# A morphism-revealing category? [closed]

Categories of sets and functions can be considered as subcategories of Set but when considered as subcategories of the category SubSet, of pairs of sets with pairs $(X,S)$, $S\subseteq X$, as objects and functions $f:X\to Y$ with $f(S)\subseteq T$ as morphisms $(X,S)\to (Y,T)$, the morphisms of SubSet seems to reveal the form of the morphisms of the subcategories. Examples:

Magmas. $R\subseteq (X\times X)\times X$ (the composition). Morphisms are functions $\alpha:(X\times X)\times X\longrightarrow(X^\prime\times X^\prime)\times X^\prime$ such that $((x,y),z)\in R \Rightarrow \alpha((x,y),z)\in R^\prime$. Functions $\alpha_1,\alpha_2,\alpha_3:X\longrightarrow X^\prime$ exists such that $\alpha((x,y),z)=((\alpha_1(x),\alpha_2(y)),\alpha_3(z))$ and if $\alpha$ is such that $\alpha_1=\alpha_2=\alpha_3$, then $\alpha_1$ correspond to magma morphisms, $((x,y),z)\in R \Rightarrow ((\alpha_1(x), \alpha_1(y)),\alpha_1(z))\in R^\prime$. That is, $xy=z\Rightarrow\alpha_1(x)\alpha_1(y)=\alpha_1(z)$.

Topological spaces. $\tau\subseteq \mathcal{P}(X)$. Morphisms are functions $\mathcal{P}(X)\overset{\alpha}{\longrightarrow}\mathcal{P}(X^\prime)$ such that $\mathcal{O}\in\tau \Rightarrow \alpha(\mathcal{O})\in \tau^\prime$. If there is a function $f:X^\prime\longrightarrow X$ such that $\alpha = \mathcal{Q}(f)$, where $\mathcal{Q}$ is the contra-variant power set functor, this correspond to the condition $\mathcal{O}\in\tau\implies f^{-1}(\mathcal{O})\in\tau^\prime$.

Hence, studying subcategories of Set yields that the morphisms are functions, while studying subcategories of SubSet also yields additional information of the morphisms. So my questions are:

Is this approach used in mathematics?

• I don't think it's not really true that concrete categories are subcategories of Set, in any sense that "subcategory" is usually meant. – Mike Shulman Feb 16 '16 at 18:03
• @MikeShulman, are there severe problems looking at Grp as a subcategory of Set? – Lehs Jul 7 '16 at 18:29
• It just isn't what people mean by "subcategory". – Mike Shulman Jul 8 '16 at 3:25
• @MikeShulman. OK! I've got a nice answer here math.stackexchange.com/questions/1852811/… – Lehs Jul 8 '16 at 6:41

$\newcommand{\SubSet}{\mathbf{SubSet}}\newcommand{\Set}{\mathbf{Set}}\newcommand{\Top}{\mathbf{Top}}$ $\newcommand{\ra}{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$
This general idea is definitely important in categorical logic and the theory of fibrations. Let me rename your category Pos = SubSet. There is an evident forgetful functor $p : \SubSet \to \Set$ given by the first projection, which has the important property that:
• $p$ is a bifibration
This means that there is an operation of pushing a subset $S \subseteq X$ along a morphism $f : X \to Y$ to obtain a subset $\mathbf{push}_fS \subseteq Y$ satisfying certain universal properties, as well as an operation of pulling a subset $T \subseteq Y$ along $f : X \to Y$ to obtain a subset $\mathbf{pull}_f T$ satisfying dual properties. In particular, these operations define functors between the fibers of $X$ and $Y$ \begin{align*} \mathbf{push}_f &: \SubSet_X \to \SubSet_Y \\ \mathbf{pull}_f &: \SubSet_Y \to \SubSet_X \end{align*} satisfying an adjunction $\mathbf{push}_f \dashv \mathbf{pull}_f$. In this case, of course, pushing and pulling are defined by image and inverse image: $$\mathbf{push}_fS = \{f(x) \mid x \in S\} \qquad \mathbf{pull}_fT = \{x \mid f(x) \in T\}$$ But the definition of bifibration itself can be seen as a categorical abstraction of these basic operations on subsets.
Now, I think the idea that you are getting at (but correct me if I'm wrong) is that many concrete categories, seen as categories $C$ equipped with a forgetful functor $q : C \to \Set$, can be embedded into $p : \SubSet \to \Set$. In particular, to take one of your examples, let $q : \Top \to \Set$ be the forgetful functor which returns the underlying set of a topological space. This functor happens to be a fibration (given a topology on $Y$ and a function $f : X \to Y$ we can form the inverse image topology on $X$) but not quite a bifibration. However, essentially as you observed, the concrete category of topological spaces can be studied using the $\SubSet$ bifibration. More precisely, there is a commuting diagram $$\begin{array}{c} \Top^{op} & \ra{} & \SubSet \\ \da{q^{op}} & & \da{p} \\ \Set^{op} & \ra{\mathcal{P}} & \Set \end{array}$$ that can be read as a morphism of functors $q^{op}\to p$, where $\mathcal{P} : \Set^{op} \to \Set$ is the contravariant powerset functor and $\Top^{op} \to \SubSet$ is the functor viewing a topological space $\tau$ over $X$ as a subset of $\mathcal{P}(X)$. Moreover, this morphism $q^{op} \to p$ is full and faithful in the sense that for any fixed function $f : X \to X'$ and topological spaces $\tau,\tau'$ such that $q(\tau) = X,q(\tau') = X'$, then $f$ is a continuous function from $\tau$ to $\tau'$ (i.e., there is a morphism $\alpha : \tau \to \tau'$ in $\Top$ such that $q(\alpha) = f$) just in case the image of $\tau'$ under $\mathcal{P}(f)$ is included in $\tau$ (i.e., there is a morphism $\beta : \tau' \to \tau$ in $\SubSet$ such that $p(\beta) = \mathcal{P}(f)$).