**Proof.**
Assume $E$ is compact. By Lemma 22.13.4 we may assume that $E$ is represented by a differential graded $A$-module $P$ with property (P). Consider the distinguished triangle

\[ \bigoplus F_ iP \to \bigoplus F_ iP \to P \xrightarrow {\delta } \bigoplus F_ iP[1] \]

coming from the admissible short exact sequence of Lemma 22.13.1. Since $E$ is compact we have $\delta = \sum _{i = 1, \ldots , n} \delta _ i$ for some $\delta _ i : P \to F_ iP[1]$. Since the composition of $\delta $ with the map $\bigoplus F_ iP[1] \to \bigoplus F_ iP[1]$ is zero (Derived Categories, Lemma 13.4.1) it follows that $\delta = 0$ (follows as $\bigoplus F_ iP \to \bigoplus F_ iP$ maps the summand $F_ iP$ via the difference of $\text{id}$ and the inclusion map into $F_{i - 1}P$). Thus we see that the identity on $E$ factors through $\bigoplus F_ iP$ in $D(A, \text{d})$ (by Derived Categories, Lemma 13.4.10). Next, we use that $P$ is compact again to see that the map $E \to \bigoplus F_ iP$ factors through $\bigoplus _{i = 1, \ldots , n} F_ iP$ for some $n$. In other words, the identity on $E$ factors through $\bigoplus _{i = 1, \ldots , n} F_ iP$. By Lemma 22.27.3 we see that the identity of $E$ factors as $E \to P \to E$ where $P$ is as in part (2) of the statement of the lemma. In other words, we have proven that (1) implies (2).

Assume (2). By Derived Categories, Lemma 13.34.2 it suffices to show that $P$ gives a compact object. Observe that $P$ has property (P), hence we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(P, M) = \mathop{\mathrm{Hom}}\nolimits _{K(A, \text{d})}(P, M) \]

for any differential graded module $M$ by Lemma 22.15.3. As direct sums in $D(A, \text{d})$ are given by direct sums of graded modules (Lemma 22.15.4) we reduce to showing that $\mathop{\mathrm{Hom}}\nolimits _{K(A, \text{d})}(P, M)$ commutes with direct sums. Using that $K(A, \text{d})$ is a triangulated category, that $\mathop{\mathrm{Hom}}\nolimits $ is a cohomological functor in the first variable, and the filtration on $P$, we reduce to the case that $P$ is a finite direct sum of shifts of $A$. Thus we reduce to the case $P = A[k]$ which is clear.
$\square$

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