Yoneda argument

I don’t want to repeat the Yoneda lemma, but one of the important corollary should be emphasized. Roughly speaking, given any object c in the category \mathcal{C}, c is determined by its morphisms up to isomorphism. That is, the natural isomorphism between hom-sets corresponds to the isomorphism between objects, i.e. hom(-, c) \cong hom(-, d) \Leftrightarrow c \cong d. How do we prove it? Well, let’s see a simpler analogy in partial orders.

Lemma. \uparrow x = \uparrow y if and only if x = y.
Proof. If \uparrow x = \uparrow y , then a \in \uparrow b and b \in \uparrow a which imply a \leq b and b \leq a. Hence a = b. The converse is quite trivial.

First, we can prove it by brute force method.

Lemma. hom( - , c ) \cong hom( - , d ) if and only if c \cong d.
Proof. Consider the following diagrams,

by replacing f with \phi^{-1}_{d}(id_{d}), we will see that id_{d} = \phi_{c}(id_c) \circ \phi^{-1}_{d}(id_{d}) and we will have the other isomorphism by replacing f with \phi_{c}(id_c) and reversing the arrows.

However, the diagram above is exactly used in Yoneda lemma, and of course we can use Yoneda lemma to prove it.

First recall that Yoneda lemma states that Nat ( hom ( - , r ), K ) \cong Kr where K is a Set-valued functor and r is an object in a locally small category \mathcal{C}. Let K be the hom-set functor hom(-, s). It gives the isomorphism between Nat(hom(-, r), hom(-, s)) \cong hom(r, s). Hence, it indeed tells you that Y(d) = hom(-, d) is fully faithful. Therefore, we have the second proof.

Proof2. Given that Y is fully faithful, thus the natural isomorphisms Y ( d ) \cong Y ( c ) actually correspond to the isomorphism d \cong c and vice verse. That’s it!


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