Problem set 1: Answers

So, suppose that i ◦ f : Z → X is continuous. Since for every α, the projection πα : X → Xα is continuous, πα ◦ (i ◦ f) : Z → Xα is continuous. The map πα ◦ (i ◦ f) = iα ◦ (ρα ◦ f). By the universal property of the subspace topology on Yα, the map iα ◦ (ρα ◦ f) being continuous implies ρα ◦ f : Z → Yα is continuous. So, we have a collection of continuous maps: ρα ◦ f : Z → Yα so by the universal property of the product topology on Y , the map f : Z → Y is continuous. On the other hand, suppose that f : Z → Y is continuous. Because the projections ρα : Y → Yα are all continuous, the compositions ρα ◦f : Z → Yα are all continuous. Since the inclusions iα : Yα → Xα are all continuous, the compotisions, iα ◦ ρα ◦ f : Z → Xα are all continuous. Note the maps iα ◦ρα ◦f = πα ◦ (i◦f), and the universal