We now show that, if a topological space X is the union of a finite collection of closed sets, and if a function from X to some topological space is continuous on each of these closed sets, then that function is continuous on X.

Lemma 1.1 Let X and Y be topological spaces, let f: X → Y be a function from X to Y , and let X = A₁∪A₂∪· · ·∪A_k, where A₁, A₂, . . . , A_k are closed sets in X. Suppose that the restriction of f to the closed set Ai is continuous for i = 1, 2, . . . , k. Then f: X → Y is continuous.

Proof A function f: X → Y is continuous if and only if f⁻¹ (G) is closed in X for every closed set G in Y (Lemma 1.6). Let G be an closed set in Y . Then f⁻¹ (G) ∩ Aiis relatively closed in Ai for i = 1, 2, . . . , k, since the restriction of f to Ai is continuous for each i. But Ai is closed in X, and therefore a subset of Ai is relatively closed in Ai if and only if it is closed in X. Therefore f⁻¹ (G) ∩ Aiis closed in X for i = 1, 2, . . . , k. Now f⁻¹ (G) isthe union of the sets f⁻¹ (G) ∩ Ai for i = 1, 2, . . . , k. It follows that f⁻¹ (G), being a finite union of closed sets, is itself closed in X. It now follows from Lemma  that f: X → Y is continuous.

Example: Let Y be a topological space, and let α: [0, 1] → Y and β: [0, 1] → Y be continuous functions defined on the interval [0, 1], where α(1) = β(0).

Let γ: [0, 1] → Y be defined by

Now γ|[0, 1/2] = α ◦ ρ where ρ: [0, 1/2] → [0, 1] is the continuous function defined by ρ(t) = 2t for all t ∈ [0, 1/2]. Thus γ|[0, 1/2] is continuous, being a composition of two continuous functions. Similarly γ [1/2, 1] is continuous. The subintervals [0, 1/2 ] and [1/2 , 1] are closed in [0,1], and [0,1] is the union of these two subintervals. It follows from Lemma 1.1 that γ: [0; 1] ! Y is continuous.

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