Let $X$ be a complex manifold. Denote by Div$(X)$ the Weil divisors group of $X$.

We have to:

Let $f : X \longrightarrow Y$ be a holomorphic map of connected complex manifolds and suppose that $f$ is dominant, i.e. $f(X)$ is dense in $Y$. Then the pull-back defines a group homomorphism $$ f^* :Div(Y) \longrightarrow Div(X). $$

Question: If $f : X \longrightarrow Y$ is an analytic isomorphism ( bi-holomorphic map), then is it true that $f^*:Div(Y)\longrightarrow Div(X)$ is a group isomorphism?



The inverse map gives rise to a map on Div's in the opposite direction. Since $\text{Div}(\cdot)$ is functorial, and composition of $f$ with its inverse gives the identity map, this means that the map on Div's is inverse to the original one.

  • $\begingroup$ $D=\sum a_iZ_i $, where $Z_i$ is irreduible hypersurfaces... $(f^{-1})^*(f^*(Z_i))=Z_i$ this is where you use that Div(⋅) is functorial? Because that must be true, to close the argument. Right? $\endgroup$ – Manoel Nov 16 '19 at 17:58
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    $\begingroup$ Correct. But you can rerphase this to work in great generality. Anytime a map is functorial, an isomorphism in the source category induces an isomorphism in the target via this argument. $\endgroup$ – RghtHndSd Nov 16 '19 at 19:44
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    $\begingroup$ Also, just to make sure, the functorality statement here is $g^* \circ f^* = (f \circ g)^*$. $\endgroup$ – RghtHndSd Nov 16 '19 at 19:46
  • $\begingroup$ Helped me a lot! Thanks :) $\endgroup$ – Manoel Nov 16 '19 at 19:48

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