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.
Products in many categories are said to possess a universal mapping property- formally, a product of two objects (say $A$ and $B$), is another object (say $P$) along with two arrows: $p_1:P \to A$ and $p_2:P \to B$ such that if $C$ is any other object (in our category), along with any pair of arrows, $f_1:C \to A, f_2: C \to B$, then there exists a UNIQUE arrow $\phi:C \to P$ such that:
$p_1\circ\phi = f_1\\p_2\circ\phi = f_2.$
The "baby example" of this construction is the cartesian product in the category $\mathbf{Set}$. Indeed, if we set, for two sets $A,B$:
$P = \{(a,b): a \in A, b \in B\}$, and take:
$p_1((a,b)) = a\\p_2((a,b)) = b,$
Then given any pair of maps $f_1:C \to A, f_2:C \to B$ for an abitrary set $C$, we can let:
$\phi(c) = (f_1(c),f_2(c))$, for any $c \in C$, and it's clear we have:
$(p_1\circ\phi)(c) = p_1(\phi(c)) = p_1(f_1(c),f_2(c)) = f_1(c)$ for ALL $c \in C$, and similarly:
$(p_2\circ \phi)(c) = f_2(c)$.
So certainly the map $\phi$ exists (for "this product"). On the other hand, if we have a map $\psi: C \to A \times B$ such that:
$p_1 \circ \psi = f_1$ and $p_2\circ \psi = f_2$, then from $p_1\circ \psi = f_1$ we know that:
$\psi(c) = (f_1(c),-)$, and similarly we know that $\psi(c) = (-,f_2(c))$, that is: $\psi$ has to be $\phi$, so $\phi$ is unique.
Now it should be pretty clear that this isn't "the only product" we can make: for example, we could take:
$P' = B \times A$ with $p_1:P' \to A$ given by $p_1'((b,a)) = a$, and similarly for $p_2'$.
In this case, we have two unique maps $\phi: P \to P'$, and $\phi': P' \to P$, and since $\phi((a,b)) = (b,a)$ and $\phi'((b,a)) = (a,b)$ fulfil the requirements, they must be the maps the universal property describes. Here, it's clear these are isomorphisms (in $\mathbf{Set}$, isomorphisms are just bijections), as these maps are inverses of each other.
The universal property invoked here is often paraphrased as: "the product map (what we are calling $\phi$, and is often written $f_1 \times f_2$) factors through the projections (what we are calling $p_1,p_2$)".
Now, it's "intuitively clearer" to regard the cartesian product of sets to be "the product" of $A$ and $B$. Note how the emphasis is on the elements in each ordered pair. In the construction above, the emphasis is on the mapping $\phi$, and the projection maps $p_1,p_2$, we don't even "look inside" the sets to see who lives there.
There is a similar factorization happening in your field of fractions example: any embedding $f$ of $D$ in any field $F$ factors through the embedding $i: D \to Q$. This captures (in this case) the sense of $Q$ being the "smallest" field we can insert $D$ into.
Best Answer
This is an interesting question as much about English language usage as mathematics.
First,
simply sounds wrong. That's reason enough to stick with the conventional "a". You don't want your readers or listeners to stumble over unfamiliar usage.
I can venture a reason. When you say "the" you are assuming existence, which is what your sentence is trying to assert. The proof usually runs along the lines "here is one, and in fact it's the only one". Using "a unique" somehow suggests the two separate steps.
As for
The same unfamiliarity holds. You could say
you could say
but it wouldn't sound exactly right.
Grammatically, I read "up to isomorphism" as an adverb phrase modifying the adjective "unique" which in turn modifies the noun that names the essentially unique thing you are writing about.