ScalingStacks

[05YV]

Proof. To show that f:A→Bf\colon A\to B is nn-connected, we need to show that the space of lifts for every square

A\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}X\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}B\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}Y,\textstyle{Y,}

in which the right vertical arrow is nn-truncated, is contractible. Applying 4.3.3 and 4.1.3, we see that this space is equivalent to the space of lifts in the square

A\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}τ≤n+1𝒞​X\textstyle{\tau_{\leq n+1}^{\mathcal{C}}X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}B\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}τ≤n+1𝒞​Y,\textstyle{\tau_{\leq n+1}^{\mathcal{C}}Y,}

which, by 4.1.4, is equivalent to the space of lifts in the adjoint square

τ≤n+1𝒞​A\textstyle{\tau_{\leq n+1}^{\mathcal{C}}A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}τ≤n+1𝒞​X\textstyle{\tau_{\leq n+1}^{\mathcal{C}}X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}τ≤n+1𝒞​B\textstyle{\tau_{\leq n+1}^{\mathcal{C}}B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}τ≤n+1𝒞​Y,\textstyle{\tau_{\leq n+1}^{\mathcal{C}}Y,}

which is contractible since the left vertical arrow is an equivalence. ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Tomer Schlank, Lior Yanovski

Original source: arXiv:1808.06006v3

    Original source page 32

    Original source · 1808.06006v3