Ugh, another dumb question but: suppose I have a monoidal, simplicially enriched category. Presumably this corresponds to what, a functor $\Delta^{op}\to sSet-Cat$? So then, can I take coherent nerves on both sides to obtain a monoidal quasicategory $N(\Delta^{op})\to Cat_\infty$?
@JonBeardsley Last year I saw a great talk directed at (very strong) high school students trying to sell the (compactly supported) Euler characteristic as the "cardinality of a space". Are you familiar with this: math.ucr.edu/home/baez/counting
The talk I saw was not by Baez, but it must have been influenced by him.
@TylerLawson: Thanks. I think that DGA is for p=3 exactly the (reduced) Cobar resolution. x \in C^1 corresponds to the "identity" $Z/p\rightarrow F_p$ and $y$ corresponds to $Z/p\mapsto F_p, x\mapsto x^2$. I think in that this case the ideal generated by $y^2$ and $d(y^2)$ is acyclic, so dividing it out, would make the DGA smaller in some sense. I don't know what is going on for other p's.
@JonBeardsley brouwers fixed point theorem isn't bad. you can reduce it to a computation that $\pi_1(\mathbb{C}\setminus \{0\})$ contains the natural numbers as a set.
@JonBeardsley I also made german hs students prove that the human number was a knot invariant. (this is the smallest number of people that you can represent the knot with)
@JonBeardsley I also have a fun talk on compactification of conic sections based at the origin that is accessible and fun but not AT.
@HenrikRüping You're right. In fact, I guess at odd primes I really have described the cobar complex; that's somewhat ridiculous that it seems to be the "minimal" resolution if you want the underlying DGA to be free. So you've described the case G = Z/3 as a graded-commutative DGA with generators x, y, satisfying y^2 = 0 and d(y) = x^2.
@JonBeardsley, yes, you can apply any functorial fibrant replacement of simplicial sets that commutes with finite products (e.g. Sing|-| or Ex^\infty).
