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have added to regular epimorphism the statement (here) that a pullback square of regular epis is also a pushout.
This must be true for effective epimorphisms in an $\infty$-topos, too. What’s the proof?
If I’m not mistaken, the $\infty$-analogue is false. For instance,
$\begin{array}{ccc} 4 & \rightarrow & 2 \\ \downarrow & & \downarrow \\ 2 & \rightarrow & 1 \end{array}$is a (homotopy) pullback square of effective epimorphisms (of sets) in $\infty Grpd$ but not a homotopy pushout square. (The homotopy pushout is a circle.)
That’s true of course, stupid me. Thanks.
I’ll need to consider some further conditions.
So concretely, I am looking at the following situation: $X$ a $V$-manifold (as here), $T_{inf}X$ its infinitesimal disk bundle (as here), $U \to X$ a formally étale 1-epi, and the pullback in question being the one that exhibits a local trivialization of the formal disk bundle
$\array{ && U \times \mathbb{D} \\ & \swarrow && \searrow \\ T_{inf}X && && U \\ & \searrow && \swarrow \\ && X }$So $T_{inf}X \to X$ as well as $U \times \mathbb{D} \to U$ are $\Re$-equivalences ($\Re$ the reduction modality). This means that applying $\Re$ to this diagram it does become a homotopy pushout. Since $\Re$ preserves homotopy pushouts, this means that the homotopy pushout $T_{inf} X \underset{U \times \mathbb{D}}{\coprod} U$ differs from $X$ at most in some infinitesimal extension.
So for achieving the desired $X \simeq T_{inf} X \underset{U\times \mathbb{D}}{\coprod} U$ I am reduced to arguing (or else to try to arrange by further assumptions) that also this infinitesimal difference vanishes. This should follow since $U \times \mathbb{D}\to T_{inf}X$ is something like a fiberwise equivalence. Hm…
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In more detail, what I am after is described around def. 4.7 of this note on parameterized WZW terms (pdf). My real goal is to get a useful condition that makes the necessary obstruction of theorem 4.5 also a sufficient obstruction, as in corollary 4.11. Possibly there is an altogether better way to do this than I have there right now.
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