Filling condition for quasi-categories I am trying to get a better understanding of how the inner horn filling condition in higher cells corresponds to higher associativity laws. For instance, I am trying to understand how the difference of the inner horn filling condition and outer horn filling condition makes the difference of invertibility of cells.
Let $\Delta^n \amalg_{\Delta^0} \Delta^m$ be the gluing of $\Delta^n$ and $\Delta^m$ along the head of one $\Delta$ and the tail of another; the map $\Delta^0 \rightarrow \Delta^n$ sends $0$ to $n$, and the map $\Delta^0 \rightarrow \Delta^m$ sends $0$ to $0$. There is an injection $\Delta^n \amalg_{\Delta^0} \Delta^m \rightarrow \Delta^{n+ m}$.
Consider the following alternative filling condition:
$$ \text{ All maps } \Delta^n \amalg_{\Delta^0} \Delta^m \rightarrow X \text{ lift along } \Delta^n \amalg_{\Delta^0} \Delta^m  \rightarrow \Delta^{n+m}$$
Is this filling condition equivalent to the inner horn filling condition?
Other insights on the matter of "why inner horns?" and filling conditions for $n \geq 3$ as higher associativities are much appreciated.
 A: There's a lot to stay here. There's a nice blog post by Emily Riehl which explores how to think about higher associativity in quasicategories.

To see why you don't want to fill outer horns, it suffices to think about what it would mean to fill an outer 2-horn. For instance, to fill $\Lambda^0[2] \to \Delta[2]$ is to say that if you have 1-morphisms $h: x \to z$ and $g: y \to z$, then there will be a 1-morphism $f: x \to y$, and a 2-simplex which is a homotopy that will say $h = g \circ f$. This doesn't hold in most 1-categories, so you wouldn't expect it to hold in $(\infty,1)$-categories. But it does hold in 1-groupoids, where we can take $f = h^{-1} \circ g$. Thinking along these lines leads to the idea that an $\infty$-groupoid, or space, will have fillers along all outer horns as well as inner horns.

Related to your ideas about filling $\Delta[m] \cup_{\Delta[0]} \Delta[n] \to \Delta[m+n]$, Joyal and Tierney showed that the spine inclusions $\Delta[1] \cup_{\Delta[0]} \dots \cup \Delta[1] \to \Delta[n]$ "generate" the lifting conditions enjoyed by quasicategories, but only in a weaker sense (see Lemma 3.5 there) -- It's not the case that lifting against spine inclusions implies that you're a quasicategory. Similarly, the lifting conditions you describe don't suffice to ensure one has a quasicategory. For example, $X = \Delta[2] \cup_{\Lambda^1[2]} \Delta[2]$ lifts against all spine inclusions, and all of your inclusions, but it's not a quasicategory -- the composable pair in the spine has two non-homotopic composites.
I believe that Dimitri Ara used Joyal and Tierney's observation to show that the Joyal model structure is the minimal model structure on simplicial sets with cofibrations the monomorphisms such that the spine inclusions are weak equivalences-- in the terminology of the Grothendieck school, the quasicategorical equivalences are generated as a localizer by the spine inclusions.

For my money, thinking about these lifting conditions is most intuitive in the setting of Segal spaces rather than quasicategories. A Segal space is a simplicial space $\mathcal C: [n] \mapsto \cup_{X_0,\dots, X_n \in Ob \mathcal C} \mathcal C(x_0,\dots ,x_n)$ such that we have canonical equivalences
$$(\ast) \qquad \mathcal C(x_0,\dots, x_n) \to \mathcal C_(x_0,x_1) \times \cdots \times \mathcal C(x_{n-1},x_n)$$
corresponding to the spine inclusions. We can think of $\mathcal C(x_0,\dots, x_n)$ as the space of chains of morphisms $x_0 \to \dots \to x_n$ together with all the data required to compose them and all subchains; the equivalence $(\ast)$ tells us that up to coherent homotopy, this is no more data than just knowing the chain of morphisms $x_0 \to \cdots \to x_n$.
Incidentally, you can think of this in the terms you mention, by factoring the map $(\ast)$ through $\mathcal C(x_0,\dots, x_k) \times \mathcal C(x_k, \dots, x_n)$; the second map in this factorization must be an equivalence and the composite is also, so the first map is an equivalence. Similarly, inner horns all induce equivalences in this way.
I tend to think of the lifing conditions for quasicategories as being a "shadow" of the equivalence conditions for (complete) Segal spaces, and the fact that they suffice to give a theory of $(\infty,1)$-categories as a stroke of luck.

Maybe a natural question to ask is: "If the equivalence conditions for a Segal space can be formulated just for spines, then why is it that, when we work with lifting conditions, we need higher-dimensional domains like the inner horns?".
I'd say that, roughly, this is because lifting conditions are weaker than equivalence conditions. You can see some of the extra power of equivalence conditions in the "2-out-of-3" argument above relating your inclusions to spine inclusions in the Segal setting -- there's nothing like that to be had when working with lifting conditions, so you need to throw in extra lifting conditions to make up for this.
Consider again the example from above, $X = \Delta[2] \cup_{\Lambda^1[2]} \Delta[2]$. Here, $X$ lifts against the inner 2-horn (which is the same as the 2-spine inclusion), but it lifts in two different ways. If you just ask for lifting against spine inclusions, then you don't have any lifting conditions which force these two lifts to become equivalent. The higher associativity conditions / higher inner horn lits will do things like forcing lower-dimensional associativity data to become equivalent. As in this case -- there will be an inner 3-horn, lifting against which would force our two composites to become homotopic. You need to more complicated "domains for your lifting problems", like the inner horns, so that you can specify more complicated configurations of lower associativity data that you want to force to become equivalent.
The higher-dimensional inner horn lifting conditions should be forcing the space of composites of any given composable chain to be more and more highly connected.
A very loose analogy I have in my head is the difference between saying "Every vector space has a basis" and "In a vector space, if you have a linearly independent subset, it can always be extended to a basis".
Okay, I am definitely rambling at this point, so I think I'll stop there!
