Universal properties and representable functors

Assume $\mathcal C$ is locally small and let $F:\mathcal C\to\mathbf{Set}$ be covariant. An object $A$ represents $F$ when there is a natural isomorphism from $\mathcal C(A,-)$ to $F$. For every target $X$, this gives a bijection between arrows out of $A$ and elements of $F(X)$. Naturality requires that changing the target by $u:X\to Y$ corresponds to applying $F(u)$ to the encoded datum. It ties all these bijections into a single construction.

The contravariant version uses $\mathcal C(-,A)$ instead. Then elements of the functor at $X$ are encoded by arrows from $X$ into the representing object. These are different variance conventions, and their universal elements have opposite factorization directions. Before attempting a representation, decide whether the intended data travel forward with a map, as ordered pairs do, or pull back against it, as subsets do under inverse image.

Given a covariant representation, evaluate it on the identity of $A$. The resulting element $a\in F(A)$ is universal: every $x\in F(X)$ is uniquely $F(f)(a)$ for some $f:A\to X$. Conversely, such a universal element defines the representing bijections by this formula. Their naturality follows from the functor law. The word unique does serious work: existence alone would describe a supply of maps without explaining whether different maps encode the same datum.

This reformulation is often easier to use than a large family of bijections. It asks us to build one prototype datum at one object and then show that every other datum is obtained from it in exactly one way. To disprove universality, it suffices to find a datum with no factorizing map or two distinct maps with the same encoded datum. Cardinality is a useful finite obstruction, but does not check compatibility across targets or determine the encoding itself.

Suppose $(A,a)$ and $(B,b)$ are two covariant universal-element presentations of the same functor. Universality gives a unique $s:A\to B$ carrying $a$ to $b$ and a unique $t:B\to A$ carrying $b$ back to $a$. Their composites preserve the corresponding universal elements, as do the identity maps. Uniqueness in the factorization problem therefore forces the composites to be identities. This proves an isomorphism without inspecting an underlying set or choosing coordinates.

The uniqueness statement includes compatibility with the universal elements. Bare representing objects may admit additional automorphisms that move the chosen element. Consequently, saying that any two representations have exactly one isomorphism, without this qualification, is too strong. A construction can be canonical through its mapping property even when its concrete realization uses names, tags, quotient classes, or another arbitrary presentation. Those choices disappear under the uniquely compatible comparison.