Repeating the Yoneda embedding Let $\cal C_1$ be a category.
Let $Y_1 : \cal C_1 \rightarrow \operatorname{Set}^{\mathcal{C}_1^{\operatorname{op}}}$ be the Yoneda embedding. Name the category on the right side $\cal C_2$.
Now do this again. Let $Y_2 : \cal C_2 \rightarrow \operatorname{Set}^{\mathcal{C}_2^{\operatorname{op}}}$ be the Yoneda embedding of $\cal C_2$ inside its functor category. Call the category on the right side $\cal C_3$.
The question is:

Is $Y_2 : \cal C_2 \rightarrow \cal C_3$ an equivalence(or isomorphism) of categories?

 A: Let $\mathcal{C}$ be any category. Then the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is never an equivalence. Indeed, consider the functor $\mathcal{C}^\mathrm{op} \to \mathbf{Set}$ that sends everything to the empty set. This is not isomorphic to any representable functor, because $\mathcal{C}(-, c)$ is non-empty at $c$. Thus the Yoneda embedding is not essentially surjective on objects.
A: THE FOLLOWING PROOF DOES NOT WORK, but is left up as a challenge for anyone who wants to think about the error (hint: size is definitely an issue).
The Yoneda embedding satisfies the following universal property: if $\mathcal{D}$ is a category with all small colimits, and $\mathcal{C}\to\mathcal{D}$ is any functor, then there is an extension of $F$ to $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$ along the Yoneda embedding, which is unique up to natural equivalence.
It follows that $Y_2$ has an inverse (namely, the extension of $Y_1$ to $\mathcal{C}_3$), so it is indeed an equivalence of categories.
