I'm trying to understand the single-set category in Categories for the Working Mathematician , specifically the 2-category version, and I'm confused about the commutation of $s_0$ and $s_1$.
"Similarly a 2-category can be considered to be a single set $X$ considered as the set of 2-cells (e.g., of natural transformations). Then the previous 1-cells (the arrows) and the 0-cells (the objects) are just regarded as special "degenerate" 2-cells. On the set $X$ of 2-cells there are then two category structures, the "horizontal" structure $(\#_0, s_0, t_0)$ and the "vertical" structure $(\#_1 , s_1, t_1)$.
Each satisfies the axioms above for a category structure and in addition (i) Every identity for the 0-structure is an identity for the 1-structure; (ii) The two category structures commute with each other. Here, the condition (ii) means, of course, that
$s_0 s_1= s_1 s_0 , s_0 t_1 = t_1 s_0 , t_0 t_1 = t_1 t_0$"
So my question it this. If I'm understanding this right, $s_0 t_0$ gives us the identities of 1-cells, which correspond to the objects, and $t_1 s_1$ gives us the identities of 2-cells, which correspond to the morphisms being mapped by the 2-cell. I'm assuming these equalities are functional, so they actually mean $s_0 (s_1 x)= s_1 (s_0 x)$ and this x is a 2-cell. When we apply $s_0$ to a 2-cell, do we directly recover the object? If so, how do we later recover the morphism from the object? Or what?
Also, when we say "Every identity for the 0-structure is an identity for the 1-structure", do we mean the identity of a 1-cell is an identity of a 2-cell too? I'm assuming from this statement we can conclude $s_0 x = x \implies s_1 (s_0 x) = s_0 x$ which again leaves me wondering how we apply $s_1$ to $s_0 x$.
TIA
The maps $s_0,s_1,t_0$ and $t_1$ are maps $X\to X$. Thinking about this data as a 2-category, we may think of an element $x\in X$ as data of morphisms $f,g\colon A\to B$ and a $2$-morphism $\alpha\colon f\to g$ between them. In this light, the element $s_0x$ can be thought of as the object $A$, but in light of the codomain of $s_0$ being $X$ as well, it is for the purposes of your question better to think about it as the data of two identity morphisms $\mathrm{id}_A,\mathrm{id}_A\colon A\to A$ and an identity $2$-morphism $\mathrm{id}_{\mathrm{id}_A}\colon\mathrm{id}_A\to\mathrm{id}_A$ between them. Likewise, $s_1x$ can be thought of as the $1$-morphism $f\colon A\to B$, but let us instead present it as the data of the two morphisms $f,f\colon A\to B$ and the identity $2$-morphism $\mathrm{id}_f\colon f\to f$.
The statement that $s_0s_1x=s_1s_0x$ for all $x\in X$ (you are correct that it is a functional equality) is therefore in agreement with how we think about $s_0x$ and $s_1x$ as $2$-morphisms: both sides correspond to the data of two identity morphisms $\mathrm{id}_A,\mathrm{id}_A\colon A\to A$ and an identity $2$-morphism $\mathrm{id}_{\mathrm{id}_A}\colon\mathrm{id}_A\to\mathrm{id}_A$ between them.
So when we apply $s_0$, we recover not directly the object, but rather the identity $2$-morphism between the identity $1$-morphisms on our object, and likewise for $s_1$, $t_0$ and $t_1$, and this allows you to compose the maps in the way they do.
The statement that every identity $0$-cell is an identity $1$-cell means that if $s_0x=x$, then $s_1x=x$ as well. This is equivalent to saying that $s_0=s_1s_0$. In the explanatory language of the first paragraph, this says that if I give you an $x\in X$ corresponding to data of morphisms $f,g\colon A\to B$ and a $2$-morphism $\alpha\colon f\to g$ between them, then the left hand side and the right hand side both are equal, since they both are the data of two identity morphisms $\mathrm{id}_A,\mathrm{id}_A\colon A\to A$ and an identity $2$-morphism $\mathrm{id}_{\mathrm{id}_A}\colon\mathrm{id}_A\to\mathrm{id}_A$ between them.