The Yoneda lemma and a natural bijection

Question

Let $S\colon\mathbf {Set}^{\cal A^{op}}\to \mathbf{ Set}$ be a functor. How does it follow from the Yoneda lemma that the following is a natural bijection:

$\underline{\hom(A,-)\to SY \quad\quad\quad }$

$\hom(\hom(,-A),-)\to S$

Here $Y\colon {\cal A} \to \mathbf{Set}^{\cal A^{op}}$ is the Yoneda embedding with $Y(A)=\hom(-,A)$.

I know that the Yoneda lemma states that there is a natural bijection for natural transformations from a $\hom(A,-)$ functor to a functor $K$ with this set:the image of $K$ under $A$: $K(A).$

Here is the statement of the above bijection in question.

Answer 1

The Yoneda Lemma is used twice. In the following I avoid the notation $\hom{}$, so that the different categories involved are easier to distinguish. As in the (edited) question, $Y\colon {\cal A} \to \mathbf{Set}^{\cal A^{op}}$ will denote the Yoneda embedding with $YA = {\cal A}(-,A)$.

By the Yoneda Lemma, applied to ${\cal A}$, a natural map

$\varphi\colon {\cal A}(A,-) \to SY$

corresponds to an element of $(SY)A$. But because $(SY)A = S(YA)$, such an element corresponds to a natural map

$\hat\varphi\colon \mathbf{Set}^{\cal A^{op}}(YA,-) \to S$

by the Yoneda Lemma, applied to $\mathbf{Set}^{\cal A^{op}}$.