2
$\begingroup$

I am reading a proof in a paper of Neeman's, Some New Axioms for Triangulated Categories. It can be found here. Let $\mathfrak{T}$ be a triangulated category. There he defines a subcategory $CT(\mathfrak{T})$ of so-called candidate triangles, consisting of sequences, $$ X \stackrel{u}{\longrightarrow} Y \stackrel{v}{\longrightarrow} Z \stackrel{w}{\longrightarrow} \Sigma X $$ with $v \circ u$, $w \circ v$, and $\Sigma u \circ w$ all being the zero morphism. Morphisms in this subcategory are defined in exactly the way that morphisms of triangles are defined: As $\Sigma$-periodic morphisms making the resulting diagram commute.

A homotopy of two such morphisms is defined in the natural way. That is, they respect this $\Sigma$-periodicity.

Finally, he define a full subcategory of $CT(\mathfrak{T})$, denoted by $T(\mathfrak{T})$ consisting of so-called true triangles, which have as objects candidate triangles which are also distinguished triangles provided by the triangulated structure on $\mathfrak{T}$.

The following is Lemma 1.4 in the paper linked above.

Let $X^{\bullet}$ and $Y^{\bullet}$ be objects in $T(\mathfrak{T})$. Then the cone on the zero map $C(X^{\bullet} \stackrel{0}{\rightarrow} Y^{\bullet})$ is in $T(\mathfrak{T})$.

To prove this, he considers two objects in $T(\mathfrak{T})$, $$ X \stackrel{u}{\longrightarrow} Y \stackrel{v}{\longrightarrow} Z \stackrel{w}{\longrightarrow} \Sigma X $$ and $$ X' \stackrel{u'}{\longrightarrow} Y' \stackrel{v'}{\longrightarrow} Z' \stackrel{w'}{\longrightarrow} \Sigma X'. $$ He then claims that we need to show that the diagram $$ X' \oplus X' \longrightarrow Y' \oplus Y' \longrightarrow Z' \oplus Z' \longrightarrow \Sigma X \oplus \Sigma X' $$ is a triangle.

I don't see how this proves the claim at all. What does the above sequence have anything to do with the mapping cone in question? The cone is the diagram $$ X' \oplus Y \longrightarrow Y' \oplus Z \longrightarrow Z' \oplus \Sigma X \longrightarrow \Sigma X \oplus \Sigma Y. $$ Can anyone explain this proof to me?

$\endgroup$
2
$\begingroup$

Recall that, in Neeman's notation, $\tilde{\Sigma}$ is the operation on candidate triangles that takes $$X\stackrel{u}{\longrightarrow}Y\stackrel{v}{\longrightarrow}Z\stackrel{w}{\longrightarrow}\Sigma X$$ to $$Y\stackrel{-v}{\longrightarrow}Z\stackrel{-w}{\longrightarrow}\Sigma X\stackrel{-\Sigma u}{\longrightarrow}\Sigma Y,$$ and that if $X^\bullet$ is a triangle, then $\tilde{\Sigma}X^\bullet$ is also a triangle.

The cone of the zero map $X^\bullet\to Y^\bullet$ between two triangles is precisely $\tilde{\Sigma}X^\bullet\oplus Y^\bullet$, so to prove that the cone of a zero map between triangles is a triangle, it suffices to prove that the direct sum of two triangles is a triangle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.