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Let $\mathcal{C}$ be a category, let's denote by $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$ the category of presheaves of sets defined on $\mathcal{C}$ and natural transformations.

I want to prove that $\mathcal{C}$ is isomorphic (or at least equivalent) to the full subcategory of $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$ given by representable functors.

The only things I know are:

1) Yoneda embedding is fully faithful;

2) Yoneda embedding is injective on objects;

3) Yoneda's Lemma.

Do you think I can prove what I want using 1), 2), 3)? Could you suggest me how?

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2 Answers 2

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Well you should know that in order to be an equivalence of category it's necessary and sufficient for a functor $F$ to be fully faithful and that each object of the target category is isomorphic to $FA$ for a suitable $A$ in the source category. This is definitely your case.

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  • $\begingroup$ so i don't need the fact that Yoneda embedding is injective on objects? Do you confirm this? $\endgroup$
    – bateman
    Jun 21, 2013 at 10:41
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    $\begingroup$ That fact gives you that $\mathcal{C}$ is actually isomorphic to a full subcategory of $Set^{C^{op}}$, namely the one made of the hom-functors. $\endgroup$ Jun 21, 2013 at 13:04
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Fully faithful functor, which is injective on objects, is an isomorphism with its image(which is full subcategory of its target category). So your knowledge is sufficient to show that they are isomorphic.

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  • $\begingroup$ this is precisely what i'm not able to prove: how injectivity on objects implies that the equivalence is actually an isomorphism $\endgroup$
    – bateman
    Jun 21, 2013 at 16:12
  • $\begingroup$ @bateman: no need to think about equivalence! it is a simple fact about fully faithful functors. $\endgroup$
    – Oskar
    Jun 21, 2013 at 16:13
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    $\begingroup$ @bateman: The first step: prove that functor is an isomorphism iff it is fully faithful and bijective on objects. The second step: prove(its obvious) that corestriction of functor on its image is surjective on objects. Now everything is done! $\endgroup$
    – Oskar
    Jun 21, 2013 at 16:16

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