# The Yoneda Lemma and “Completely Characterize”

I am reading Roman's new book "An Introduction to the Language of Category Theory" (2015). On p.60 he starts to describe the Yoneda Lemma. In short he says:

Let $C$ be a category and $a \in C$ is an object. Given a functor $H:C \longrightarrow Set$ and a natural transformation $\lambda: hom_C(a,-) \Longrightarrow H$. Then there is an element $p \in Ha$ such that $\lambda _x (g) = Hg(p)$ for all arrows $g:a \longrightarrow x$ in $C$. The element $p \in Ha$ "completely characterizes" the natural transformation $\lambda$.

I do not understand the last sentence, and I want to ask: 1) what does "completely characterizes" means in mathematics in general, and 2) what do I have to do to prove this statement of Roman?

I hope someone can help me with this.

Here is something that will hopefully help. If you plug in $a$ to the functors $hom_C(a,-)$ and $H$, then the natural transformation $\lambda$ gives a map

$$\lambda_a:hom_C(a,a)\rightarrow Ha$$

There is a natural element to consider in $hom_C(a,a)$, namely the identity map. Let $p = \lambda_a(id\rvert_a)$. Then write down the commutative square coming from $\lambda$, for an arrow $g:a\rightarrow x$ and you will see the property $\lambda_x(g) = Hg(p)$.

The sentence that $p$ completely characterizes the natural transformation $\lambda$, comes from the equation $\lambda_x(g) = Hg(p)$. That is, we can recover the natural transformation $\lambda$ from knowing $p$ in the following sense. Given $p$, we can define a natural transformation, $\hat{\lambda}:hom_C(a,-)\rightarrow H$ by the rule for $x$ in $C$ and $g: a\rightarrow c$, define

$$\hat{\lambda}_x(g) := Ha(p)$$

Then $\hat{\lambda}$ is the natural transformation $\lambda$. In this sense, $\lambda$ is determined by $p$.

• Did you mean: “Given $p$, we can define a natural transformation $\hat {\lambda}: hom_C(a,-) \rightarrow H$ by the rule for $x$ in $C$ and $g:a \rightarrow x$ define $\hat{\lambda}_x(g) := Ha(p)$. Then $\hat{\lambda}$ is the natural transformation $\lambda$. In this sense $\lambda$ is determined by $p$.” ? – Steenis Feb 7 '17 at 18:15
• In my former comment I meant "Given any $p$ in $Ha$," ? Sorry about that. But, I am sorry, I do not understand why $\hat \lambda$ equals $\lambda$, because your $p$ is different from Roman's $p$ which is $p=\lambda _a (1_a)$ and your $p$ is any element of $Ha$. – Steenis Feb 7 '17 at 18:52
• I'm not saying given any $p$ in $Ha$, I'm saying the $p$ that is $\lambda_a(1_a)$. If you have this exact $p$ then you can recover $\lambda$. – Will Dukeminier Feb 7 '17 at 22:41
• I think that I get it now. I proved that $\hat{\lambda} = \lambda$ as follows. For all $x \in C$ and all $k:a \longrightarrow x$ we have $\hat{\lambda}_x(k) = \hat{\lambda}_x(k \circ 1_a) = \hat{\lambda}_x \circ k(1_a) = Hk \circ \lambda_a(1_a) = Hk(p) = \lambda_x(k)$. This is valid for all $k \in hom_C(a,x)$, thus $\hat{\lambda}_x = \lambda_x$ for all $x \in C$, and thus $\hat{\lambda} = \lambda$. I hope this is correct. Thank you for all your help. – Steenis Feb 10 '17 at 14:26
• Rethinking my former comment, this is maybe a better proof: $\hat{\lambda}_x(k) = Hk(p)$ by definition and we already have $\lambda_x(k) = Hk(p)$. Thus $\hat{\lambda}_x(k) = \lambda_x(k)$ for all $k \in hom_C(a,x)$ and all $x \in C$, thus $\hat{\lambda}_x = \lambda_x$ for all $x \in C$, and thus $\hat{\lambda} = \lambda$. I hope this is correct. – Steenis Feb 11 '17 at 10:24