AH, I think I know what's going on: When they write $\mathrm{const}_1(\mathcal{C})$, they don't mean your $\mathrm{const}_1$, instead they mean the simplicial diagram left Kan extended from the functor $\Delta_{\leq 1}\to \mathrm{Cat}$ taking $[0]\mapsto 0$ and $[1]\mapsto \mathcal{C}$. This should realize to the suspension of $\mathcal{C}^\simeq$, and a map $\Sigma\mathcal{C}^\simeq \to |Q(\mathcal{C})|$ corresponds to a map $\mathcal{C}^\simeq \to \Omega|Q(\mathcal{C})|$ by adjunction.

* at least two infinite orbits, of course

More generally, following exactly the same approach as in the answer to your previous question: The $f$ for which there exist $S$ which are infinite and co-infinite and satisfy $f(S)=S$ are exactly those with infinitely many orbits or more than two infinite orbits.

@mathworker21 I don't think that's true: Writing $\sin(n)$ as imaginary part of $e^{in}$, the partial sums evaluate to the imaginary part of a geometric sum $e^i \cdot \frac{e^{iN}-1}{e^i-1}$, where the numerator stays bounded by $2$ no matter how large $N$ gets.

For two abstract abelian varieties, what does it mean for them to share torsion points? This seems to inherently rely on a choice of coordinates.

You can lift polynomial rings ($\mathbb{S}[x_1,\ldots,x_n]$ can be written as suspension spectrum of the monoid $\mathbb{N}^n$), but not all maps between them. For example already $\mathbb{S}[x] $ does not admit an $E_\infty$ endomorphism taking $x\mapsto x+1$.

I'd think that something like the following works, but I'm not sure if I can make it 100% rigorous. Given any abstract connected cell complex with the specified structure (consisting a priori of "standard dodecahedra"), it admits a covering map to the boundary of the 120-cell, uniquely specified by fixing it on one face and then extending compatibly with the combinatorial structure. Now topology enters: this boundary is homeomorphic to $S^3$, in particular simply-connected, and so any connected covering is a homeomorphism.

Yes, if $\pi_1(M)$ is free, then $H_1(M;\mathbb{Z})$ is free abelian (of the same rank), and then UCT determines $H^1(M;R)$.

One version of Poincare duality for unoriented Manifolds is that the cup-product pairing $H^i(M;\mathbb{F}_2)\otimes H^{n-i}(M;\mathbb{F}_2) \to H^n(M;\mathbb{F}_2)\cong \mathbb{F}_2$ is nondegenerate. I think this is for example discussed in Hatcher.

Ah, I see. Then let me try the following: $H^1(M;\mathbb{F}_2)$ is nontrivial since $\pi_1$ is free and nontrivial, and by Poincare duality there exists a class in $H^3(M;\mathbb{F}_2)$ such that their cup product is nontrivial, so the cup-length is $2$, which should imply LS category $\geq 2$

Well, compact manifolds (without boundary) are not contractible, for example since they have nontrivial mod 2 homology

You don't really have to say it with derived categories in this instance, as $\mathrm{RHom}(C, D)$ for $C$ a bounded below complex of projectives is just computed by the internal Hom of chain complexes. And this applies to $C_*(X)$, of course.