section 18: 13) Suppose A ⊂ X. Let f : A → Y be continuous where Y is T_{2}. If g_{i} : \overline{A} → Y is continuous for i = 1, 2 and if g_{1}|_{A} = g_{2}|_{A} = f, then g_{1} = g_{2}

Pf: Suppose g_{1} ≠ g_{2}. Then ∃x ∈ \overline{A} − A such that g_{1}(x) ≠ g_{2}(x). Y T_{2} implies ∃V_{1}, V_{2} open in Y such that g_{i}(x) ∈ V_{i} and V_{1} ∩ V_{2} = ∅

x ∈ g_{i}^{−1}(V_{i}) implies x ∈ g_{1}^{−1}(V_{1}) ∩ g_{2}^{−1}(V_{2}). g_{i} continuous implies g_{i}^{−1}(V_{i}) open in \overline{A} and hence g_{1}^{−1}(V_{1}) ∩ g_{2}^{−1}(V_{2}) open in \overline{A}. Thus there exists U open in X such that g_{1}^{−1}(V_{1}) ∩ g_{2}^{−1}(V_{2}) = U ∩ \overline{A}. Note x ∈ U.

x ∈ \overline{A} − A implies x ∈ A′. Thus ∃a ∈ U ∩ A − \{x\} ⊂ g_{1}^{−1}(V_{1}) ∩ g_{2}^{−1}(V_{2}). But f(a) = g_{i}(a) ∈ g_{i}(g_{i}^{−1}(V_{i})) ⊂ V_{i}. But V_{1} ∩ V_{2} = ∅.

NOTE: It may have been simpler to WLOG assume X = \overline{A}, but you must state why this WLOG works.