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## Comments

Apparently there is a difference between paths and morphisms. I believe the morphisms form an equivalence group over the paths.

For example, to which morphisms do the following paths belong?

$$ id_z $$ $$ s $$ $$ s.s $$ $$ s.id_z.s.s $$

`Apparently there is a difference between paths and morphisms. I believe the morphisms form an equivalence group over the paths. For example, to which morphisms do the following paths belong? $$ id_z $$ $$ s $$ $$ s.s $$ $$ s.id_z.s.s $$`

Since the set of paths is isomorphic to \(\mathbb{N}\), let's see what the equation means in \(\mathbb{N}\). \(s^4=s^2\) corresponds to \(n\equiv n-2\) for \(n\geq4\). This means that every even number except 0 is equivalent to 2, and every odd number except 1 is equivalent to 3. So, there are 4 morphisms in \(\mathcal{D}\): \(\mathrm{id}_z,s,s^2,\) and \(s^3\).

Fredrick Eisele wrote in #1:

They certainly form equivalence classes. They don't quite form a group, though the paths of length at least 2 have a \(\mathbb{Z}/2\mathbb{Z}\)-like structure (with composition corresponding to addition modulo 2, and the even numbers corresponding to the identity).

`Since the set of paths is isomorphic to \\(\mathbb{N}\\), let's see what the equation means in \\(\mathbb{N}\\). \\(s^4=s^2\\) corresponds to \\(n\equiv n-2\\) for \\(n\geq4\\). This means that every even number except 0 is equivalent to 2, and every odd number except 1 is equivalent to 3. So, there are 4 morphisms in \\(\mathcal{D}\\): \\(\mathrm{id}_z,s,s^2,\\) and \\(s^3\\). Fredrick Eisele wrote in [#1](https://forum.azimuthproject.org/discussion/comment/18569/#Comment_18569): >I believe the morphisms form an equivalence group over the paths. They certainly form equivalence classes. They don't quite form a group, though the paths of length at least 2 have a \\(\mathbb{Z}/2\mathbb{Z}\\)-like structure (with composition corresponding to addition modulo 2, and the even numbers corresponding to the identity).`