"If an element has any positive period, it must have a smallest period. In fact, all its periods are multiples of this smallest one" (p. 177). Since 5 and 7 are coprime, their only positive common divisor is 1, and so the only element having both period 5 and period 7 is an element having period 1—that is, a fixed point. We proceed via Bézout's identity, {7 \times 3 − 5 \times 4 = 1}.

We define C_4 as follows:

Then there are four distinct maps from \mathbb{N}^{↻\sigma} to C_4. We give these maps below, in each case showing that we can form a valid morphism in the category 𝑺↻.

TODO Exercise 15.3

Essentially, the exercise is asking for a proof that α ⊚ α = α ⊚ α, which is trivially true.

cf. Session 12, Exercise 3.

We define the gender map g and show that it commutes (i.e., is a map in the category 𝑺↻).

Equivalently, taking B as an object in the category 𝑺↻, we can show that g forms a valid morphism.

The authors' tongue-in-cheek comment alerts us to the likelihood that there are not, in fact, 14 maps. Following the algorithm given on pp. 182–185, we find only 12 distinct maps. The 12 maps are given below, in each case with a proof that the map forms a valid morphism {X^{↻\alpha} \rightarrow Y^{↻\beta}}.