From ed2efdfca0a5a057a2bdb5623e047bdef0bb7b93 Mon Sep 17 00:00:00 2001 From: Aise Johan de Jong Date: Sat, 13 Apr 2024 15:12:04 -0400 Subject: [PATCH] lower becomes upper THanks to Elias Guisado https://stacks.math.columbia.edu/tag/014R#comment-8370 --- derived.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/derived.tex b/derived.tex index 2e51f6ca..a4e76ea9 100644 --- a/derived.tex +++ b/derived.tex @@ -3248,7 +3248,7 @@ \section{Distinguished triangles in the homotopy category} and $Q_3^\bullet$ for the ``quotient'' complexes. In other words, $Q_1^n = \Ker(\pi_1^n)$, $Q_3^n = \Ker(\pi_3^n)$ and $Q_2^n = \Ker(\pi_2^n)$. Note that the kernels exist. Then -$B^n = A^n \oplus Q_1^n$ and $C_n = B^n \oplus Q_3^n$, where we think of $A^n$ +$B^n = A^n \oplus Q_1^n$ and $C^n = B^n \oplus Q_3^n$, where we think of $A^n$ as a subobject of $B^n$ and so on. This implies $C^n = A^n \oplus Q_1^n \oplus Q_3^n$. Note that $\pi_2^n = \pi_1^n \circ \pi_3^n$ is zero on both $Q_1^n$ and $Q_3^n$. Hence