Are there obstructions to refining partially coherent natural transformations up-to-homotopy between quasicategories?

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?

• This is only possible in general when $R$ is already weakly equivalent to a 1-category. – Kyle Ferendo Jan 30 '17 at 17:49
• @KyleFerendo Thanks! Any pointers as to why? – Kevin Arlin Jan 30 '17 at 17:56
• Notice my $R^{\sim\Delta 1}$ is not the nerve of the homotopy category of $R$-perhaps my terminology makes this unclear. – Kevin Arlin Jan 30 '17 at 20:48
• It is possible that I have misinterpreted you, but I did not misinterpret you in the way you thought I might have. My understanding of your definition of $R^{\sim\Delta_1}$ is that it is the functor category $N(Ho(R))^{\Delta_1}$. There is a natural map $G:R\to N(Ho(R))$, and you define $H$ to be $G^{\Delta_1}$. So $H$ will generally only be a trivial fibration if $G$ is (in which case $R$ is weakly equivalent to its homotopy category). – Kyle Ferendo Jan 31 '17 at 7:09
• Right, I wasn't clear. That's not what I meant. Please see the edit if you're still interested. – Kevin Arlin Jan 31 '17 at 15:00