# Upper semicontinuity of the push-forward of plurisubharmonic function

How to construct a surjective holomorphic map $$F:X\to Y$$ between (connected) complex manifolds with a real-valued function $$u$$ on $$Y$$ such that $$u{\circ}F$$ is plurisubharmonic on $$X$$, but $$u$$ is not upper semicontinuous on $$Y$$.

• It is easy to construct an example when $X$ is disconnected. Let $Y$ be the complex plane, $u(0)=0$, $u(z)=1, z\neq 0$. Let $X$ be a disjoint union of two complex planes, and $F(z)\equiv 0$ on one of them, while on the other one $F(z)=e^z$. Commented Dec 12, 2019 at 12:52
• Yes, you are right. I think the difficulty of this question is when X is connected. Commented Dec 12, 2019 at 13:45
• When $X$ is connected, I see no difficulty: there is no such function. Commented Dec 12, 2019 at 18:33
• Dear Prof. Eremenko, could you kindly sketch a proof of your claim? I can give one when F is a proper map, but I can not do for general F. Thank you very much. Commented Dec 13, 2019 at 0:20

Suppose that $$X$$ is connected. If $$u$$ is not u.s.c, then $$u(a)<\limsup_{z\to a} u(z)$$ for some $$a$$. Since $$F$$ is surjective, there is $$b$$ such that $$F(b)=a$$. Consider the set $$E=\{ z\in X:F(z)=a\}$$. This set is not a neighborhood of $$b$$, since otherwise $$F$$ will be constant in a neighborhood of $$b$$, thus, since $$X$$ is connected, constant on the whole $$X$$. So there is a sequence $$z_k\to b$$ such that $$F(z_k)\neq a$$. Then $$F(z_k)\to a$$ since $$F$$ is continuous, and thus $$\limsup_{k\to\infty}u\circ F(z_k)< u\circ F(b)$$ which the assumption that $$u\circ F$$ is u. s. c.
• It seems not to be enough for making a contradiction. Could we deduce that $\lim\sup_{k\to\infty}u\circ F(z_k)=\lim\sup_{z\to a}u(z)$? Commented Dec 13, 2019 at 4:48