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Saying (see here): "Tell me who your friends are, and I will tell you who you are."

Yoneda lemma: Let $X, Y$ be two objects of a category $C$ and for each object $A\in C$, let $f_A\colon \mathsf{mor}(A, X)\to \mathsf{mor}(A, Y)$ be a function. Furthermore, assume that this condition holds:$$ f_B(x\circ h) = f_A(x)\circ h\text{ for all $A, B \in C$, $h\colon B\to A$ and $x\colon A\to X$}.$$ Then there is exactly one morphism $f\colon X\to Y$ such that $f_A(x) = f\circ x$ for all objects $A$ and morphisms $x\colon A\to X$.

Question: What has the Yoneda lemma to do with the above saying? If my formulation of the Yoneda lemma hasn't much to do with the saying, how can one reasonable reformulate the Yoneda lemma so that its relation to the above saying is apparent?

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To each object $A$ of a category $\mathcal{C}$, we can define a presheaf (a functor $\mathcal{C}^\text{op} \to \mathbf{Set}$) $\mathbf{y}A$ by the formula

$$ \mathbf{y}A(x) = \hom(x, A) $$

Similarly, for an arrow $f : A \to B$ of $\mathcal{C}$, the same formula would define a natural transformation $\mathbf{y}f : \mathbf{y}A \to \mathbf{y}B$.

In this way, $\mathbf{y}$ is a functor $\mathcal{C} \to \mathbf{Set}^{\mathcal{C}^\text{op}}$ called the Yoneda embedding.

The content of the yoneda lemma is that this embedding is full and faithful. If we restrict to just the representable presheaves (which are precisely the ones in the image of $\mathbf{y}$) then we get an equivalence of categories

$$ \mathcal{C} \equiv \text{RepresentablePresheaves}(\mathcal{C}) $$

In light of this equivalence, we have the principle

An object $A$ is basically the same thing as the representable presheaf $\mathbf{y}A = \hom(-, A)$

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  • $\begingroup$ Of course, this can also be said in a more elementary, maybe more consumable, way, but the fact is "as stated". If one reads that statement as being "too un-elementary", that fine, and understandable. Nevertheless, it is not so hard to parse it, and it is exactly correct... and worth parsing, even if one wants to rewrite it in more immediately-elementary terms. $\endgroup$ – paul garrett Mar 21 '17 at 23:45

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