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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!

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  • $\begingroup$ It seems to me that, thinking in terms of the associated long exact sequence, it's rather unusual for one of the maps to be uniquely determined by the other two, even at the level of homotopy groups... Isn't the indeterminacy parameterized by $Map(Y, \Omega Z')$? $\endgroup$
    – Tim Campion
    Sep 18, 2021 at 18:20

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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 $X→Z'$: $$0≃γ∘0≃γ∘g∘f≃g'∘β∘f≃g'∘f'∘δ≃0∘δ≃0.$$

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