #6118 by Owen Barrett on

Comment #6200 by Johan on

There are also: