# The difference between weak Kan complexes and Kan complexes

Let $X$ be a simplicial set and let $\wedge_i^2$ be the $i-th$ horn of the simplicial set $\Delta^2$. The Kan condition is the horn filling condition for $i=0,1,2$ and the weak Kan condition is the horn filling condition for $i=1$. In the setting of the diagram

we let $f$, $g$, $h$ be assignments of 1-simplices in $X$ satisfying the obvious compatibility conditions. The Kan condition for $i=0$ says that $g \circ f$ can be defined.

Question: Taking $h$ to be the null degenerate 'identity' assignment, and using the Kan condition for $i=2$, we get that there is a candidate for an inverse of $g$. Is there any sense in which this implies that there is a unique extension of any horn?(In other words, is the difference between weak and regular Kan complexes the uniqueness of the extension of a horn.)

I am already seeing some difficulties with my characterization in the case of the singular set of a topological space. But I just want to find out if the difference between weak and regular Kan complexes can be summarized in terms of something like this.

If I recall and understand your question correctly, horn fillers are not necessarily unique. However, for (weak) Kan complexes they are unique up to contractible choice.

One way to think about weak Kan complexes and Kan complexes is as follows:

• Kan complexes are $\infty$-groupoids, i.e., $(\infty, 0)$-categories. All morphisms are invertible, but "composition" is not unique. One can also think of them as topological spaces, and there are multiple ways of composing paths.

• A weak Kan complex with unique inner horn fillers is isomorphic to the nerve of some category. Not all morphisms are invertible, but there is a unique composition law.

• In general, weak Kan complexes are $(\infty, 1)$-categories (tautologically, depending on your model). Not all $1$-morphisms are invertible (but all $n$-morphisms, $n > 1$, are) and composition is not unique.

In a sense, weak Kan complexes are a simultaneous generalization of topological spaces and ordinary categories.

• I understand how Kan complexes are generalizations of the groupoid $\pi_{k \geq 0}$ of a topological space. How is it a generalization of a topological space itself? – user062295 Jan 26 '17 at 0:10
• Thank you so much for helping me organize this in my head. Would you be willing to elaborate on why composition is uniquely defined – user062295 Jan 26 '17 at 1:17
• Kan complexes are topological spaces. More precisely, there is a Quillen equivalence between the category of topological spaces and the category of simplicial sets (with the Quillen model structure), and the bifibrant objects in $\mathbf{sSet}$ are precisely the Kan complexes. I don't understand your second comment: by definition in an ordinary category, if $f: X \to Y$ and $g: Y \to Z$ are two morphisms, then their composition $g \circ f: X \to Z$ is uniquely specified. – JHF Jan 26 '17 at 14:51
• Thanks. So I understand that composition is unique up to homotopy in weak kan complexes and kan complexes. What do you mean when you say that there are multiple ways of composing paths in Kan complexes and a unique composition law in weak kan complexes? – user062295 Jan 26 '17 at 15:37
• Consider just the fundamental groupoid for a topological space. If I have two composable paths, there are many ways to "compose" them, e.g., a reparametrization of any composition gives another composition. Also, weak Kan complexes don't have a unique composition law - composition is only unique up to contractible choice. Another way to put this is to say that if $\mathcal{C}$ is a quasicategory, $\Delta$ a simplex and $\Lambda \subset \Delta$ a fillable horn, the morphism $\mathcal{C}^\Delta \to \mathcal{C}^\Lambda$ is an acyclic Kan fibration. – JHF Jan 26 '17 at 17:01