If induced homomorphism of homologies of complex projective space is non zero then the map is surjective
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Let $f colon mathbb{CP}^n to mathbb{CP}^n$ be a continuous function which induces a non-zero map $f_*$ on every Homology group $H_{2k} (mathbb{CP}^n)$. Show that $f$ is surjective.
As we all know $H_{2k} (mathbb{CP}^n) cong mathbb{Z}$, if $n geq k$.
Thus if we look at the $2k$-th homology, $f_* : mathbb{Z} xrightarrow{} mathbb{Z}$ must be a multiplication by a scalar. i.e. $f_*(x) = c_k x$ (for $x$ the generator of $H_{2k} (mathbb{CP}^n)$). By the assumption $c_k neq 0$ for all $k leq n$.
What else can we say?
general-topology algebraic-topology homotopy-theory
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
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add a comment |
$begingroup$
Let $f colon mathbb{CP}^n to mathbb{CP}^n$ be a continuous function which induces a non-zero map $f_*$ on every Homology group $H_{2k} (mathbb{CP}^n)$. Show that $f$ is surjective.
As we all know $H_{2k} (mathbb{CP}^n) cong mathbb{Z}$, if $n geq k$.
Thus if we look at the $2k$-th homology, $f_* : mathbb{Z} xrightarrow{} mathbb{Z}$ must be a multiplication by a scalar. i.e. $f_*(x) = c_k x$ (for $x$ the generator of $H_{2k} (mathbb{CP}^n)$). By the assumption $c_k neq 0$ for all $k leq n$.
What else can we say?
general-topology algebraic-topology homotopy-theory
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
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1
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You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
1
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Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
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– Rob Arthan
Dec 20 '18 at 20:08
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I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30
add a comment |
$begingroup$
Let $f colon mathbb{CP}^n to mathbb{CP}^n$ be a continuous function which induces a non-zero map $f_*$ on every Homology group $H_{2k} (mathbb{CP}^n)$. Show that $f$ is surjective.
As we all know $H_{2k} (mathbb{CP}^n) cong mathbb{Z}$, if $n geq k$.
Thus if we look at the $2k$-th homology, $f_* : mathbb{Z} xrightarrow{} mathbb{Z}$ must be a multiplication by a scalar. i.e. $f_*(x) = c_k x$ (for $x$ the generator of $H_{2k} (mathbb{CP}^n)$). By the assumption $c_k neq 0$ for all $k leq n$.
What else can we say?
general-topology algebraic-topology homotopy-theory
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
$endgroup$
Let $f colon mathbb{CP}^n to mathbb{CP}^n$ be a continuous function which induces a non-zero map $f_*$ on every Homology group $H_{2k} (mathbb{CP}^n)$. Show that $f$ is surjective.
As we all know $H_{2k} (mathbb{CP}^n) cong mathbb{Z}$, if $n geq k$.
Thus if we look at the $2k$-th homology, $f_* : mathbb{Z} xrightarrow{} mathbb{Z}$ must be a multiplication by a scalar. i.e. $f_*(x) = c_k x$ (for $x$ the generator of $H_{2k} (mathbb{CP}^n)$). By the assumption $c_k neq 0$ for all $k leq n$.
What else can we say?
general-topology algebraic-topology homotopy-theory
general-topology algebraic-topology homotopy-theory
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
edited Dec 22 '18 at 15:43
Arnon Hod
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
asked Dec 20 '18 at 19:59
Arnon HodArnon Hod
62
62
1
$begingroup$
You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
1
$begingroup$
Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:08
$begingroup$
I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30
add a comment |
1
$begingroup$
You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
1
$begingroup$
Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:08
$begingroup$
I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30
1
1
$begingroup$
You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
$begingroup$
You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
1
1
$begingroup$
Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:08
$begingroup$
Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:08
$begingroup$
I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30
$begingroup$
I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30
add a comment |
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$begingroup$
You'll get better answers if you show what you've already considered and where you've got stuck.
$endgroup$
– postmortes
Dec 20 '18 at 20:05
1
$begingroup$
Hint: you haven't made it clear what tools you have available, but you shouldn't need any special properties of $Bbb{C}Bbb{P}^n$ other than that it's a connected $2n$-manifold or of $f$ other than$H_{2n}(f)$ is non-zero.
$endgroup$
– Rob Arthan
Dec 20 '18 at 20:08
$begingroup$
I think you can probably try arguing by contradiction. If $f$ missed a point, could it really be nonzero on all homology groups?
$endgroup$
– Hempelicious
Dec 22 '18 at 17:30