https://file.aiquickdraw.com/mathbot/file/1729691221126zyj6wysl.jpg

Let's break down and solve the problem step by step. The image contains two parts of a mathematical proof related to set theory and cardinality.

### Part 1a:
We are asked to prove the following equivalences related to sets:
\[
(B \times C)^A \sim B^A \times C^A \quad \text{and} \quad (C^B)^A \sim C^{A \times B}
\]
where \( B^A = \{ f \mid f: A \to B \} \), and \( \sim \) denotes equinumerosity (sets having the same cardinality).

#### Step-by-step solution:

1. **First equivalence: \( (B \times C)^A \sim B^A \times C^A \)**

   Consider the meaning of \( (B \times C)^A \). By definition, \( (B \times C)^A \) is the set of all functions from \( A \) to \( B \times C \). A function \( f: A \to (B \times C) \) assigns to each element \( a \in A \) a pair \( (b_a, c_a) \in B \times C \). In other words, it can be thought of as two functions:
   - \( f_1: A \to B \)
   -

Prove that (B × C)^A ~ B^A × C^A and (C^B)^A ~ C^(A × B), where B^A = {f | f: A → B} and ~ denotes equinumerosity. Also, prove that if there is a surjection from A to B and a surjection from B to A, then sets A and B have the same cardinality.

This problem explores set theory by proving the equinumerosity between function spaces and Cartesian products, as well as applying surjections to determine the cardinality of sets. Suitable for advanced undergraduate math students.

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