Totalizing a complex in triagulated category

Question

I am self-studying homotopy theory and trying to understand a proof in this paper on page 218

Let

$$ ... \to X_n \xrightarrow{{f_n}} X_{n-1} \xrightarrow{{f_{n-1}}} ... \xrightarrow{{f_2}} X_1 \to 0$$

be a sequence in trianguated category $\mathcal{I}$.

(sequence or more conventionally a complex means $f_{i+1} \circ f_i=0$).

Complete $X_2 \xrightarrow{{f_2}} X_1$ to a triangle

$X_2 \xrightarrow{{f_2}} X_1 \xrightarrow{{f_y}} Y_1 \xrightarrow{{f_{s2}}} \Sigma X_2 $.

Then the text says " Because the composite $X_3 \xrightarrow{{f_3}} X_2 \xrightarrow{{f_2}} X_1$ is zero, we can lift to $\Sigma X_3 \to Y_1$".

I don't understand why it's possible to make this lift.

Answer 1

The answer is that exact triangles have the property that each morphism is a weak kernel/weak cokernel for the following/previous morphism in the following precise sense.

Lemma In a (pre-)triangulated category, if $$ \newcommand\toby\xrightarrow X\toby{f} Y \toby{g} Z\toby{h} \Sigma X$$ is an exact triangle, and $k:A\to Y$ is such that $gk=0$, then there exists a (not unique!) map $\tilde{k}:A\to X$ such that $k=f\tilde{k}$, and dually, if $l:Y\to A$ is such that $lf=0$, then there exists a map $\tilde{l}:Z\to A$ such that $l=\tilde{l}g$.

Proof.

Apply the morphism axiom to the following diagram $$ \require{AMScd} \begin{CD} A @>1_A>> A @>>> 0 @>>>\Sigma A\\ @. @VkVV @V0VV @. \\ X @>f>> Y @>g>> Z @>h>> \Sigma X\\ \end{CD} $$ $\blacksquare$

Now, how this gets applied to your question. We have $f_2f_3=0$, and we have an exact triangle $$\Sigma^{-1}Y_1 \to X_2 \toby{f_2} X_1\to Y_1,$$ so we can apply the lemma to make the lift.