Coherence of shift functors

Warning 4.2.11. While the shift notation is convenient, it should be used with care to avoid sign issues.

For integers \(n\) and \(m\), one may identify the composite \([n] \circ [m]\) with \([n + m]\) by explicitly checking each of the four cases. In particular, this shows that \([n] \circ [m]\) is equivalent to \([m] \circ [n]\): both are shifting by the sum \(m + n\).

The subtlety is that there are several inequivalent ways to exhibit this identification. For example, since \([n]\colon C \to C\) is exact, it commutes with \([m]\), giving another identification \([n] \circ [m] \simeq [m] \circ [n]\). To see that this identification differs from the previous one, let us consider the case \(n = m = 1\), so that both shift functors are simply the suspension functor. The first identification is then nothing but the identity on \(\Sigma ^2\colon C \to C\). However, the second identification \(\Sigma ^2 \simeq \Sigma ^2\) is the one coming from commutation of \(\Sigma \) with itself. To see that these are not the same, consider the case \(C = \An _*\), for which the commutation map is obtained by forming smash products with the swap map \(S^1 \wedge S^1 \iso S^1 \wedge S^1\) from Definition 2.4.10. Since this map has degree \(-1\), it is not homotopic to the identity.