# when is the pullback along the shift (decalage) morphism a direct product, in sSets

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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• 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). – Tim Campion yesterday