The graded multiplication on topological $K$-theory In every reference I have looked at (the books by Atiyah, Karoubi, Lawson--Michelsohn, Hatcher's unpublished book) the exterior multiplication on (reduced, negative) $K$-theory is given by the following recipe:
$$\tilde{K}^{-i}(X) \otimes \tilde{K}^{-j}(Y) = \tilde{K}^0(S^i \wedge X) \otimes \tilde{K}^0(S^j \wedge Y) \to \tilde{K}^0(S^i \wedge X \wedge S^j \wedge Y) \cong \tilde{K}^0(S^i \wedge S^j \wedge X \wedge Y) = \tilde{K}^{-i-j}(X \wedge Y),$$
where the arrow is the exterior product on $\tilde{K}^0$ and the isomorphism is given by the swap homeomorphism $X \wedge S^j \to S^j \wedge X$.
In a multiplicative, commutative, reduced generalised cohomology theory $h^{\ast}(-)$ (i.e. one given by a homotopy commutative ring spectrum) the suspension isomorphism $h^{\ast}(X) \to h^{\ast+1}(S^1 \wedge X)$ is effected by exterior product $\sigma \wedge -$ with the class $\sigma \in h^{1}(S^1)$ corresponding to $1 \in h^0(S^0)$ under the suspension isomorphism. So for classes $a \in h^{-i}(X)$ and $b \in h^{-j}(Y)$ following the recipe above gives
$$\sigma^i \wedge a \wedge \sigma^ j \wedge b \in h^0(S^i \wedge X \wedge S^j \wedge Y) $$
which under the swap map gives
$$(-1)^{ij} \sigma^{i} \wedge \sigma^j \wedge a \wedge b \in h^0(S^i \wedge S^j \wedge X \wedge Y),$$
and so seems to correspond to $a \otimes b \mapsto (-1)^{ij} a \wedge b$. In other words, this recipe does not give the product in the theory $h^{\ast}(-)$, and ought to be corrected by $(-1)^{ij}$.
Question: Is this really correct, and does anybody know somewhere where this point is discussed?
 A: This is really correct: the stabilization isomorphism isn't immediately compatible with the multiplication. This is a problem even in ordinary homological algebra and is largely a consequence of pretending that integer indices are sufficient. Let me try to illustrate.
Suppose that $K$ is a homotopy-commutative differential graded algebra and that $X$ and $Y$ are chain complexes. For any integer $i$ we can define
$$
K^{-i}(X) = [\Bbb Z[i] \otimes X, K]
$$
by chain homotopy classes of maps. This gives an external product
$$
K^{-i}(X) \otimes K^{-j}(Y) \to [\Bbb Z[i] \otimes X \otimes \Bbb Z[j] \otimes Y, K \otimes K] \to [\Bbb Z[i] \otimes \Bbb Z[j] \otimes X \otimes Y, K]
$$
using the product of $K$. This points out that we need to choose an identification of $\Bbb Z[i] \otimes \Bbb Z[j]$ with $\Bbb Z[i+j]$ to proceed further. Choosing one makes it land in $K^{-(i+j)}(X \otimes Y)$.
The reason why I'm going to this length is that, when you describe the exterior product stability isomorphism $h^\ast X \to h^{\ast + 1}(S^1 \wedge X)$, it corresponds in homological algebra to choosing an element
$$
\sigma \in K^1(\Bbb Z[1]) = [\Bbb Z[-1] \otimes \Bbb Z[1], K] \cong [\Bbb Z, K] = H_0 K
$$
lifting the unit. Again, there is this last casual identification of $\Bbb Z[-1] \otimes \Bbb Z[1]$ with $\Bbb Z[0]$. Without making casual identifications early, the exterior-product stabilization isomorphism you describe is most naturally the exterior product map
$$
\begin{align*}
K^{-i}(X)
&= [\Bbb Z[i] \otimes X, K] \\
&\to [\Bbb Z[-1] \otimes \Bbb Z[1] \otimes \Bbb Z[i] \otimes X, K \otimes K] \\
&\to [\Bbb Z[-1] \otimes \Bbb Z[i] \otimes \Bbb Z[1] \otimes X, K] \\
&\cong K^{-(-1 +i)}(\Bbb Z[1] \otimes X).
\end{align*}
$$
Note in particular the swap of the factors $\Bbb Z[1]$ and $\Bbb Z[i]$ at the second-last step. This introduces a sign of $(-1)^i$ into the standard identifications that we should probably account for in the suspension isomorphism. If you repeat this $i$ times to land in $K^0$, you introduce a sign of $(-1)^{\binom{i+1}{2}}$, and so the net sign discrepancy introduced by this on the source and target has parity
$$
\binom{i+1}{2} + \binom{j+1}{2} - \binom{i+j+1}{2} \equiv ij \mod 2.
$$
In homotopy theory we have the same issue except that we are permuting sphere factors, and have less canonical justification for suspending vs desuspending. Adams wrote a rather scathing criticism of these types of casual identification in section 6 of "Prerequisites for Carlsson's lecture". Schwede's book on symmetric spectra also discusses systematically keeps track of the order of indices in terms like $i+j$  to make these types of issues more transparent.
