Proposition 5.43.18A7090F92CA39

Proof. Let $S\subset X$ be a proper analytic subset over whose complement $\pi $ is an isomorphism, and put $D=\pi ^{-1}(S)$. By lemma 5.43.2, pullback preserves the curvature inequality, and the pullback of a $\theta $-plurisubharmonic function is not identically $-\infty $, since $\pi $ is an isomorphism off $D$. The pullback map is injective because $\pi $ is surjective.

To prove surjectivity, let $b\in \mathrm{PSH}(Y,\pi ^*\theta )$, and define $u=b\circ \pi ^{-1}$ off $S$. On a coordinate neighborhood in $X$, choose a smooth function $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \theta $. Then $b+\rho \circ \pi $ is plurisubharmonic on the inverse image of this neighborhood. By properness, its local upper bounds give a local upper bound for $u+\rho $ near $S$, so the removable-singularity theorem extends $u+\rho $ plurisubharmonically across $S$. The resulting quasi-plurisubharmonic extensions of $u$ agree almost everywhere on overlaps, hence everywhere, and glue to a function on $X$.

The current $\theta +\mathrm{dd}^{\mathrm{c}}u$ is positive off $S$. Locally it is the sum of the positive current $\mathrm{dd}^{\mathrm{c}}(u+\rho )$ and the smooth form $\theta -\mathrm{dd}^{\mathrm{c}}\rho $. The smooth form gives $S$ zero mass, so the restriction of $\theta +\mathrm{dd}^{\mathrm{c}}u$ to $S$ is positive as well. Thus $u\in \mathrm{PSH}(X,\theta )$. The plurisubharmonic functions $b+\rho \circ \pi $ and $(u+\rho )\circ \pi $ agree off $D$, hence almost everywhere, hence everywhere. Therefore $b=u\circ \pi $. $\square$