@AndreaMarino In general it is false (e.g. let C'→C be the first face map Δ°→Δ¹ and D→C be ðΔ¹→Δ¹).

:(

Maybe it is the homotopy pullback that gives the correspondence $sSet_{/C} \simeq sSet_{/C'}$

whenever $C,C'$ are equivalent.

Indeed, it seems to me that in order for $\Delta^0 \to \Delta^1$ to be an equivalence, one must invert the only arrow in $\Delta^1$ (otherwise it is not essentially surjective), getting $I$. In that case the homotopy fiber of $1$ in $\partial \Delta^1 \to I$ is equivalent to $\partial \Delta^1$ by the formula $\partial \Delta^1 \times_{I} I^{\simeq}_{/1}$.

If I am not wrong, the homotopy pullback varies up to homotopy equivalence if we vary one factor up to homotopy equivalence, and $C' \to C$ can be susbtituted by $1_C$.

@AndreaMarino What do you mean with homotopy equivalence then?

You certainly have a map Δ¹→Δ° such that the two compositions are Δ¹-homotopic to the identity...

But recall that in $Cat_{\infty}$ map sets are given by $Fun(C,D)^{\simeq}$, so that being homotopic to the identity means via a componentwise equivalence! So $1_{\Delta^1}$ and $const_{0}$ are not homotopic.. Am i wrong?

Well, it was not obvious you were working in Cat_∞. You just said "simplicial sets" :)

So you really meant an equivalence of ∞-categories?

(In any case my counterexample works with Δ¹ replaced by the nerve of the contractible groupoid with two elements)

@DenisNardin you are right! Yes, the counterexample works the same.