Can distinct morphisms between curves induce the same morphism on singular cohomology? The 2019 Stack Overflow Developer Survey Results Are InGeneric fiber of morphism between non-singular curvesWhat are unramified morphisms like?Does the normalization of a projective morphism determine the line bundle?Field of definition of canonical morphism between (congruence) modular curvesEtale covers of a hyperelliptic curvepushing out families of curvesWhat happens to the gonality under a finite morphism of curvesCan a birational morphism between two smooth varieties of the same betti numbers exist?Examples of endomorphisms of complex curvesDoes there exist trace maps between $ell$-adic cohomology groups for finite flat morphisms?
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Can distinct morphisms between curves induce the same morphism on singular cohomology?
The 2019 Stack Overflow Developer Survey Results Are InGeneric fiber of morphism between non-singular curvesWhat are unramified morphisms like?Does the normalization of a projective morphism determine the line bundle?Field of definition of canonical morphism between (congruence) modular curvesEtale covers of a hyperelliptic curvepushing out families of curvesWhat happens to the gonality under a finite morphism of curvesCan a birational morphism between two smooth varieties of the same betti numbers exist?Examples of endomorphisms of complex curvesDoes there exist trace maps between $ell$-adic cohomology groups for finite flat morphisms?
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 4 hours ago
rj7k8rj7k8
180117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 4 hours ago
rj7k8rj7k8
180117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 4 hours ago
rj7k8rj7k8
180117
$endgroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
ag.algebraic-geometry
asked 4 hours ago
rj7k8rj7k8
180117
asked 4 hours ago
rj7k8rj7k8
180117
asked 4 hours ago
rj7k8rj7k8
180117
asked 4 hours ago
rj7k8rj7k8
180117
asked 4 hours ago
rj7k8rj7k8
180117
180117
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
$endgroup$
add a comment |
$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
$endgroup$
add a comment |
$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
$endgroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
answered 4 hours ago
Piotr AchingerPiotr Achinger
8,49712854
8,49712854
add a comment |
add a comment |
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