I know that it's common to write $S$ when you're referring to an object in the category of sets. But I'm attempting to solve a problem that states:
Prove that $S^{2(op)}$ has all finite colimits
So I'm pretty sure $S^2$ has to be a category (in order to actually obtain its opposite), but I´m not sure what it's objects or mapps would be.
Thanks in advance for any help provided.
On
$S^2$ is the arrow category of a category $S$. It is a special case of the comma category of $S$.
Objects of $S^2$ are the morphisms of $S$
A morphism in $S^2$ between objects $f:a \to b$ and $h:c \to d$ of $S^2$ is a pair $g=(g_1, g_2)$ of morphisms in $S$: $g_1:a \to c$, $g_2:b \to d$, such that:
$$g_2 \circ f = h \circ g_1$$.
$S^2$ is also denoted as $S^{\rightarrow}$
Awodey - Category Theory (2nd edition,2010), p.15-16
Adamek - Abstract and Concrete Categories; The Joy of Cats (edition July 9,2006), p.43
The most likely interpretation is that $S$ is supposed to be a category, and $S^2$ is the category of functors $2\to S$, where $2$ is the poset with two elements $0,1$ such that $0\leq 1$, seen as a category. This is also called the arrow category or category of arrows of $S$; its objects are morphisms of $S$, and morphisms $f\to g$ are pairs $(\alpha,\beta)$ of maps between domains and codomains that make the square $$\require{AMScd}\begin{CD}A@>{\alpha}>> C\\ @V{f}VV @VV{g}V\\ B@>>{\beta}>D\end{CD}$$ commute.