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DamienC
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I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I don't think that you need $f$ to be biholomorphic for this to be true.

EDIT: $\Gamma$ -equivariance shall be understood in the following sense. There exists a group morphism $\varphi:\Gamma\to\Gamma$ such that $\tilde{f}(\gamma\cdot x)=\varphi(\gamma)\cdot\tilde{f}(x)$. Then the pulled-back factor of automorphy shall be $J\big(\varphi(\gamma),\tilde{f}(x)\big)$.

EDIT2: here $\cdot$ means $+$.

I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I don't think that you need $f$ to be biholomorphic for this to be true.

I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I don't think that you need $f$ to be biholomorphic for this to be true.

EDIT: $\Gamma$ -equivariance shall be understood in the following sense. There exists a group morphism $\varphi:\Gamma\to\Gamma$ such that $\tilde{f}(\gamma\cdot x)=\varphi(\gamma)\cdot\tilde{f}(x)$. Then the pulled-back factor of automorphy shall be $J\big(\varphi(\gamma),\tilde{f}(x)\big)$.

EDIT2: here $\cdot$ means $+$.

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DamienC
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I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorpicholomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I don't think that you need $f$ to be biholomorphic for this to be true.

I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorpic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.

I don't think that you need $f$ to be biholomorphic for this to be true.

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DamienC
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I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorpic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).

Hence a factor of automorphy giving $f^*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.