A question about cofiber diagrams in stable $\infty$-categories

Question

My question is as follows say I have a commutative diagram $\require{AMScd}$ \begin{CD} X @>f>> Y @>g>> Z\\ @V \alpha V V @VV \beta V @VV \gamma V\\ X’ @>>f’> Y @>>g’> Z’ \end{CD}

in a stable $\infty$-category, where the two horizontal rows are fiber/cofiber sequences. Suppose I have another morphism $\delta: X \rightarrow X’$ such that the diagram $\require{AMScd}$ \begin{CD} X @>f>> Y\\ @V \delta V V @VV \beta V\\ X’ @>>f’> Y’ \end{CD}

commutes.

$\textbf{Question:}$ What extra information do I need to conclude that $ \delta \simeq \alpha$ up to some contractile space of homotopies?

Presumably I need something like there should also be a commutative diagram of homotopies: $\require{AMScd}$ \begin{CD} \gamma gf @>\simeq>> g’f’\delta\\ @V \simeq V V @VV \simeq V\\ 0 @>>id> 0 \end{CD}

But I am having a little bit trouble wrapping my head around why this doesn’t “obviously” commute.

Thanks!

Answer 1

Working in the stable ∞-category of morphisms, whose objects are morphisms and morphisms are commutative squares with a choice of homotopy, the statement

$δ≃α$ up to some contractile space of homotopies

amounts to the morphism $(f,f')\colon δ→β$ being the kernel of $(g,g')\colon β→γ$, since then we get a canonical (up to a contractible choice) morphism $(h,h')\colon δ→α$, whose $h$ and $h'$ must be homotopic to identity morphisms because kernels in the morphism category are computed objectwise.

Thus, the only data we need is a nullhomotopy for the morphism $(g,g')∘(f,f')$. Since we already have nullhomotopies for $g∘f$ and $g'∘f'$, such a nullhomotopy amounts to a choice of a null-2-homotopy for the composition of the following homotopies between morphisms of the form $X→Z'$: $$0≃γ∘0≃γ∘g∘f≃g'∘β∘f≃g'∘f'∘δ≃0∘δ≃0.$$