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?
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
rj7k8rj7k8
180117
asked 1 hour ago
rj7k8rj7k8
180117
asked 1 hour ago
rj7k8rj7k8
180117
asked 1 hour ago
rj7k8rj7k8
180117
asked 1 hour 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 1 hour ago
Piotr AchingerPiotr Achinger
8,49712854
$endgroup$
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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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Piotr AchingerPiotr Achinger
8,49712854
answered 1 hour ago
Piotr AchingerPiotr Achinger
8,49712854
answered 1 hour ago
Piotr AchingerPiotr Achinger
8,49712854
answered 1 hour ago
Piotr AchingerPiotr Achinger
8,49712854
8,49712854
add a comment |
add a comment |
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