I am attending a course on Introduction to Complex Geometry, and the definition they have given me of a holomorphic function between complex manifolds is as follows:
Let $X,Y$ be complex manifolds. A map $f: X \rightarrow Y$ is $\textbf{holomorphic}$ if 1. f is continuous and 2. $\forall p\in X$, there exist charts $(U,\phi)$ in $X$, $(V,\psi)$ in $Y$ so that $\psi \circ f \circ \phi^{-1} : \phi(U\cap f^{-1}(V)) \rightarrow \psi(V)$ is holomorphic.
My question is: Is the first hypothesis (f continuous) necessary? Does not the second one imply the first one? This question also applies to Differential Geometry in the real case. Does anyone know if there is a counterexample if the first hypothesis is removed? Thanks.
Let $f : A \rightarrow \Bbb{C}$ be a complex function, with $A \subseteq \Bbb{C}$. Usually, to even say that $f$ is holomorphic, we require that $A$ be an open subset of $\Bbb{C}$, just like with differentiability of real functions. It's not entirely obvious why, but I think it's standard.
So, in order to be able to say $\psi \circ f \circ \phi^{-1}$ is continuous, we need its domain to be open. This requires $f$ to be continuous, because we need $f^{-1}(V)$ to be open.