TITLE: Is there a general notion of a subobject generated by a subset in any concrete category?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $f(\langle S \rangle)=\langle T \rangle$.