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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!

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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.

Here every corresponds with surjectivity and unique with injectivity.

Does this make things more clear for you?

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  • $\begingroup$ Perfect! This is exactly what I wanted to know. Thanks a lot. $\endgroup$ – Martin Adam May 8 '19 at 16:49
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab May 8 '19 at 20:01

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