Free functor from $Set$ to $Set^*$ I would like to find a left adjoint functor to forgetful functor $U:Set^* \rightarrow Set$. I think it must be ,,free'' functor $F:Set \rightarrow Set^*$, but I have problem to construct this and I can't find it in books and Internet. Could anyone give me a hint? I would be very grateful.
PS:Is there also a right adjoint?
 A: Pick a set $s$. We want to construct a functor $F\colon \mathrm{Set}\rightarrow\mathrm{Set}^\ast$ such that for all pointed sets $(t,t_0)$ the isomorphism
$$\hom_{\mathrm{Set}^\ast}(F(s),t)\cong \hom_{\mathrm{Set}}(s,U(t))$$
holds. Denoting by $\mathrm{pt}$ the terminal object in $\mathrm{Set}$, consider the construction
$$F(s):=(s\sqcup \mathrm{pt},\mathrm{pt})$$
(where I denote by $\mathrm{pt}$ both the one-point set and the unique element it contains).
Now the above isomorphism is given by
$$f\mapsto f\vert_s$$
with inverse 
$$g\mapsto g\sqcup[\mathrm{pt}\mapsto t_0].$$
I'll let you check by yourself that this construction is natural in $s$ and $t$.
Edit: This proves that $U$ admits a left adjoint $F$ (which was asked for at the time this answer was created). As for a right adjoint, see Kevin's answer.
A: $U$ admits no right adjoint, as it does not preserve colimits. For instance, the coproduct of pointed sets is the so-called wedge product, $(S,s)\vee (T,t)=S\sqcup T/(s\sim t)$, which does not map to the disjoint union under $U$.
