# The Yoneda lemma and a natural bijection

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 $$A$$: $$K(A).$$

• That doesn't make sense, $\hom(A,-)$ is a functor and $SY$ a set, so what is $\hom(A,-)\to SY$ ? – Maxime Ramzi May 5 '19 at 16:04
• What is $Y{}{}$? – Eric Wofsey May 5 '19 at 16:04
• I gave a reference in the last line of my question. I have just rewritten that from that page 82 in [Adamek Rosicky] book. – user122424 May 5 '19 at 16:10
• 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. – Marc Olschok May 8 '19 at 13:27

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 $$\mathbf{Set}^{\cal A^{op}}$$.