spectral sequence example diagonal map confusion
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I'm attempting to wrap my head around spectral sequences, so constructed a really basic example to apply the definitions and go through the motions. My filtered chain complexes are:
$F_2C_*: 0 rightarrow mathbb{Z} xrightarrow{i_1} mathbb{Z_1}oplus mathbb{Z_2} xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_1C_*: 0 rightarrow 0 rightarrow mathbb{Z}_2 xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_0C_*:$ The zero complex.
Here $i_1$ and $p_2$ are the inclusion and projection maps, and the sub-indices on the $mathbb{Z}s$ are just for labeling purposes.
With $Z^r_{p,q}$ defined as ${x in F_pC_q : partial x in F_{p-r}C_{q-1}}/F_{p-1}C_q$, the differential is meant to restrict to well-defined maps $d^r:Z^r_{p,q} rightarrow Z^r_{p-r,q-1}$.
But in my example the map $Z^1_{2,2} rightarrow Z^1_{1,1}$ does not appear to be well-defined:
$Z^1_{2,2} := {x in mathbb{Z}oplus mathbb{Z} : p_2 x in mathbb{Z} }/mathbb{Z}_2 cong mathbb{Z}_1$,
$Z^1_{1,1} := mathbb{Z}$.
However $[(a,b)]$ and $[(a,b')]$ in $Z^1_{2,2}$ get sent to $b$ and $b'$ in $mathbb{Z}$ respectively, even though $[(a,b)] = [(a,b')]$.
I cannot see where I've made a mistake, but I'm pretty sure something's not right. Where have I gone wrong?
Given a map $A xrightarrow{f} C$ with $B subset A$ and $D subset C$, in order for the quotient map $A/B xrightarrow{bar f} C/D$ to be defined we must have $f(B) subset D$.
But it this case we have a map $(Z_1 oplus Z_2) / Z_2 xrightarrow{bar p_2} Z/{0}$, but $p_2(Z_2) = Z_2$ which is not contained in ${0}!$
homological-algebra spectral-sequences
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
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$begingroup$
I'm attempting to wrap my head around spectral sequences, so constructed a really basic example to apply the definitions and go through the motions. My filtered chain complexes are:
$F_2C_*: 0 rightarrow mathbb{Z} xrightarrow{i_1} mathbb{Z_1}oplus mathbb{Z_2} xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_1C_*: 0 rightarrow 0 rightarrow mathbb{Z}_2 xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_0C_*:$ The zero complex.
Here $i_1$ and $p_2$ are the inclusion and projection maps, and the sub-indices on the $mathbb{Z}s$ are just for labeling purposes.
With $Z^r_{p,q}$ defined as ${x in F_pC_q : partial x in F_{p-r}C_{q-1}}/F_{p-1}C_q$, the differential is meant to restrict to well-defined maps $d^r:Z^r_{p,q} rightarrow Z^r_{p-r,q-1}$.
But in my example the map $Z^1_{2,2} rightarrow Z^1_{1,1}$ does not appear to be well-defined:
$Z^1_{2,2} := {x in mathbb{Z}oplus mathbb{Z} : p_2 x in mathbb{Z} }/mathbb{Z}_2 cong mathbb{Z}_1$,
$Z^1_{1,1} := mathbb{Z}$.
However $[(a,b)]$ and $[(a,b')]$ in $Z^1_{2,2}$ get sent to $b$ and $b'$ in $mathbb{Z}$ respectively, even though $[(a,b)] = [(a,b')]$.
I cannot see where I've made a mistake, but I'm pretty sure something's not right. Where have I gone wrong?
Given a map $A xrightarrow{f} C$ with $B subset A$ and $D subset C$, in order for the quotient map $A/B xrightarrow{bar f} C/D$ to be defined we must have $f(B) subset D$.
But it this case we have a map $(Z_1 oplus Z_2) / Z_2 xrightarrow{bar p_2} Z/{0}$, but $p_2(Z_2) = Z_2$ which is not contained in ${0}!$
homological-algebra spectral-sequences
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
$endgroup$
add a comment |
$begingroup$
I'm attempting to wrap my head around spectral sequences, so constructed a really basic example to apply the definitions and go through the motions. My filtered chain complexes are:
$F_2C_*: 0 rightarrow mathbb{Z} xrightarrow{i_1} mathbb{Z_1}oplus mathbb{Z_2} xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_1C_*: 0 rightarrow 0 rightarrow mathbb{Z}_2 xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_0C_*:$ The zero complex.
Here $i_1$ and $p_2$ are the inclusion and projection maps, and the sub-indices on the $mathbb{Z}s$ are just for labeling purposes.
With $Z^r_{p,q}$ defined as ${x in F_pC_q : partial x in F_{p-r}C_{q-1}}/F_{p-1}C_q$, the differential is meant to restrict to well-defined maps $d^r:Z^r_{p,q} rightarrow Z^r_{p-r,q-1}$.
But in my example the map $Z^1_{2,2} rightarrow Z^1_{1,1}$ does not appear to be well-defined:
$Z^1_{2,2} := {x in mathbb{Z}oplus mathbb{Z} : p_2 x in mathbb{Z} }/mathbb{Z}_2 cong mathbb{Z}_1$,
$Z^1_{1,1} := mathbb{Z}$.
However $[(a,b)]$ and $[(a,b')]$ in $Z^1_{2,2}$ get sent to $b$ and $b'$ in $mathbb{Z}$ respectively, even though $[(a,b)] = [(a,b')]$.
I cannot see where I've made a mistake, but I'm pretty sure something's not right. Where have I gone wrong?
Given a map $A xrightarrow{f} C$ with $B subset A$ and $D subset C$, in order for the quotient map $A/B xrightarrow{bar f} C/D$ to be defined we must have $f(B) subset D$.
But it this case we have a map $(Z_1 oplus Z_2) / Z_2 xrightarrow{bar p_2} Z/{0}$, but $p_2(Z_2) = Z_2$ which is not contained in ${0}!$
homological-algebra spectral-sequences
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
$endgroup$
I'm attempting to wrap my head around spectral sequences, so constructed a really basic example to apply the definitions and go through the motions. My filtered chain complexes are:
$F_2C_*: 0 rightarrow mathbb{Z} xrightarrow{i_1} mathbb{Z_1}oplus mathbb{Z_2} xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_1C_*: 0 rightarrow 0 rightarrow mathbb{Z}_2 xrightarrow{p_2} mathbb{Z} rightarrow 0 $
$F_0C_*:$ The zero complex.
Here $i_1$ and $p_2$ are the inclusion and projection maps, and the sub-indices on the $mathbb{Z}s$ are just for labeling purposes.
With $Z^r_{p,q}$ defined as ${x in F_pC_q : partial x in F_{p-r}C_{q-1}}/F_{p-1}C_q$, the differential is meant to restrict to well-defined maps $d^r:Z^r_{p,q} rightarrow Z^r_{p-r,q-1}$.
But in my example the map $Z^1_{2,2} rightarrow Z^1_{1,1}$ does not appear to be well-defined:
$Z^1_{2,2} := {x in mathbb{Z}oplus mathbb{Z} : p_2 x in mathbb{Z} }/mathbb{Z}_2 cong mathbb{Z}_1$,
$Z^1_{1,1} := mathbb{Z}$.
However $[(a,b)]$ and $[(a,b')]$ in $Z^1_{2,2}$ get sent to $b$ and $b'$ in $mathbb{Z}$ respectively, even though $[(a,b)] = [(a,b')]$.
I cannot see where I've made a mistake, but I'm pretty sure something's not right. Where have I gone wrong?
Given a map $A xrightarrow{f} C$ with $B subset A$ and $D subset C$, in order for the quotient map $A/B xrightarrow{bar f} C/D$ to be defined we must have $f(B) subset D$.
But it this case we have a map $(Z_1 oplus Z_2) / Z_2 xrightarrow{bar p_2} Z/{0}$, but $p_2(Z_2) = Z_2$ which is not contained in ${0}!$
homological-algebra spectral-sequences
homological-algebra spectral-sequences
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
edited Jan 15 at 11:08
Cookie Monster
asked Jan 12 at 10:34
Cookie MonsterCookie Monster
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asked Jan 12 at 10:34
Cookie MonsterCookie Monster
17617
asked Jan 12 at 10:34
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17617
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$Z_{p,q}^r$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := {x in F^pC_q: dx in F^{p-r}C_{q-1}}$$ check eg wikipedia.
answered Jan 15 at 12:01
BenBen
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$begingroup$
$Z_{p,q}^r$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := {x in F^pC_q: dx in F^{p-r}C_{q-1}}$$ check eg wikipedia.
answered Jan 15 at 12:01
BenBen
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$begingroup$
$Z_{p,q}^r$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := {x in F^pC_q: dx in F^{p-r}C_{q-1}}$$ check eg wikipedia.
answered Jan 15 at 12:01
BenBen
3,423616
$endgroup$
add a comment |
$begingroup$
$Z_{p,q}^r$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := {x in F^pC_q: dx in F^{p-r}C_{q-1}}$$ check eg wikipedia.
answered Jan 15 at 12:01
BenBen
3,423616
$endgroup$
$Z_{p,q}^r$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := {x in F^pC_q: dx in F^{p-r}C_{q-1}}$$ check eg wikipedia.
answered Jan 15 at 12:01
BenBen
3,423616
answered Jan 15 at 12:01
BenBen
3,423616
answered Jan 15 at 12:01
BenBen
3,423616
answered Jan 15 at 12:01
BenBen
3,423616
3,423616
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