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Let $\mathfrak{T}$ be a triangulated category and $\mathcal{A}$ an abelian category. We say that an additive functor $H: \mathfrak{T} \rightarrow \mathcal{A}$ is a homological functor if, for every distinguished triangle, $$ X \stackrel{u}{\longrightarrow } Y \stackrel{v}{\longrightarrow} Z\stackrel{w}{\longrightarrow} \Sigma X, $$ the sequence, $$ H(X) \stackrel{H(u)}{\longrightarrow} H(Y) \stackrel{H(v)}{\longrightarrow} H(Z) $$ is exact in $\mathcal{A}$. Moreover, we say that a homological functor $H: \mathfrak{T} \rightarrow \mathcal{A}$ is a decent homological functor if the abelian category $\mathcal{A}$ satisfies Grothendieck's AB4* axiom and $H$ respects products.

Finally, we say that a candidate triangle, $$ X \stackrel{u}{\longrightarrow } Y \stackrel{v}{\longrightarrow} Z\stackrel{w}{\longrightarrow} \Sigma X, $$ is a pretriangle if for every decent homological functor, the long sequence, $$ H(\Sigma^{-1}(Z)) \stackrel{H(\Sigma^{-1}(w))}{\longrightarrow} H(X) \stackrel{H(u)}{\longrightarrow} H(Y) \stackrel{H(v)}{\longrightarrow} H(Z) \stackrel{H(w)}{\longrightarrow} H(\Sigma X) $$ is exact.

Why is any direct summand of a pretriangle also a pretriangle?

From what I have read, this is supposed to be immediate, but I can't see why it is true. The obvious approach is to draw up the degree-wise split short exact sequence of long sequences and do a diagram chase, or take the long exact sequence of homology, but I can't actually get to the result this way.

Also to clarify, I'm assuming here that the phrase "direct summand of a pretriangle" means that you have a direct sum of candidate triangles making a pretriangle, and a direct summand is one of the candidate triangles right? Or does it mean that the pretriangle can be made up of a direct sum of literally any diagram of four terms?

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If $T$ is a candidate triangle and $H$ a “decent homological functor”, let $H(T)$ be the long sequence obtained by applying $H$ to $T$. Since the composition of two consecutive maps in $T$ is zero, we can regard $H(T)$ as a chain complex, and $H(T)$ is a long exact sequence exactly when this chain complex is acyclic. So $T$ is a pretriangle iff $H(T)$ is acyclic for all $H$.

If $T=T’\oplus T’’$, then $H(T)=H(T’)\oplus H(T’’)$, which is acyclic if and only if both $H(T’)$ and $H(T’’)$ are acyclic.

So $T$ is a pretriangle iff both $T’$ and $T’’$ are pretriangles.

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