A Kan complex is a simplicial set satisfying the horn-filler condition.
What examples are there of simplicial sets which do not satisfy the horn-filler condition? In particular, what simplicial sets are not Kan complexes?
Simplicial sets which are not Kan complexes
algebraic-topologycategory-theorycombinatoricsgroupoidssimplicial-stuff
Related Solutions
Nerves of categories are 2-coskeletal, i.e. right orthogonal to the boundary inclusion $\partial \Delta^n \hookrightarrow \Delta^n$ for all $n > 2$. On the other hand, the singular set of a topological set $X$ is right orthogonal to $\partial \Delta^n \hookrightarrow \Delta^n$ if and only if $X$ itself is right orthogonal to the geometric realisation of $\partial \Delta^n \hookrightarrow \Delta^n$. This almost never happens. The existence of many continuous endomaps of $\left| \Delta^n \right|$ that restrict to the identity on $\left| \partial \Delta^n \right|$ means you can easily manufacture more continuous maps $\left| \Delta^n \right| \to X$ with the same restriction to $\left| \partial \Delta^n \right|$ – unless the map is constant on $\left| \Delta^n \right| \setminus \left| \partial \Delta^n \right|$.
Similarly, the existence of many continuous endomaps of $\left| \Delta^2 \right|$ that restrict to the identity on $\left| \Lambda^2_1 \right|$ makes it difficult for $X$ to be right orthogonal to the geometric realisation of $\Lambda^2_1 \hookrightarrow \Delta^2$, unless every continuous map $\left| \Delta^1 \right| \to X$ is constant. This implies every continuous map $\left| \Delta^n \right| \to X$ is constant, for all $n \ge 0$. Therefore:
Proposition. The singular set of $X$ is (isomorphic to) the nerve of a category if and only if, for every $n \ge 0$, every continuous map $\left| \Delta^n \right| \to X$ is constant.
For example, this happens if $X$ is discrete, or more generally, totally disconnected.
No. This cannot be true, since neither Kan complexes nor quasicategories are closed under limits or colimits in simplicial sets. For limits, this can be simply seen by the fact that not every sub-simplicial set of a Kan complex or of a quasicategory has the same property. For colimits, the quotient identifying the vertices of the nerve of the groupoid representing an isomorphism is not a Kan complex or a quasicategory. The problem with your proposed reflector is that it gives a construction unique only up to weak equivalence, not up to isomorphism.
What is true is that Kan complexes are reflective and coreflective in the $(\infty,1)$-category of quasicategories. This is directly analogous to the situation for categories and for groupoids. Less intuitively, the completion of a simplicial set into a quasicategory or a Kan complex can be seen as the fibrant replacement functor in the relevant model categories.
Best Answer
If you write down some small simplicial set with an easy geometric description (e.g., the simplicial set corresponding to some simplicial complex), it will almost never be a Kan complex. This should be evident if you get some practice working with Kan complexes and using the horn filling condition: it implies your simplicial set must be very "rich" in simplices and that you can "combine" simplices in it similar to the way you can combine singular simplices in a topological space. This is highly incompatible with how simplices behave in a typical nice triangulation of, say, a manifold.
For a really simple example, consider the $1$-simplex $\Delta^1$. Note that there is a horn $\Lambda^2_0\to\Delta^1$ given by mapping the vertices $0,1,2$ to $0,1,0$, respectively ($\Lambda^2_0$ only has the edges $0,1$ and $0,2$, and this map sends them the the edge $0,1$ and the degenerate edge $0,0$). This horn cannot be filled, since there is no edge from $1$ to $0$ in $\Delta^1$.
This example more generally shows that in any Kan complex, edges must be "reversible": if there is an edge $e$ beween two vertices, there is another edge $\bar{e}$ between the same vertices in the opposite order, such that $e$ and $\bar{e}$ together form a 2-simplex where the third edge is degenerate. This property can never hold in a $\Delta$-complex with any edges, since if $e$ was nondegenerate, the 2-simplex will also be nondegenerate, but a nondegenerate simplex in a $\Delta$-complex cannot have a degenerate face. So no non-discrete $\Delta$-complex is a Kan complex. On the other hand, this property is familiar from the context of arbitrary continuous paths in a topological space, which can be reversed.