Basic question about Yoneda's lemma Out of curiosity, I am trying to self teach myself a bit of category theory and I have a question about Yoneda's lemma.
I recall the notation that I am using. I hope to not get it (too) wrong. Given a category $\mathbf{A}$ to every object $A$ we can assign a presheaf $H_A : A^{op} \rightarrow Set$, which is the set of maps $\mathbf{A}(-,A)$.
The Yoneda lemma says that, if we consider the category of preshaves $[\mathbf{A}^{op},Set]$ where morphisms are the natural transformations between functors, we have for every presheaf $X$ and every object $A$:
$[\mathbf{A}^{op},Set](H_A,X) \sim X(A)$ [1]
where I do not recall the (functorial) meaning of $\sim$, but as sets there is a bijection.
I hope the setting is correct, otherwise please correct me . Now my question, probably trivial:
Question: apart from functiorial properties, relation [1] says that if $X(A)$ is not the empty set, than there is always a natural transformation between $H_A$ and $X$. Told another way, for every $X : \mathbf{A}^{op} \rightarrow Set$ and for every $A$ such that $X(A)$ is not the empty set,there is always a natural transformation between $X$ and $H_A$. Is it correct ? Is it maybe trivial that it is so easy for a presheaf to be "connected naturally" (this expression is naif) with a representable $H_A$ ?
 A: The meaning of your $\sim$ is that there is a "natural isomorphism" between the two sides, that is, it is compatible with the maps induced by a natural transformation between two functors $X$ and $X'$, and with the maps induced by a function between two objects $A$ and $A'$.
Yes, it's correct that every element of $X(A)$ gives rise to a natural transformation between $\mathbf{A}(-, A)$ and $X$. It helps to think of what such a natural transformation consists of: a family of morphisms $\beta_a : \mathbf{A}(a, A) \to X(a)$ for each object $a$ of $\mathbf{A}$ which is "compatible with composition" in the sense that if $f : a \to b$ then $f^*\circ \beta_a = \beta_b \circ X(f)$, where $f^*$ is pre-composition with $f$.  When you do that it doesn't seem strange for there to be a connection between these natural transformations and elements of $X(A)$. After all, every $g : a \to A$ gives rise to $X(g): X(A) \to X(a)$, so there's an easy way to produce elements of $X(a)$, namely, evaluate at an element of $X(A)$.
You can see precisely how this works by tracing through a proof of the (dual) Yoneda lemma. Every $r \in X(A)$ gives rise to the natural transformation $\beta$ such that $\beta_a(g:a \to A) = X(g)(r)$.  Conversely every such natural transformation $\beta$ gives rise to the element $\beta_A(\operatorname{id}_A) \in X(A)$.  Yoneda is telling you that the only "natural" (informal sense) way to produce elements of $X(a)$ from morphisms in $\mathbf{A}(a, A)$ is to apply $X$ to get a function $X(A) \to X(a)$ and then evaluate somewhere.
