Let X_1 and X_2 be sets. Define functions  pi_1: X_1 x X_2 -> X_1, pi_2:X_1 x X_2 -> X_2 by pi_1(x_1,x_2)=x_1, pi_2(x_1, x_2)=x_2 (x_1 ∈ X_1, x_2 ∈ X_2). Also let A be a set, and let f_1: A -> X_1, f_2: A -> X_2 be functions. Prove that there is exactly one function f: A -> X_1 x X_2 such that pi_1 ∘ f = f_1 and pi_2 ∘ f = f_2.

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