I agree with you that this is not about “concrete examples.” More about language. I apologize if my story is elementary, but there is really nothing complex.
Maybe you do not realize that “$A$ has universal property” is the same as “$A$ is a universal object” is the same as “an object $A$ is universal.” These are different names of the same term. So the definition of a universal object also defines universal property.
Consider the definition of an initial object. “…an initial object is an object… such that…” “Initial” is a property of objects, thus this definition defines a property. Properties are named not only by adjectives (e.g. “transitive”, “injective”), but also by nouns (e.g. “equivalence”, “injection”; “a function $f$ is an injection” is the same as “a function $f$ is injective”). In contrast, the definition of average, i.e. $(x, y)\mapsto \frac{x+y}{2}$, defines not a property.
Consider the shorter definition in Wikipedia which you did not cite:
An initial morphism from $X$ to $U$ is an initial object in the category
$X \downarrow U$.
This definition defines a property because it uses the definition of an initial object. The longer definition in Wikipedia which you cited is the shorter definition with the terms “initial object” and “comma category” unfolded.
“The universal property of the quotient group” is not a definition, it is a theorem which says that the quotient group $G/N$ is an initial object in a category defined as:
- object: $(X, f)$ where $X$ is a group and $f:G\to X$ and $N\subseteq ker(f)$;
- morphism of type $(X_0, f_0)\to (X_1, f_1)$: $g:X_0\to X_1$ such that $g\circ f_0 = f_1$.
I have essentially seconded lhf's answer, but he/she did not construct the category. I just can not find explicit construction of this category in textbooks.
Wikipedia's definition of the universal property does not include the universal property of the quotient group as a particular case. The problem is that in Wikipedia's definition $f$ is a morphism, but in the case of groups $f$ is a homomorphism such that $N\subseteq ker(f)$. IMHO Wikipedia's definition is not general enough.
P. S. I prefer “initial” and “terminal” over “universal”. A universal object is an initial object or a terminal object depending on context. Therefore, any text involving “universal” forces a reader to guess a precise meaning.
Almost all monomorphisms in $\mathbf{Set}$ are split (hence are preserved by any functor whatsoever), the exceptions being maps with empty domain. So it's just a question of what happens with those maps.
We consider an adjunction
$$F \dashv U : \mathcal{C} \to \mathbf{Set}$$
where the "forgetful" functor $U : \mathcal{C} \to \mathbf{Set}$ reflects monomorphisms. (If $U$ is faithful, then $U$ reflects monomorphisms. In particular this holds when $U$ is monadic, e.g. $U : R\mathbf{-Mod} \to \mathbf{Set}$, $U : \mathbf{CRing} \to \mathbf{Set}$ etc.) Now consider $U F(\emptyset \to X)$ for a non-empty set $X$. There are two cases:
- If $U F \emptyset = \emptyset$, then $U F (\emptyset \to X)$ is an injective map (trivially), so $F (\emptyset \to X)$ is a monomorphism in $\mathcal{C}$.
- If $U F \emptyset \ne \emptyset$, then there exists a map $X \to U F \emptyset$, hence a morphism $F X \to F \emptyset$ by adjoint transposition. But $F \emptyset$ is the initial object in $\mathcal{C}$, so there is a unique morphism $F \emptyset \to F X$; thus the composite $F \emptyset \to F X \to F \emptyset$ must be the identity, i.e. the morphism $F (\emptyset \to X)$ is split monic.
So $F : \mathbf{Set} \to \mathcal{C}$ indeed preserves all monomorphisms.
Best Answer
Let's take the following example:
Fix a field $k$. $\mathcal C :=\ _k\underline{\mathrm{Mod}}$ the category of vector spaces$_{/k}$, $\mathcal D := \underline{Set}$ the category of sets and $F : \mathcal C \to \mathcal D$ the forgetful functor, i.e. $F: V \mapsto V$ as a set and $(f : V \to W) \mapsto (f : V \to W)$ as plain mapping of sets.
Take some set $X$ in $\mathcal D$. What does it mean for a vector space$_{/k}$ $U(X)$ together with a morphism $L_X : X \to FUX$ to be free over $X$?
Choose a vector space $W$ and a set-theoretic mapping $f : X \to W$. If $(U(X), L_X)$ is free over $X$ there is a unique morphism $g : U(X) \to W$ such that $F g \circ L_X = f$. So put more concretely:
There exists a unique $k$-linear mapping $g : U(X) \to W$ such that $Fg \circ L_X = f$. But keep in mind that $Fg$ still is the same mapping - we just "forgot" that is actually is $k$-linear.
$L_X$ is a set theoretic mapping that associates to each $x \in X$ a vector $v_x$ in $U(X)$.
$Fg \circ L_X = f$ just says that $g : v_x \mapsto f(x) \in W$.
The fact that $g$ is unique with this property means: If any $k$-linear map $h$ sending $v_x \mapsto f(x)$, then $g = h$. Hence $g$ is uniquely determined by the images of $v_x$, i.e. $\{v_x | x \in X\}$ is a basis of $U(X)$.
Hence being free over $X$ in this particular case means that $L_X$ identifies $X$ with a $k$-basis of $U(X)$.
This explains the terminology: $U(X)$ is free with basis $L_X(X)$.
As a generaly philosophy behind this you should think of $X$ as some kind of "basis" of $U(X)$ in the sense that any morphism $U(X) \to A$ arises in a unique way by choosing a (potentially simpler morphism) $X \to FA$ and "extending" that to the whole of $U(X)$.
In many cases (like the examples you mentioned), you'll want $\mathcal D = \underline{Set}$ and $F : \mathcal C \to \mathcal D$ some kind of "forgetful" functor, that associates to the objects and morphisms of $\mathcal C$ their "set-theoretic" versions, where one "forgets" the "extra-structure" on $\mathcal C$.