I think this is a silly question, but I'm quite confused. In Lurie's "Rotation Invariance In Algebraic K-Theory" Notation 3.2.4. he defines a filtered spectrum $\mathbb{A}$ given by
$$\mathbb{A}\_n = \begin{cases}
\mathbb{S} &\text{ if } n =0 \\
0 &\text{ otherwise}\end{cases} $$


He then notes (example 3.2.9) that $$\mathbb A \otimes \mathbb A$$ takes the form
$$
\cdots \to 0 \to 0 \to \mathbb S \to \Sigma \mathbb S \to 0 \to \cdots
$$
It seems to me like this should (as a filtered spectrum) be the same as $\mathbb A + \Sigma \mathbb A(1)$, since there are no nontrivial maps $\mathbb S \to \Sigma \mathbb S$. But later (Construction 3.4.4) he writes a fiber sequence
$$
\mathbb A \otimes \mathbb A \to \mathbb A \to\_\beta \Sigma^2 \mathbb A (1)
$$
and gives $\beta$ a name ("the anchor map"). But $\beta$ corresponds to a square
$\require{AMScd}$
\begin{CD}
\mathbb{S} @>>> 0\\
@VVV @VVV \\
0 @>>> \Sigma^2\mathbb S
\end{CD}
and therefore a map of spectra $\mathbb S \to \Sigma \mathbb S$, and therefore (to my eyes) should equal zero.


What am I missing here? If this map is zero, why does it get a name? And if it's nonzero, why doesn't this produce a nontrivial element of $\pi\_{-1}\mathbb{S}$?



Why isn't the anchor map in Lurie's "Rotation Invariance in Algebraic K-Theory" zero?
It's zero as a map of filtered spectra, but it is considered as a map of $\mathbb{A}$-bimodules.


