Definition of the evaluation map as a universal element of a representation This is exercise Exercise 2.3.iii from Riehl's "Category Theory in Context":

The set $B^A$ of functions from a set $A$ to a set $B$ represents the contravariant functor ${\rm Set}(-\times A,B):{\rm Set}^{\rm op}\to{\rm Set}$. The universal element for this representation is a function
$${\rm ev}:B^A\times A\to B$$
called the evaluation map. Define the evaluation map and describe its universal property, in analogy with the universal bilinear map $\otimes$ of Example $2.3.7$ (the tensor product).

I got the description of the universal property, but I'm having trouble with the definition of the map itself. Searching online, I found that it is given by:
$${\rm ev}(g,a) = g(a)$$
for some function $g: A \rightarrow B$ and $a \in A$.
What I'm not sure about is if this definition is forced by the information given. For instance, in the example Riehl mentions, the universal bilinear map $\otimes: V \times W \rightarrow V \otimes_k W$ is never defined explicitely; it's just stated that it exists for a given natural isomorphism and the universal property is described.
My question is then: Should I be able to recover the definition of ${\rm ev}$ from Yoneda (or something else)?
 A: Consider $B$ and $A$ to be fixed for a moment. Then an element $x\in \operatorname{Hom}_{\mathrm{Set}}(B^A\times B,B)$ is exactly the same thing as a transformation $$\Phi(x)_C \colon \operatorname{Hom}_{\mathrm{Set}}(C,B^A)\to \operatorname{Hom}_{\mathrm{Set}}(C\times A,B)$$
naturally in $C$. This is because of the Yoenda lemma. Now the transformation $\Phi(x)$ associted to $x$ will be a natural isomorphism, precisely when the function $x$ satisfies the correct universal property of an evaluation map.
For your question: Fixing an evaluation $\mathrm{ev} \colon B^A\times B\to B$ with the correct universal property is the same as fixing a bijection $\operatorname{Hom}_{\mathrm{Set}}(C,B^A)= \operatorname{Hom}_{\mathrm{Set}}(C\times A,B)$ which varies naturally in $C$ (the sets $A$ and $B$ are fixed). For example, postcompising $\mathrm{ev}$ with an automorphism of $B$ will give you a different evaluation map, and also a different natural isomorphism between the $\operatorname{Hom}$-sets.
Once the natural bijection $\Phi \colon \operatorname{Hom}(-,B^A)=\operatorname{Hom}(-\times A,B)$ is fixed, you do have enough information to find $\mathrm{ev}$. It is then the element  $\Phi_{B^A}(\mathrm{id})\in \operatorname{Hom}(B^A\times A,B)$. There is a standard natural bijection between the $\operatorname{Hom}$-sets called currying. What happens, when you curry the identity function of $B^A$?
