**Exercise 4.1.29**

**Theorem**:
The forgetful functor is representable.
In particular .

*Proof*:
First, we establish the following fact.
For a commutative ring and element , define the *evaluation map* by , the map that evaluates the given polynomial at inside .
Note that .
This is indeed the *unique* homomorphism that sends .
This is straightforward to see:
Suppose were a homomorphism with .
Let .
Then,

and thus .

The preceding fact is most of the content of the proof. Indeed, define the natural transformation with components taking an element to the map . We verify naturality: Let be a ring homomorphism. We must have . Again is just left-composition by , and is just . So, let . Applying the right side to yields . Applying the left side to yields . But of course,

and . So is natural and .