Lemma 13.37.4. With assumptions and notation as in Lemma 13.37.3. If $C$ is a compact object and $C \to X_ n$ is a morphism, then there is a factorization $C \to E \to X_ n$ where $E$ is an object of $\langle E_{i_1} \oplus \ldots \oplus E_{i_ t} \rangle $ for some $i_1, \ldots , i_ t \in I$.

**Proof.**
We prove this by induction on $n$. The base case $n = 1$ is clear. If $n > 1$ consider the composition $C \to X_ n \to Y_{n - 1}[1]$. This can be factored through some $E'[1] \to Y_{n - 1}[1]$ where $E'$ is a finite direct sum of shifts of the $E_ i$. Let $I' \subset I$ be the finite set of indices that occur in this direct sum. Thus we obtain

By induction the morphism $C' \to X_{n - 1}$ factors through $E'' \to X_{n - 1}$ with $E''$ an object of $\langle \bigoplus _{i \in I''} E_ i \rangle $ for some finite subset $I'' \subset I$. Choose a distinguished triangle

then $E$ is an object of $\langle \bigoplus _{i \in I' \cup I''} E_ i \rangle $. By construction and the axioms of a triangulated category we can choose morphisms $C \to E$ and a morphism $E \to X_ n$ fitting into morphisms of triangles $(E', C', C) \to (E', E'', E)$ and $(E', E'', E) \to (Y_{n - 1}, X_{n - 1}, X_ n)$. The composition $C \to E \to X_ n$ may not equal the given morphism $C \to X_ n$, but the compositions into $Y_{n - 1}$ are equal. Let $C \to X_{n - 1}$ be a morphism that lifts the difference. By induction assumption we can factor this through a morphism $E''' \to X_{n - 1}$ with $E''$ an object of $\langle \bigoplus _{i \in I'''} E_ i \rangle $ for some finite subset $I' \subset I$. Thus we see that we get a solution on considering $E \oplus E''' \to X_ n$ because $E \oplus E'''$ is an object of $\langle \bigoplus _{i \in I' \cup I'' \cup I'''} E_ i \rangle $. $\square$

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