Proof:
Consider $(f(a), g(a)) \in <f, g>$. By the diagram $x = \pi_1(x, y) = f(a)$ and $y = \pi_2(x, y) = g(a)$ and so $(f(a), g(a)) \in X \times Y$ for all $a \in A$. But that’s the definition of $<f, g>: A \to X \times Y$.
Does that make sense?
Proof:
Consider $(f(a), g(a)) \in <f, g>$. By the diagram $x = \pi_1(x, y) = f(a)$ and $y = \pi_2(x, y) = g(a)$ and so $(f(a), g(a)) \in X \times Y$ for all $a \in A$. But that’s the definition of $<f, g>: A \to X \times Y$.
Does that make sense?
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