About argument of forgetful functor 
Let $(\{0,1\},0) \in \text{Set}_*$ be the set $\{0,1\}$, equipped with the base point $0 \in \{0,1\}$. Prove that $(\{0,1\},0)$ represents the forgetful functor $U:\text{Set}_* \rightarrow \text{Set}$.

Forgetful functor is used for any functor that forgets structure, and whose codomain is the category of sets. For example, $U: \text{Group} \rightarrow \text{Set}$ sends a group to its underlying set and a group homomorphism to its underlying function.
The only thing I can think of for argument is the codomain does not contain base point, so $(\{0,1\},0)$ represents the forgetful functor. How to improve on this? Thanks.
 A: To show that $(\{0,1\},0)$ represents $U$, we want to show that
$$\text{Hom} \big ((\{0,1\},0), - \big ) \cong U(-).$$
But notice what a hom $f : (\{0,1\},0) \to (X,x)$ looks like! To preserve base points, we need to know that $f(0)= x$. But where can $1$ go? The answer is "anywhere"!
So homs $(\{0,1\},0) \to (X,x)$ are in bijection with a choice of $f(1) \in X$! That is, with $UX$!
We can make this isomorphism explicit by looking at

*

*$f \mapsto f(1)$
$\quad$ which takes $\text{Hom} \big ((\{0,1\},0), (X,x) \big )$ to $UX$

*$x' \mapsto \begin{cases} f(0) = x \\ f(1) = x' \end{cases}$ $\quad$which takes $UX$ to $\text{Hom} \big ((\{0,1\},0), (X,x) \big)$
Can you show this isomorphism is natural?

If more abstract is more your speed, you might consider the functor
$$P : \mathsf{Set} \to \mathsf{Set}_*$$
where $PX = (X \cup \{ 0 \}, 0)$ is the functor which "freely" adds a basepoint to $X$ (here we must assume $0 \not \in X$. If $0 \in X$ then we should rename it). Then you might check that $P \dashv U$ is an adjoint pair.
Do you see how you might use this to show that $(\{0,1\}, 0) = P \{1\}$ represents $U$?

I hope this helps ^_^
A: Given any pointed set $(S, *)$, what are the morphisms $(\{0, 1\}, 0)\to(S, *)$? They are precisely the functions $f\colon\{0, 1\}$ with $f(0)=*$ and with $f(1)\in S$ arbitrary. We may define a bijection $\varphi\colon\text{Hom}_*(\{0, 1\}, S) \to S$ (where the subscript signals that these are pointed functions) the following way: given any pointed function $f$, let $\varphi(f)=f(1)$. The inverse is given by $\theta(s)=f_s\colon \{0, 1\}\to S$, where $f_s(0)=*, f_s(1)=s$. I'll let you check the naturality as an exercise.
