I have a question about Yoneda lemma etc. In class, we proved the following lemma:
Lemma: Let $F: \mathcal{C} \rightarrow \text{Sets}$ be a functor. There is a bijection between the natural transformations from $h^{X}$ to $F$ and the elements of $F(X)$.
Surjectivity is shown as: let $a \in F(X)$. We prove there exists some natural transformation $\eta: h^{X} \rightarrow F$ sending $1_{X}$ in $h^{X} (X)$ to $a$. Let $Y$ be any object in $\mathcal{C}$ and define $$ \eta_{Y} : h^{X} (Y) \rightarrow F(Y): f \mapsto F(f)(a). $$
I understand the proof, but there is a remark following the proof, saying:
Remark: From the proof of Yoneda lemma, it is easy to see that if $F$ is a representable functor, i.e. it is naturally isomorphic to some $h^{X}$, then there is no need to write down a natural isomorphism $\eta: h^{X} \rightarrow F$ explicitly. It is determined by $a : = \eta_{X} (1_X).$ So we will often say $F$ is represented by $(X,a)$ instead.
I'm not sure if I understand this remark. Is the $a$ in the $(X,a)$ unique? Let's say I have the following problem:
problem: Consider the forgetful functor. Show that $\text{Forget}: Ab \rightarrow \text{Sets}$ is represented by $(\mathbb{Z}, 1)$. Why? What is the natural equivalence?
Attempt: I would define a natural transformation $\eta: h^{\mathbb{Z}} \rightarrow \text{Forget}.$ Then for any $m \in \text{Forget}(\mathbb{Z})$ and any $G \in \text{Ob}(Ab)$ (this is an abelian group) I would define $$\eta_G : h^{\mathbb{Z}} (G) \rightarrow \text{Forget} (G): f \mapsto \text{Forget}(f)(m) = f(m). $$ Then I easily see that $\eta_{\mathbb{Z}} (1_{\mathbb{Z}}) = m. $
So I would say that the forgetful functor can be represented by $(\mathbb{Z}, m)$ where $m$ is any integer.
My question is: what is the $a$ in $(X,a)$? Is this supposed to be a unique element? And how does one in general show that a functor is representable on certain categories?
The Yoneda Lemma asserts that to each natural transformation $\eta: h^{X} \rightarrow F$, there corresponds a unique element, $a$, of $F(X)$, via the natural bijection $\theta:\eta\mapsto a:=\eta_X(1_X).$ Now, if $\eta$ happens to be an isomorphism; that is, if $F$ is representable, then it is determined by $a$ because $\eta=\theta^{-1}(a).$