category-theoryexamples-counterexamplesforgetful-functorsfunctorsmorphism
A concrete category is a category $C$ endowed with a faithful functor $U:C\rightarrow Set$. And if $a$ is an object in $C$, then a subobject of $a$ is an isomorphism class of monomorphisms with codomain $a$.
I’m wondering if subobjects can be related to subsets using $U$. My question is, given a subobject of $a$, does there always exist a monomorphism $f$ in the subobject such that $U(f)$ is an inclusion map from a subset of $U(a)$ to $U(a)$?
If not, does anyone know of a counterexample?
The question is ancient, but IMHO the most convincing answer is missing:
The definition of "subobject in a category $\mathcal{C}$" is chosen in such a way that it generalize the notion of $k$-vector subspaces (when $\mathcal{C} = \mathsf{Vect}_k$), the notion of subsets (when $\mathcal{C} = \mathsf{Set}$), the notion of subrings (when $\mathcal{C} = \mathsf{Ring}$), and lots of other classical "sub-something" notions that appear throughout mathematics (probably not all of them, though). If you were to define "subobjects of $A \in \mathcal{C}$" to mean "monomorphisms with codomain $A$", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub-something". For instance, in $\mathsf{Set}$, the monomorphisms $\emptyset \to \left\{1\right\}$, $\left\{1\right\} \to \left\{1\right\}$ and $\left\{2\right\} \to \left\{1\right\}$ would be three different subobjects of the object $\left\{1\right\}$, but there are only two subsets of the set $\left\{1\right\}$. So this would be a bad definition.
However, if you define subobjects of $A \in \mathcal{C}$ to be isomorphism classes of monomorphisms with codomain $A$ (where "isomorphism" is to be correctly interpreted: an isomorphism between two monomorphisms $\alpha : S \to A$ and $\alpha^{\prime} : S^{\prime} \to A$ means a morphism $s : S \to S^{\prime}$ satisfying $\alpha = \alpha^{\prime} \circ s$), then, in all of the examples listed above, the subobjects of $A$ are in a canonical bijection with the "sub-somethings" (i.e., the $k$-vector subspaces, or the subsets, or the subrings). For instance, in $\mathsf{Set}$, the subobjects of a set $A$ are the isomorphism classes of monomorphisms with codomain $f$. The isomorphism class of such a monomorphism $f : S \to A$ can be identified with the subset $f\left(S\right)$ of $A$. Thus, the subobjects of $A$ are in bijection with the subsets of $A$ here. The same construction works for rings and for $k$-vector spaces.
The good definition of a subobject also has the advantage (compared with the bad definition) that the subobjects of a given object $A \in \mathcal{C}$ often form a set (as opposed to just a class). I don't personally find this vital; it is not usually true in constructive mathematics anyway, and I don't believe that a definition is necessarily bad just because it sometimes returns proper classes.
Monomorphisms are embeddings functors (i.e. functors whose both the object-function and the arrow-function are injective).
It easily seen that embeddings are monomorphisms because if $F \colon \mathbf C \to \mathbf D$ is an embedding then for every pair of fucntors $G,H \colon \mathbf X \to \mathbf C$ such that $F \circ G=F \circ H$ then
for every $x \in \mathbf X$ we should have $F \circ G(x)=F \circ H(x)$ and since $F$-object part is injective it would follow that $G(x)=H(x)$;
in exactly the same way for every $f \colon x \to x'$ in $\mathbf X$ we have that $F \circ G(f)=F\circ H(f)$ and for the injectivity of $F$-arrow part we also have that $G(f)=H(f)$;
so $G=H$.
Now let's suppose $F \colon \mathbf C \to \mathbf D$ is monomorphism.
For every object $c \in \mathbf C$ there's a unique functor $\bar x \colon 1 \to \mathbf C$ (where $1$ is the category with one object ($0$) and one arrow) such that $\bar x(0)=x$.
In similar way for every morphism $f \in \mathbf C(c,c')$ there's a unique functor $\bar f \colon 2 \to \mathbf C$ (where $2$ is the category with two objects, $0$ and $1$, and only one non identity morphism $0 \to 1$) such that $\bar f(0)=c$, $\bar f(1)=c'$ and $\bar f(0 \to 1)=f$.
Then it follows that for every pair of objects $c,c' \in \mathbf C$ if we have $F \circ \bar c(0)=F(c)=F(c')=F \circ \bar c'(0)$ then $F \circ c=F \circ c'$ and so by monomorphism property $\bar c=\bar c'$. From this follows that $$c=\bar c(0)=\bar c'(0)=c'$$ which gives the injectivity of the object part of $F$.
If $f,g \colon c \to c'$ are morphisms of $\mathbf C$ such that $F(f)=F(g)$ it can be proven similarly (using the functor $\bar f$ and $\bar g$) that $f=g$.
From this it follows that also the arrow part of $F$ is injective.
Edit I see that I didn't address the part about epimorphisms in $\mathbf{Cat}$. Let's try to make ammend.
For start it can be shown that every functor that is not sujective on objects cannot be an epimorphism. Indeed if $F \colon \mathbf C \to \mathbf D$ is not surjective we can consider the full subcategory $\mathbf D' \hookrightarrow \mathbf D$ that is spanned by the object in $\text{Im} F$. The consider the pushout of the diagram
where the morphism from $\mathbf D'$ are the embeddings. Since $F$ factors through $i \colon\mathbf D' \hookrightarrow \mathbf D$ (since it's image is contained in the full subcategory $\mathbf D'$) we have that $G \circ F = H \circ F$.
More in detail there's a functor $\mathcal F \colon \mathbf C \to \mathbf D'$ such that $F=i \circ \mathcal F$ and so
$$G \circ F = G \circ i \circ \mathcal F$$ and $$H \circ F = H \circ i \circ \mathcal F\, .$$
Since $G \circ i= H \circ i$, because of the pushout property, we have that $G \circ F=H \circ F$, nonetheless $G \ne H$.
This basically prove that epimorphism should at least be sujective on objects. Clearly this condition is not sufficient, it's easy to find counter examples.
What can be proven is that $F \colon \mathbf C \to \mathbf D$ is sujective on objects and the graph $\text{Im}F$ contains a set of morphism which are generators for the category $\mathbf D$ then this functor is an epimorphism.
By generators for the category $\mathbf D$ I mean a family of morphisms such that every other morphism in $\mathbf D$ is a composite of these morphisms.
The proof of this is no different from the analog proofs for other categories. Nontheless this is just a sufficient condition as the following example show.
Consider the category $2$ defined as above and $\bar 2$ obtained by $2$ just adding an inverse to the morphism $0 \to 1$.
The embedding $j \colon 2 \to \bar 2$ is an epimorphism since for every $G,H \colon \bar 2 \to X$ such that $G(0\to 1)=G(0 \to 1)$ have to send $1 \to 0$ in *the inverse of $G(0 \to 1)=H(0 \to 1)$.
Of course from this example it becomes clear that one can perfect the previoius said condition by proving that $F \colon \mathbf C \to \mathbf D$ is an epimorphism iff it's surjective on object and every morphism in $\mathbf D$ is a composite of morphism in $\text{Im} F$ and their inverse (if they exist).
Nonetheless I'm not aware if this is a full characterization of epimorphisms in $\mathbf{Cat}$.
Hope this helps.
Best Answer
Consider the full subcategory $C$ of Set on exactly two objects: the natural numbers and the singleton set $b$ whose element is my coffee mug (if you prefer, let $b=\{\pi\}$.) $C$ comes equipped with the obvious forgetful functor $U$ which is just the inclusion. There are infinitely many distinct ways in which $b$ is a subobject of $\mathbb N$ in $C$, but there are no proper subsets of $\mathbb N$ in the image of $U$.
For a more general answer, observe that the existence of a faithful functor into Set is invariant under equivalence of categories. So $U$ can't possibly impose any on-the-nose conditions like yours.