Let $k$ be a base commutative ring, and let $A$ be a (unital but not necessarily commutative) $k$-algebra. The *cone* on $A$ is the ring $CA$ of infinite matrices $(a_{ij})_{i,j \geq 1}$ that are finitely supported in each row and column. Then, the *suspension* of $A$ is $CA/\mathcal{M}_\infty(A)$, the quotient by the ideal $\mathrm{colim}_n \mathcal{M}_n(A)$ of finitely-supported matrices.

Here are some facts (lifted from section 1.4 of Loday's *Cyclic Homology*).

- $HH_*(CA) = 0$ and $HH_*(SA) \cong HH_{*-1}(A)$.
- There is a map $A[x^\pm] \to SA$ (selecting an infinite string of just-above-the-diagonal 1's) inducing a surjection $HH_*(A[x^\pm]) \twoheadrightarrow HH_*(SA)$.
- An external product with an element $\chi = [x] \in HH_1(k[x^\pm])$ induces an inverse $HH_*(SA) \cong HH_{*-1}(A) \xrightarrow{\cdot \chi} HH_*(A[x^\pm])$.

Recall that Hochschild homology can be defined as $HH_*(A) = \pi_*|N^{cyc}(A)|$, the geometric realization of the cyclic nerve. Thus, I would expect that fact 1 is actually a consequence of an equivalence $|N^{cyc}(A)| \simeq \Omega | N^{cyc}(SA)|$. Is this true? More generally, is there a sense in which I can read "$\mathrm{Spec}(SA) \to \mathrm{Spec}(CA) \to \mathrm{Spec}(A)$" as a (co?)fiber sequence with contractible middle term, at least "to the eyes of Hochschild homology"?

The second and third facts, however, I find yet more mysterious: they seem to assert that taking an "algebro-geometric sphere" (read: $\mathbb{G}_m$) over $A$ witnesses this delooping at the level of Hochschild homology. I have no idea why this should be true, but would love to have some intuitive (algebro-)geometric explanation.