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

1$\begingroup$ 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 $F(z)=e^z$. $\endgroup$– Alexandre EremenkoCommented Dec 12, 2019 at 12:52

$\begingroup$ Yes, you are right. I think the difficulty of this question is when X is connected. $\endgroup$– BLANKCommented Dec 12, 2019 at 13:45

$\begingroup$ When $X$ is connected, I see no difficulty: there is no such function. $\endgroup$– Alexandre EremenkoCommented Dec 12, 2019 at 18:33

$\begingroup$ 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. $\endgroup$– BLANKCommented Dec 13, 2019 at 0:20
1 Answer
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.

$\begingroup$ It seems not to be enough for making a contradiction. Could we deduce that $\lim\sup_{k\to\infty}u\circ F(z_k)=\lim\sup_{z\to a}u(z)$? $\endgroup$– BLANKCommented Dec 13, 2019 at 4:48