Given an object $X$, we can define an equivalence relation on the monomorphisms with range $X$: $u:S\to X,v:T\to X$ are equivalent iff exists an isomorphism $\phi:S\to T$ such that $u=v\circ \phi$. By the definition in wikipedia, the equivalence classes of monomorphisms with range $X$ are the subobjects of $X$.
In the section "interpretation", it says that when the objects are sets with additional structure, the equivalence classes of monomorphisms are determined by the image of the monomorphisms. And indeed if $u,v$ are equivalent monomorphisms then ($F$ is the forgetful functor to Set) $F(u),F(v)$ are functions with $F(u)=F(v)\circ F(\phi)$, where $F(\phi):F(S)\to F(T)$ is surjective, therefore $F(u),F(v)$ really have the same image.
But the other direction is what made me doubt this definition of subobject, because it's not true that every two monomorphisms with the same image are equivalent. For example: in the category Top we can take $Id_{S^1}:S^1\to S^1$ and $f:[0,1)\to S^1,f(x)=(cos(2\pi x),sin(2\pi x))$, they are monomorphisms with the same image but are not equivalent.
How can the equivalence class of such $f$ even be considered a subobject of $S^1$? It sounds weird to me, like there should have been additional conditions other than being a monomorphism.