So here's a possibly easy/dumb question: given a Kan complex $X$ and a sub-Kan-complex $Y\subset X$, and a functor of quasicategories $F:X\to C$, we can take the colimit of the composition $Y\to X\overset{F}\to C$. This should still determine a diagram $X\to C$ right? What's the indexing simplicial set of this new diagram with part of the colimit already taken?
let X, Y be pointed spaces, let [X, Y] be the space of maps X -> Y, and let [X, Y]_{\ast} be the space of pointed maps X -> Y. what's the homotopy fiber of the natural map [X, Y]_{\ast} \to [X, Y]? I think it's Omega Y... does that sound right?
The homology of the cubical construction of an abelian group A gives you the homology of the corresponding Eilenberg-Mac Lane spectrum. Is there a construction which would somehow give the unstable homology of the Eilenberg-Mac Lane spaces?
it gives a compact and clear description of them, for sure, but it's not obtained from something of different nature done algebraically to the abelian group
@AaronMazel-Gee yeah i think so. but that's just the cone over $Y$. I want to do this in a more structured way, and I think I need the operadic Kan extension framework.
@JonBeardsley well, this isn't a kan complex anymore -- it's got non-invertible arrows from $y \in Y$ to $\infty \in \triangleright$. you might call it a "directed" cofiber. anyways, i'm not sure what you mean by "more structured" here, if X and Y are just kan complexes
@AaronMazel-Gee ah that's a good point. i guess i hadn't really thought about the fact that it wasn't a kan complex. yeah... hm. and ultimately i worked it out, but by more structured i mean being n-fold loop spaces.
Hey all, in Remark 3.1.3.15 of Higher Algebra, Lurie asks for a certain morphism $h'':M^\otimes\to C^\otimes\times\Delta^1$ to be an operadic q-left Kan extension. However, his usual notation for such things is $M^\otimes\to C^\otimes$. Is it possible he means something like $h''$ composed with projection onto $C^\otimes\times\{1\}$ or something?
The reason is, the set-up is really about extending along a map of $\infty$-operads $A^\otimes\to B^\otimes$, given a map $A^\otimes\to C^\otimes$. So one should expect the Kan extension to land in $C^\otimes$ rather than $C^\otimes\times\Delta^1$... maybe?
In other words, it doesn't quite look like Definition 3.1.2.2.