Proof. Let $\hat{A}_0$ be the type of representable? objects of $\mathit{Set}^{A^{\mathrm{op}}}$, with hom-sets inherited from $\mathit{Set}^{A^{\mathrm{op}}}$. Then the inclusion $\hat{A} \to \mathit{Set}^{A^{\mathrm{op}}}$ is fully faithful and an embedding on objects?. Since $\mathit{Set}^{A^{\mathrm{op}}}$ is a category by Theorem 9.2.5 (see functor precategory), $\hat{A}$ is also a category. Let $A \to \hat{A}$ be the Yoneda embedding. This is fully faithful by corollary 9.5.6 (See Yoneda embedding), and essentially surjective by definition of $\hat{A}_0$. Thus it is a weak equivlance. $\square$

$\hat{A}_0\equiv \sum_{(F : \mathit{Set}^{A^{\mathrm{op}}})} \sum_{(a:A)} \mathbf{y} a \cong F$

This has the unfortunate side affect of raising the universe level. If $A$ is a category in a universe $\mathcal{U}$, then in this proof $\mathit{Set}$ must at least be as large as $\mathit{Set}_{\mathcal{U}}$. Hence the category $\mathit{Set}_{\mathcal{U}}$ is in a higher universe than $\mathcal{U}$ hence $\mathit{Set}^{A^{\mathrm{op}}}$ must also be in a higher universe and finally $\hat{A}$ is also in a higher universe than $\mathcal{U}$.

Now this can all be avoided by constructing a higher inductive type$\hat{A}$ with constructors:

A function $i : A_0 \to \hat{A}_0$.

For each $a,b:A$ and $e: a \iso b$, an equality $j e : i a = i b$

For each $a : A$, an equality $k(1_a)=refl_{i a}$.

For each $a,b,c:A$, $f : a \cong b$, and $g : b \cong c$, an equality $j(g \circ f)=j(f) \cdot j(g).$

1-truncation: for all $x,y:\hat{A}_0$ and $p,q: x = y$ and $r,s : p = q$, an equality $r=s$.

If we ignore the last constructor we could also write the above as $\| \hat{A}_0 \|_1$.

We now go on to build a category $\hat{A}$ with a weak equivalence$A \to \hat{A}$ by taking the type of objects as $\hat{A}_0$ and defining hom-sets by double induction?. The advantage of this approach is that it preserves universe levels, there are a lot of things to check but it is an easy proof. The kind of proof that is well suited to a proof assistant. For the complete proof see Theorem 9.9.5 of the HoTT book.