是否存在满足特定复合等式与不等关系的非恒等函数或函数类别？

Hey there! Let’s tackle your two function-related questions one by one, with concrete examples where possible.

Question 1: Existence of Functions/Categories Meeting Unspecified Conditions #

Since you didn’t spell out the exact conditions for the first question, I’ll cover a few common, non-trivial function categories that fit interesting criteria:

• Idempotent functions: Functions where applying them twice gives the same result as applying them once (e.g., the floor function for real numbers, or projecting a 3D vector onto its x-axis component).
• Involutions: Functions that are their own inverse (so $f(f(x)) = x$) but aren’t the identity (e.g., $f(x) = -x$ for integers, or swapping the first and last elements of a list).
• Injective/surjective functions: Functions that are one-to-one (injective) or onto (surjective) but not both (e.g., $f(x) = 2x$ on natural numbers is injective but not surjective).

If you had a specific condition in mind, feel free to clarify!

Question 2: Non-Identity S, U Satisfying Composition Conditions #

The short answer: Yes, such function categories exist (and we can get close with concrete functions if we relax the "for all T" rule slightly).

Strict Categorical Example (Meets "For All T") #

To satisfy the strict requirement of working for any function (or functor, in categorical terms), we can use category theory constructs:

• Let’s work with the category of ordinary sets ($\text{Set}$) and the category of pointed sets ($\text{Set}_*$, where each set has a designated "base point" like $(0,0)$ in a 2D plane).
• Let $U$ be the inclusion functor: it takes a plain set $X$ and turns it into a pointed set by picking an arbitrary base point (e.g., mapping $\mathbb{N}$ to $(\mathbb{N}, 0)$). $U$ is non-identity because it adds structure to sets.
• Let $S$ be the forgetful functor: it takes a pointed set and strips away the base point, returning the plain underlying set. $S$ is non-identity because it removes structure.

Now, for any functor $T$ that maps sets to sets:

• $S(T(U(X))) = S(T((X, x_0))) = T(X)$ (since $T$ preserves set structure, and $S$ discards the base point), which matches the requirement $S(T(U(k))) = T(k)$.
• $U(T(S((X, x_0)))) = U(T(X)) = (T(X), t_0)$ (adding a base point back to $T(X)$). This is not equal to $T(X)$ (the plain set), so $S(T(U(k))) \neq U(T(S(k)))$.

Concrete Function Example (Relaxes "For All T" to a Large Class) #

If you want plain functions (not categorical functors), we can’t satisfy the "for all T" rule strictly (it would force $S$ or $U$ to be identity). But for a large class of functions $T$, we can construct non-identity $S$ and $U$:

• Let $K = \mathbb{N}$ (natural numbers), $U(k) = 2k$ (maps every number to an even number, non-identity).
• Let $S(m) = m/2$ if $m$ is even, and $S(m) = 0$ if $m$ is odd (non-identity, and $S(U(k)) = k$ for all $k$).

For any linear function $T$ (like $T(k) = ak + b$ where $b$ is even), $S(T(U(k))) = S(T(2k)) = S(2ak + 2b) = ak + b = T(k)$, which meets the first condition. For these $T$:

• If $k$ is even: $U(T(S(k))) = U(T(k/2)) = 2*(a*(k/2) + b) = ak + 2b$, which is not equal to $T(k) = ak + b$.
• If $k$ is odd: $U(T(S(k))) = U(T(0)) = 2b$, which is not equal to $T(k) = ak + b$ for most $k$.

This satisfies the inequality requirement for a large class of functions.

内容的提问来源于stack exchange，提问作者Meekaa Saangoo