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What is the meaning of the following condition on a morphism in sSets or simplicial topological spaces:

a morphism becomes a direct product after the pullback along the shift(decalage) morphism ?

Is it related to being locally trivial ? Is such a morphism necessarily a fibration (say provided the base is fibrant; I understand the decalage morphism is a fibration under this assumption.) ?

Let me explain this in notation. Let $[+1]:\Delta\to \Delta$ be the endomorphism of $\Delta$, $[+1](n):=1+n$ adding a new least element, and for a simplicial set $B$ let $B\circ [+1] \to B$ be the shift(decalage) morphism "forgetting the first coordinate". Let $f:X\to B$ be a morphism in sSets.

What is the meaning of the condition that $B\circ [+1]\times_B X \to B\circ [+1]$ is a direct product, i.e. there is $F$ and an isomorphism $B\circ [+1]\times F \approx B\circ[+1]\times_B X$ over $B\circ [+1]$?

I understand that if $B$ is fibrant, then $B\circ [+1]\to B$ is a fibration, and if $F$ is fibrant, the map $B\circ[+1]\times F\to B\circ [+1]$ is a fibration, and so perhaps $X\to B$ would also be a fibration in this case ?

Any references would be appreciated.

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  • $\begingroup$ On the face of it, this seems like a very strange property for a map $f: X \to B$ to have. What's an example where this happens? Note that $B\circ [+1]$ is weakly contractible. In good cases ( maybe when $B$ is a Kan complex?) $B \circ [+1] \to B$ is a fibration, so your pullback should model the fiber of $f: X \to B$. The condition that your pullback splits as a product is always true up to homotopy (since $B \circ [+1]$ is contractible). $\endgroup$ – Tim Campion yesterday

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