Is there a higher analog of "category with all same side inverses is a groupoid"?

Question

There is a (most likely folklore) theorem - if in a category every morphism has a right inverse then that category is a groupoid. The proof is an honest oneliner: for $x:A\to B$ find $x':B\to A$ with $xx'=\operatorname{id}_B$; now for $x'$ find $x'':A\to B$ with $x'x''=\operatorname{id}_A$. Then $x''=xx'x''=x$, so $x'x=\operatorname{id}_A$ too.

I wonder if this admits a generalization showing that only part of the horn filling conditions for Kan complexes suffice. Let me recall that a simplicial set $X_\bullet$ is a Kan complex iff any morphism $\Lambda_i[n]\to X_\bullet$ extends along the inclusion $\Lambda_i[n]\hookrightarrow\Delta[n]$, for all $0\leqslant i\leqslant n$. Here $\Lambda_i[n]$ is the boundary of the generic $n$-simplex $\Delta[n]$ with the $i$th facet removed.

So for a starter, let me ask this. Suppose I require on $X_\bullet$ all those extension conditions except for $\Lambda_2[2]$. Is $X_\bullet$ still a Kan complex?

Looks like this should be true but I cannot finish the argument even for the particular case of $\Lambda_2[2]$ when $d_1$ is degenerate. Thus for a $1$-simplex $x$ with $d_0(x)=A$, $d_1(x)=B$ we have to find (using all $\Lambda_i[n]$ except for $\Lambda_2[2]$) a $2$-simplex $\tau$ with $d_0(\tau)=x$ and $d_1(\tau)=s_0(A)$.

I begin imitating the groupoid statement. Using $\Lambda_0[2]$, I find a $2$-simplex $\sigma$ with $d_1(\sigma)=s_0(B)$ and $d_2(\sigma)=x$; let $x'=d_0(\sigma)$. Again using $\Lambda_0[2]$, find another $2$-simplex $\rho$ with $d_1(\rho)=s_0(A)$ and $d_2(\rho)=x'$. Let $x''=d_0(\rho)$. Now using $\Lambda_1[2]$, find $\pi$ with $d_0(\pi)=x$ and $d_2(\pi)=x'$, and let $i=d_1(\pi)$. The problem is that there is no reason for $i$ to be $s_0(A)$. We certainly can use $\Lambda_0[2]$ again to find $\iota$ with $d_2(\iota)=i$, $d_1(\iota)=s_0(A)$, but so what?

I feel like there is lots of reserve unused as I have all the higher Kan conditions available but still cannot figure out how to do it.

And in fact, I would like to omit each of the $\Lambda_n[n]$ if possible...

Answer 1

Yes, this is possible. The following is a classical result of the theory of quasi-categories (You'll find it in the early part of Lurie's Higher topos theory or in Joyal notes on quasi-categories - where they discuss equivalences in quasi-categories):

Proposition: A quasi-category $X$ is a Kan complex if and only its homotopy category is a groupoid.

So if you have the lifting property against the $\Lambda_i[n] \to \Delta[n]$ for $0<i<n$ and the lifting property against $\Lambda_0[2] \to \Delta[2]$ then you can show that every arrow in the homotopy category has a one sided inverse using this last lifting property, and hence the homotopy category is a groupoid, which shows that $X$ is a Kan complex by the proposition.