Free functor from $Set$ to $Set^*$

Question

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?

Answer 1

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.

Answer 2

$U$ 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 $U$.