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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.

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  • $\begingroup$ That doesn't make sense, $\hom(A,-)$ is a functor and $SY$ a set, so what is $\hom(A,-)\to SY$ ? $\endgroup$ – Maxime Ramzi May 5 '19 at 16:04
  • $\begingroup$ What is $Y{}{}$? $\endgroup$ – Eric Wofsey May 5 '19 at 16:04
  • $\begingroup$ I gave a reference in the last line of my question. I have just rewritten that from that page 82 in [Adamek Rosicky] book. $\endgroup$ – user122424 May 5 '19 at 16:10
  • $\begingroup$ You forgot to give the meaning of $Y$. I happen to have a copy of the book, where the notation is introduced on page 3, but for most people it is not an obvious guess and a reference to page 82 is usually not enough; also Google may restrict access, depending on the IP. So it is much better to explain all relevant notation in the question itself. $\endgroup$ – Marc Olschok May 8 '19 at 13:27
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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 $\mathbf{Set}^{\cal A^{op}}$.

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