Are hammock localizations locally truncated?

Question

Let us take a relative category $(\mathcal{C},\mathcal{W})$, and consider its hammock localization $L_H \mathcal{C}$. It seems to me that for every two objects $X,Y \in \mathcal{C}$ the mapping simplicial set $L_H \mathcal{C}(X,Y)$ has the (strict) right lifting property against all inclusions $\partial \Delta^n \hookrightarrow \Delta^n$ for $n \geq 3$. The reason is the following: let us consider the case $n=3$ for simplicity, the general case is analogous. A diagram of shape $\partial \Delta^n$ means a collection of $n+1$ hammocks of width $n-1$ such that some selected $(n-2)$-faces of these correspond. Therefore, up to unreducing these hammocks, they can all be assumed to be of the same length, so let us consider only one column at a time. For each $i=0,1,2,3$ we have four columns of the form

$$\require{AMScd} \begin{CD} \bullet @>f_1^0>> \bullet\\ @Vv_{12}^0VV @VVw_{12}^0V\\ \bullet @>f_2^0>> \bullet\\ @Vv_{23}^0VV @VVw_{23}^0V\\ \bullet @>f_3^0>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0^1>> \bullet\\ @Vv_{02}^1VV @VVw_{02}^1V\\ \bullet @>f_2^1>> \bullet\\ @Vv_{23}^1VV @VVw_{23}^1V\\ \bullet @>f_3^1>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0^2>> \bullet\\ @Vv_{01}^2VV @VVw_{01}^2V\\ \bullet @>f_1^2>> \bullet\\ @Vv_{13}^2VV @VVw_{13}^2V\\ \bullet @>f_3^2>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0^3>> \bullet\\ @Vv_{01}^3VV @VVw_{01}^3V\\ \bullet @>f_1^3>> \bullet\\ @Vv_{12}^3VV @VVw_{12}^3V\\ \bullet @>f_2^3>> \bullet \end{CD} $$

I'm omitting names of objects, a superscript index $i$ just refers to the $i$-th labeled 2-face, a subscript index $i$ refers to the $i$-th row, and a subscript index $ij$ refers to the arrow going from the $i$-th row to the $j$-row. Now let us look at what it means for these wannabe $2$-faces to be assembled into a diagram of shape $\partial \Delta^3$.
For instance, the 0-th face of the 0-th diagram and the 0-th face of the 1st diagram should coincide, and similarly for the other compatibility condtions. This allows us to remove the superscript indices from all $f$'s and from all $v_{ij}$'s and $w_{ij}$'s whenever $i$ and $j$ are consecutive.
Moreover, the 1st face of the 0-th diagram and the 0-th face of the 2nd diagram should coincide, and similarly for others. This means that $v_{13}^2 = v_{23} \circ v_{12}$ and analogously for $w_{13}^2$. This means that we can also remove the subscript indices $ij$ when they are not consecutive and write the corresponding $v$'s and $w$'s in terms of those with consecutive indices. In other words, the above columns may be rewritten as

$$\require{AMScd} \begin{CD} \bullet @>f_1>> \bullet\\ @Vv_{12}VV @VVw_{12}V\\ \bullet @>f_2>> \bullet\\ @Vv_{23}VV @VVw_{23}V\\ \bullet @>f_3>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0>> \bullet\\ @Vv_{12}\circ v_{01}VV @VVw_{12} \circ w_{01}V\\ \bullet @>f_2>> \bullet\\ @Vv_{23}VV @VVw_{23}V\\ \bullet @>f_3>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0>> \bullet\\ @Vv_{01}VV @VVw_{01}V\\ \bullet @>f_1>> \bullet\\ @Vv_{23} \circ v_{12}VV @VVw_{23} \circ w_{12}V\\ \bullet @>f_3>> \bullet \end{CD} \hspace{30pt} \begin{CD} \bullet @>f_0>> \bullet\\ @Vv_{01}VV @VVw_{01}V\\ \bullet @>f_1>> \bullet\\ @Vv_{12}VV @VVw_{12}V\\ \bullet @>f_2>> \bullet \end{CD} $$

Now consider the column

$$\require{AMScd} \begin{CD} \bullet @>f_0>> \bullet\\ @Vv_{01}VV @VVw_{01}V\\ \bullet @>f_1>> \bullet\\ @Vv_{12}VV @VVw_{12}V\\ \bullet @>f_2>> \bullet\\ @Vv_{23}VV @VVw_{23}V\\ \bullet @>f_3>> \bullet \end{CD} $$

whose faces are clearly the four columns above. Reasoning like this for each column, we have constructed a diagram of shape $\Delta^3$ whose restriction to $\partial \Delta^3$ is what we started with.

There should be a mistake in the above argument, in that we know that there are relative categories with arbitrary mapping simplicial sets, but I fail to find it. Could someone please point it out?

Answer 1

Your calculation is correct. For every two objects $X, Y \in \mathcal{C}$, the hom space $L^H\mathcal{C}(X,Y)$ has the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$.

First note that the nerve of any category has the strict right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$. The space $L^H\mathcal{C}(X,Y)$ is not the nerve of a category, but it is a quotient of the nerve a category (See here). Call this nerve $N\mathcal{D}(X,Y)$. Your argument can be summarized by saying that any map $\partial \Delta^n \to L^H\mathcal{C}(X,Y)$ can be factored through $N\mathcal{D}(X,Y)$, where we can solve the lifting problem and then project back into $L^H\mathcal{C}(X,Y)$. (This is the "unreducing" part of your argument).

Note that what I just said does not yet prove that the lift to $L^H\mathcal{C}(X,Y)$ is unique. This would require a further argument about what happens when you have two different lifts $\partial \Delta^n \to N\mathcal{D}(X,Y)$. This might be possible - I am not 100% sure though.

In any case, the simplicial set $L^H\mathcal{C}(X,Y)$ is not a Kan complex (usually). So, as pointed out in the comments by Zhen Lin, having the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$ does not tell us about the higher homotopy groups of $L^H\mathcal{C}(X,Y)$. Indeed, as you correctly observe, there are relative categories which allow you to realize $L^H\mathcal{C}(X,Y)$ as any homotopy type you like.