Talk:Identity theorem for Riemann surfaces

How can this theorem be true? Suppose that X=R^2, Y=R, f(x_1,x_2)=x_1 x_2, g(x_1,x_2)=0, A=\{(x_1,0):x_1\in R\}. Then conditions of the theorem are satisfied, but f \ne g on the whole of X. — Preceding unsigned comment added by 194.166.199.15 (talk) 07:05, 2 February 2025 (UTC)

Y is not a Riemann surface and f is not holomorphic Gumshoe2 (talk) 16:45, 2 February 2025 (UTC)