Show that \(\pi_{i}\) is a ring homomorphism
A map is a ring homomorphism if it preserves the ring operations (addition and multiplication) and the multiplicative identity. We need to show that for any \((a_1, \ldots, a_k), (b_1, \ldots, b_k) \in R_1 \times \cdots \times R_k\) the following conditions hold:
1. \(\pi_i \left((a_1,\ldots, a_k) + (b_1,\ldots, b_k)\right) = \pi_i(a_1,\ldots, a_k) + \pi_i(b_1,\ldots, b_k)\)
2. \(\pi_i \left((a_1,\ldots, a_k) \cdot (b_1,\ldots, b_k)\right) = \pi_i(a_1,\ldots, a_k) \cdot \pi_i(b_1,\ldots, b_k)\)
3. \(\pi_i(1_{R_1 \times \cdots \times R_k}) = 1_{R_i}\)
For addition:
$$\pi_i\left((a_1,\ldots, a_k) + (b_1,\ldots, b_k)\right) = \pi_i\left((a_1+b_1, \ldots, a_k+b_k)\right) = a_i + b_i = \pi_i(a_1,\ldots, a_k) + \pi_i(b_1,\ldots, b_k)$$
For multiplication:
$$\pi_i\left((a_1,\ldots, a_k) \cdot (b_1,\ldots, b_k)\right) = \pi_i\left((a_1\cdot b_1, \ldots, a_k\cdot b_k)\right) = a_i \cdot b_i = \pi_i(a_1,\ldots, a_k) \cdot \pi_i(b_1,\ldots, b_k)$$
For the multiplicative identity, note that \((1_{R_1},\ldots,1_{R_k})\) is the multiplicative identity of \(R_1 \times \cdots \times R_k\):
$$\pi_i(1_{R_1 \times \cdots \times R_k}) = \pi_i(1_{R_1},\ldots,1_{R_k}) = 1_{R_i}$$
Since \(\pi_i\) preserves addition, multiplication, and the multiplicative identity, it is a ring homomorphism.