# $f : X\longrightarrow Y$ isomorphism $\Rightarrow (?)\, \, f^*:Div(Y)\longrightarrow Div(X)$ isomorphism.

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?

Thanks

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.
• $D=\sum a_iZ_i$, 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? – Manoel Nov 16 '19 at 17:58
• Also, just to make sure, the functorality statement here is $g^* \circ f^* = (f \circ g)^*$. – RghtHndSd Nov 16 '19 at 19:46