求遗忘函子U:Set^*→Set的左伴随函子构造及右伴随存在性问题

Great question! Let's break this down step by step.

First, let's align on definitions: $Set^$ (the pointed set category) has objects that are pairs $(X, x_0)$ where $X$ is a set and $x_0 \in X$ (the "basepoint"). Morphisms are functions $f: (X, x_0) \to (Y, y_0)$ that preserve the basepoint: $f(x_0) = y_0$. The forgetful functor $U: Set^ \to Set$ maps each pointed set $(X, x_0)$ to its underlying set $X$, and each basepoint-preserving function to its raw underlying function.

You're correct that $U$ has a left adjoint—the free pointed set functor $F: Set \to Set^*$. Here's how to construct it clearly:

• On objects: For any set $S$, define $F(S)$ as the pointed set $(S \sqcup {*}, )$, where $\sqcup$ means disjoint union, and $$ is a formal element not present in $S$ (to avoid conflicts, you can use something like $(S, )$ instead of a generic $$ if needed). In simple terms, we're adding one new basepoint to the set $S$.
• On morphisms: For any function $f: S \to T$ in $Set$, define $F(f): F(S) \to F(T)$ by:
  $$
  F(f)(s) = \begin{cases}
  f(s) & \text{if } s \in S, \
  • & \text{if } s = .
    \end{cases}
    $$
    This preserves the basepoint because $F(f)() = *$, which is the basepoint of $F(T)$.

Verifying the Adjunction #

To confirm this is a left adjoint, we need a natural bijection:
$$
\text{Hom}{Set^*}(F(S), (X, x_0)) \cong \text{Hom}{Set}(S, U(X, x_0))
$$

• Forward direction: Take any basepoint-preserving function $g: F(S) \to (X, x_0)$. Since $g$ must map $*$ to $x_0$, its restriction to $S$ (i.e., $g|S: S \to X$) is a valid function in $\text{Hom}{Set}(S, X)$.
• Reverse direction: Take any function $h: S \to X$. Extend it to a basepoint-preserving function $g: F(S) \to (X, x_0)$ by setting $g(s) = h(s)$ for $s \in S$ and $g() = x_0$. This satisfies the basepoint rule, so it's a valid morphism in $Set^$.

This bijection works naturally for all $S$ and $(X, x_0)$, so $F \dashv U$ holds.

Short answer: No, the forgetful functor $U: Set^* \to Set$ does not have a right adjoint. Here's a straightforward proof by contradiction:

Suppose there exists a right adjoint $R: Set \to Set^$. By adjunction rules, we'd have a natural bijection:
$$
\text{Hom}{Set}(U(X, x_0), S) \cong \text{Hom}{Set^}((X, x_0), R(S))
$$
for all $(X, x_0) \in Set^*$ and $S \in Set$.

Now take $(X, x_0)$ as the terminal object of $Set^$: the singleton pointed set $({}, *)$. Then:

• $\text{Hom}{Set}(U({*}, *), S) = \text{Hom}{Set}({*}, S)$, which has exactly $|S|$ elements (one function for each element of $S$).
• $\text{Hom}_{Set^}(({}, ), R(S))$ has exactly 1 element: the only basepoint-preserving function maps the basepoint of $({}, *)$ to the basepoint of $R(S)$.

For these sets to be in bijection, we'd need $|S| = 1$ for every set $S$, which is obviously false (e.g., take $S = {1,2}$). This contradiction proves no such right adjoint can exist.

内容的提问来源于stack exchange，提问作者jpatrick