Homotopy Theory

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Jonathan Beardsley

01:21

@MikeMiller right, I think writing down the generators/relations POV for bialgebras or Hopf algebras in homotopy theory is effectively impossible

Or at the very least, extraordinarily tedious

Daniel D.

01:51

@MikeMiller Oh sorry my bad I was not careful, thanks

9 hours later…

Bruno Stonek

10:58

do you guys have some easy or intuitive way to explain to people not used to infinity-categories why, if X is a space, then the colimit of {*}:X\to Spaces is equivalent to X?

Bastiaan Cnossen

11:33

What about the following: classically in a 1-category, any cocomplete category C is tensored over sets: for a set X, we can let X x c be the disjoint union of X many copies of C. Alternatively, it is the colimit of the constant functor X -> C with value c.
In higher category, sets should be replaced by spaces, but we still have that any cocomplete \infty-category is tensored over spaces, and the tensor X x c is still given by the colimit of the constant diagram X -> c.
Applying this to C = spaces and c = * gives that X is the colimit over the constant diagram on a point.

Maxime Ramzi

@BrunoStonek i would start with the case of Set; then explain why "colimits glue" (so a colimit along $X \cup_Y Z$ would be the pushout of the colimit along $X$ and along $Z$ along the one along $Y$); so you're left with explaining why pushouts work the way they do (because for coproducts it should be clear), and the intuition there is that "identifying x and y" in homotopy theory is "adding a path x -> y"

Alternatively, if they're comfortable with classical Yoneda, using the universal property : a map from $colim_X *$ to $Z$ is for each point of $X$, a map $*\to Z$. Then for any path in $X$, you have a naturality square, which amounts to a path in $Z$. Then for any $2$-path, the naturality squares must be compatible, which amounts to a $2$-path in $Z$. And so on and so forth, so that you end up getting "$map(colim_X *, Z) = map(X,Z)$"

Bruno Stonek

12:19

thanks for the suggestions!

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