证明复合函数的满射性及特定映射F的满射性

Let's work through these two surjectivity proofs step by step—they're great examples of applying core function theory concepts!

First, let's recap the definition of a surjective (onto) function: A function $f: A \to B$ is surjective if for every $b \in B$, there exists at least one $a \in A$ such that $f(a) = b$.

For composite functions, the key result here is:

If $f: A \to B$ is surjective and $g: B \to C$ is surjective, then their composition $g \circ f: A \to C$ is surjective.

Proof: #

1. Take any arbitrary element $c \in C$.
2. Since $g$ is surjective, there must exist some $b \in B$ where $g(b) = c$.
3. Since $f$ is surjective, for this $b$, there exists some $a \in A$ where $f(a) = b$.
4. Now compute the composition: $(g \circ f)(a) = g(f(a)) = g(b) = c$.

We've shown that for every $c \in C$, there's an $a \in A$ mapping to it via $g \circ f$. Thus, $g \circ f$ is surjective.

Note: If either $f$ or $g$ fails to be surjective, the composite function will not be surjective (this is a common edge case to keep in mind).

First, let's restate all given definitions clearly:

• $\mathbb{N}$ denotes the set of natural numbers (we'll assume $\mathbb{N} = {1,2,3,...}$ here—adjusting for $\mathbb{N} = {0,1,2,...}$ doesn't change the core logic).
• $H(x) = 2x + 1$ (maps $\mathbb{N}$ to the set of all odd natural numbers, so $H$ is injective but not surjective).
• $G(x) = \begin{cases} x - 1 & \text{if } x > 1 \ 3 & \text{if } x = 1 \end{cases}$
• $F(f) = G \circ f \circ H$, where $\mathbb{N}^\mathbb{N}$ is the set of all functions from $\mathbb{N}$ to $\mathbb{N}$.

To prove $F$ is surjective, we need to show: for every function $k \in \mathbb{N}^\mathbb{N}$, there exists some $f \in \mathbb{N}^\mathbb{N}$ such that $F(f) = k$ (i.e., $G(f(H(x))) = k(x)$ for all $x \in \mathbb{N}$).

Proof: #

1. Define $f$ strategically:

   • For odd numbers (since $H(x)$ only outputs odd numbers):
     • If $k(x) \neq 3$, set $f(2x+1) = k(x) + 1$. This works because $k(x)+1 > 1$, so $G(f(2x+1)) = (k(x)+1) - 1 = k(x)$.
     • If $k(x) = 3$, set $f(2x+1) = 1$. This works because $G(1) = 3 = k(x)$.
   • For even numbers (which are never in the image of $H$), we can define $f$ arbitrarily—for simplicity, set $f(y) = 1$ for all even $y \in \mathbb{N}$ (any natural number choice here is valid, since these values don't affect $F(f)$).

2. Verify $F(f) = k$:
   For any $x \in \mathbb{N}$:

   • If $k(x) \neq 3$: $F(f)(x) = G(f(H(x))) = G(k(x)+1) = (k(x)+1)-1 = k(x)$.
   • If $k(x) = 3$: $F(f)(x) = G(f(H(x))) = G(1) = 3 = k(x)$.

Since $k$ was an arbitrary function in $\mathbb{N}^\mathbb{N}$, we've shown that every $k$ has a preimage $f$ under $F$. Thus, $F$ is surjective.

内容的提问来源于stack exchange，提问作者Mat Research