Is there a general notion of a subobject generated by a subset in any concrete category?

Question

Let $A$ be an object of a concrete category $C$ with a forgetful functor $F\rightarrow Set$ and let $S \subseteq F(A)$. Is there a general construction that gives us a subobject $\left<S\right>$ of $A$ in $C$?

$S$ should, for example, satisfy the universal property that if $B$ is a subobject of $A$ in $C$ such that $S \subseteq F(B)$, then $\left< S \right>$ is a subobject of $B$ in $C$.

I know that in the category of groups it's true that if $f$ is a homomorphism of groups, then $f(\left< S \right>)=\left< f(S) \right>$. Is there a corresponding theorem here?

Answer 1

In a category $Alg$ of algebraic structures, such as groups, the forgetful functor $U:Alg \to Set$ has a left adjoint $F$. Thus given a subset $S \subseteq U(A)$ we have the adjoint map $FS \to A$ and now the substructure $\langle S \rangle$ of $B$ generated by $S$ is obtained by factorising $FS \to A$ through its image as a surjective homomorphism followed by an injective homomorphism $FS \to \langle S \rangle \to A$.

The universal property follows from this construction: namely, given a substructure $T \hookrightarrow A$, then to say that $S \subseteq UT$ as a subobject of $UA$ is, by adjointness, to give a map $FS \to T$ such that the square $$\begin{matrix}FS\rightarrow&<S>\\\\\downarrow&\downarrow\\\\T\rightarrow&A\end{matrix}$$ commutes. Because $FS \to \langle S \rangle$ is surjective and $T \to A$ injective there exists a unique factorisation $\langle S \rangle \to T$ making both triangle commute. (This relationship between surjections and injections is called orthogonality - see the notion of factorisation system.)

Your theorem can then be established using the orthogonality and composability properties of surjections and injections. A natural generalisation of the above is to a "concrete category" $U:C \to D$ over a general $D$ where $U$ has a left adjoint and where $C$ admits a (strong epi/mono)-factorisation system.

Answer 2

Let's denote the forgetful functor by $U : C \to \mathsf{Set}$. (Usually its left adjoint is denoted by $F$.)

If $C$ is well-powered, has intersections (which are special limits), and $U$ preserves monomorphisms and intersections (for example when $U$ preserves limits), we may define $\langle S \rangle = \bigcap_{i : B \to A \text{ mono}, S \subseteq \mathrm{im}(U(i))} B$ with the evident monomorphism $\langle S \rangle \to A$.

Now for the second question: If $f : A \to B$ is a morphism in $C$ and $S \subseteq U(A)$, then let $T:=U(f)(S) \subseteq U(B)$. We ask ourselves if $f(\langle S \rangle)=\langle T \rangle$ holds. But what does $f(A')$ mean for a monomorphism $A' \to A$? It should be the smallest subobject of $B$ such that $A' \to A \to B$ factors through $f(A')$ (this exists again under the mentioned hypothesis). But when $A'=\langle S \rangle$, the condition reduces by definition of $\langle S \rangle$ to that $S \to U(A) \to U(B)$ factors through $U(f(A'))$, i.e. that $T \subseteq U(f(A'))$. This shows $f(\langle S \rangle)=\langle T \rangle$.