# Universal elements of a functor and representability

A functor $$F : \mathcal{C} \rightarrow \mathcal{Set}$$ is said to representable if it is naturally isomorphic to $$\mathcal{C}(A,–)$$ for some object $$A$$ of $$\mathcal{C}$$. By the Yoneda lemma, we know that natural transformations from $$\mathcal{C}(A,–)$$ to $$F$$ are in one-to-one correspondence with the elements of $$F A$$.

According to this Wikipedia article, the natural transformation induced by an element $$u \in F A$$ is an isomorphism if and only if $$(A,u)$$ is a universal element of $$F$$. A universal element of a functor $$F : \mathcal{C} \rightarrow \mathcal{Set}$$ is a pair $$(A,u)$$ consisting of an object $$A$$ of $$\mathcal{C}$$ and an element $$u \in F A$$ such that for every pair $$(X,v)$$ with $$v \in F X$$ there exists a unique morphism $$f : A \rightarrow X$$ such that $$(F f)\ u = v$$.

If the natural transformation induced by $$u \in F A$$ is indeed an isomorphism, is this due to the uniqueness property of the morphism $$f : A → X$$?

To clarfiy, in my head, we have:

$$α_X : \mathcal{C}(A,X) \rightarrow F X\\ α_X (f) = (F f)\ u$$

and we need to define the inverse of $$α_X$$, call it $$β_X : F X → \mathcal{C}(A,X)$$, but this can only be defined if $$α_X$$ is injective (i.e., $$f$$ satisfying the above equation is unique). Is this correct?

Many thanks!

Observe that under the prescription of $$\alpha_X$$ in your question the following statements are equivalent:
• for every $$v\in FX$$ there is unique $$f\in\mathcal C(A,X)$$ such that $$(Ff)(u)=v$$.
• for every $$v\in FX$$ there is unique $$f\in\mathcal C(A,X)$$ such that $$\alpha_X(f)=v$$.
• function $$\alpha_X:\mathcal C(A,X)\to FX$$ is bijective.