open mapping theorem(complex analysis)

By Zorn leung

Does open mapping theorem holds in {\mathbb{C}}^n for n>1?

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2 Responses to open mapping theorem(complex analysis)

  1. KKK says:

    Zorn:

    What is the target space? If f: \mathbb{C}^2 \rightarrow \mathbb{C}^2. Trivially no. Example: f(z, w)=(z, 0).

    For f: \mathbb{C}^2 \rightarrow \mathbb{C} (say). Trivially yes.
    The reason is, let U\subset \mathbb{C}^2 be open and p\in f(U). Let’s say f(0, 0)=p. Then it is easy to see that f(z, 0) is nonconstant OR f(0, w) is non-constant (by Taylor’s expansion, say). Say g(z)=f(z, 0) is non-constant, then it maps a small (1-dim!) neighborhood of 0 to an open set around p, by 1-D open mapping theorem. So there is an open set V around p which is contained in f(U).
    [My argument is wrong. ]

  2. KKK says:

    It turns out that I can correct my argument. Let my clarify it here.
    For convenience, let’s assume f: \mathbb{C}^2\rightarrow \mathbb{C} be a non-constant holomorphic (entire) function. Let U\subset \mathbb{C}^2 be open and p\in f(U). Wlog, we assume (0,0)\in U and f(0, 0)=p
    . We claim that there is an open set V\subset f(U) which contains p. To see this, since f is non-constant, there is (0, 0)\neq (a, b)\in \mathbb{C}^2 with f(a, b)\neq f(0, 0). Define g(z):=f(az, bz) to be an entire non-constant holomorphic function. Then by open mapping theorem, g maps arbitrary small neighborhood of 0 onto an open set V around p. So by choosing this neighborhood to be small enough, we see that there is an open set V\subset f(U) containing p.

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