**EDIT:** It looks like I didn't define the simplicial set $R^{\sim \Delta^1}$ clearly below. It's not simply $\mathrm{Ho}(R)^{\Delta^1}$. Rather, an $n$-simplex of $R^{\sim \Delta^1}$ is a 1-simplex of $\mathrm{Ho}(R^{\Delta^n})$, explicitly, a pair of $n$-simplices $X_0,X_1$ together with a homotopy class $[\alpha]$ of an edge between them. So $R^{\sim \Delta^1}$ is a partially coherent simplicial set of arrows of $R$.**END EDIT**

Let $Q$ be a simplicial set. Suppose I have a natural transformation up-to-homotopy $\alpha$ between two morphisms $f,g:Q\to R$, where $R$ is a quasicategory, that is, $R$ admits fillers for inner horns. By this, I mean that $\alpha$ maps $Q$ to the simplicial set $R^{\sim \Delta^1}$ which has as $n$-simplices the homotopy classes of arrows between $n$-simplices of $R$. Thus $R^{\sim \Delta^1}$ admits a map $H:R^{\Delta^1}\to R^{\sim \Delta^1}$ sending an $n$-simplex of $R^{\Delta^1}$ to its homotopy class when viewed as a morphism of $R^{\Delta^n}$.

**Question: Can $\alpha$ be lifted to $R^{\Delta^1}$?**

It's obviously sufficient, and at least apparently equivalent, to prove that $H$ is a trivial fibration. Are there any efficient techniques relevant here? Given an $(n-1)$-sphere $x$ in $R^{\Delta^1}$ restricted from an $n$-simplex $X$ of $R^{\sim\Delta^1}$, I think I may see an unpleasant argument that we can lift $X$ to $R^{\Delta^1}$ without messing $x$ up, along the following lines: to give such a lift $\tilde X$ is to give the $n+1$ $(n+1)$-simplices of the prism $\Delta^n\times \Delta^1\to R$ respresented by $\tilde X$, which is straightforward up till we get to the last $(n+1)$-simplex, by which time we've already chosen all its faces and can't directly use the horn filling property; but we can fill the last face up to changing one of *its* faces, which should be fine granting a lemma on replacing faces with equivalent ones which I can't really write down a proof for.

Surely there's a better way-or, plausibly, I'm missing something and the result isn't even true! To finish the sketch in the previous paragraph, the following subquestion would suffice:

**Possible lemma towards an answer:** Isn't it true that, given $k$-simplices $A,B$ in a quasicategory $R$ with the same boundaries (i.e. all faces of $A$ and $B$ are literally equal) and a $(k+1)$ simplex $C$ with $A$ as its $i$th face, there's a $(k+1)$-simplex $C$ with $B$ as its $i$th face and the same $i$-horn as $C$? This is easy to show for the first couple of dimensions, but in general the combinatorics of constructing a $(k+2)$-simplex with $C$ and $C'$ as two faces eludes me-I'm not sure which faces are which, and it appears to be important to keep track of which edges are isomorphisms to fill some outer horns. Is there, perhaps, an efficient route to this lemma involving defining some contractible space of solutions $C'$ to the problem?