Let $d(n)$ be the number of positive divisors of a positive integer $n$.  Let $\mathbb{N}$ be the set of all positive integers.  Say that a function $F$ from $\mathbb{N}$ to $\mathbb{N}$ is [i]divisor-respecting[/i] if $d(F(mn)) = d(F(m)) d(F(n))$ for all positive integers $m$ and $n$, and $d(F(n)) \le d(n)$ for all positive integers $n$.  Find all divisor-respecting functions.  Justify your answer.
1. **Initial Claim**: We claim that the only divisor-respecting function \( F: \mathbb{N} \rightarrow \mathbb{N} \) is \( F(x) = 1 \). We need to verify that this function satisfies the given conditions.

2. **Verification for \( F(1) \)**: 
   - Since \( d(F(1)) \leq d(1) = 1 \), we must have \( d(F(1)) = 1 \).
   - The only positive integer with exactly one divisor is 1, so \( F(1) = 1 \).

3. **Verification for Prime Numbers**:
   - For any prime \( p \), \( d(F(p)) \leq d(p) = 2 \).
   - Suppose, for contradiction, that there exists a prime \( q \) such that \( d(F(q)) = 2 \). This implies \( F(q) \) is a prime or a product of two primes.
   - Then \( d(F(q^2)) = d(F(q)) \cdot d(F(q)) = 4 \), but \( d(F(q^2)) \leq d(q^2) = 3 \), which is a contradiction.
   - Therefore, \( d(F(p)) = 1 \), implying \( F(p) = 1 \) for all primes \( p \).

4. **Induction Hypothesis**:
   - We will use strong induction to show that \( F(x) = 1 \) for all positive integers \( x \).
   - Base case: We have already shown that \( F(1) = 1 \).

5. **Inductive Step**:
   - Assume \( F(x) = 1 \) for all \( 1 \leq x < k \).
   - We need to show \( F(k) = 1 \).
   - If \( k \) is prime, we have already shown \( F(k) = 1 \).
   - If \( k \) is composite, there exist positive integers \( a \) and \( b \) such that \( k = ab \) and \( 1 < a, b < k \).
   - By the induction hypothesis, \( F(a) = 1 \) and \( F(b) = 1 \).
   - Therefore, \( d(F(k)) = d(F(a)) \cdot d(F(b)) = 1 \cdot 1 = 1 \), implying \( F(k) = 1 \).

6. **Conclusion**:
   - By induction, \( F(x) = 1 \) for all positive integers \( x \).

Thus, the only divisor-respecting function is \( F(x) = 1 \).

The final answer is \( \boxed{ F(x) = 1 } \).