Proving that $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$ using excision
$begingroup$
If $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$
If I use the Mayer-Vietoris sequence I can prove this easily, however the book I'm reading hinted at using excision to prove this.
So I tried to prove this using excision but I ended up getting stuck, this is what I've done so far.
Since $B subseteq X$ we obtain a long exact sequence
$$ dots to H_n(B) to H_n(X) to H_n(X, B) to H_{n-1}(B) to dots$$
and observe that because $X = A cup B$ by Excision we get $H_n(A) cong H_n(X, B)$ and so we obtain the following long exact sequence
$$ dots to H_n(B) xrightarrow{f} H_n(X) xrightarrow{g} H_n(A) to H_{n-1}(B) to dots$$
This is where I get stuck, because if $f$ was injective and $g$ was surjective I could use exactness and the first isomorphism theorem to get the desired result, however I have no way of knowing if $f$ is injective and if $g$ is surjective.
I'm not sure how to proceed and prove this using excision alone.
algebraic-topology
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
$endgroup$
add a comment |
$begingroup$
If $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$
If I use the Mayer-Vietoris sequence I can prove this easily, however the book I'm reading hinted at using excision to prove this.
So I tried to prove this using excision but I ended up getting stuck, this is what I've done so far.
Since $B subseteq X$ we obtain a long exact sequence
$$ dots to H_n(B) to H_n(X) to H_n(X, B) to H_{n-1}(B) to dots$$
and observe that because $X = A cup B$ by Excision we get $H_n(A) cong H_n(X, B)$ and so we obtain the following long exact sequence
$$ dots to H_n(B) xrightarrow{f} H_n(X) xrightarrow{g} H_n(A) to H_{n-1}(B) to dots$$
This is where I get stuck, because if $f$ was injective and $g$ was surjective I could use exactness and the first isomorphism theorem to get the desired result, however I have no way of knowing if $f$ is injective and if $g$ is surjective.
I'm not sure how to proceed and prove this using excision alone.
algebraic-topology
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
$endgroup$
1
$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
1
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
1
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39
add a comment |
$begingroup$
If $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$
If I use the Mayer-Vietoris sequence I can prove this easily, however the book I'm reading hinted at using excision to prove this.
So I tried to prove this using excision but I ended up getting stuck, this is what I've done so far.
Since $B subseteq X$ we obtain a long exact sequence
$$ dots to H_n(B) to H_n(X) to H_n(X, B) to H_{n-1}(B) to dots$$
and observe that because $X = A cup B$ by Excision we get $H_n(A) cong H_n(X, B)$ and so we obtain the following long exact sequence
$$ dots to H_n(B) xrightarrow{f} H_n(X) xrightarrow{g} H_n(A) to H_{n-1}(B) to dots$$
This is where I get stuck, because if $f$ was injective and $g$ was surjective I could use exactness and the first isomorphism theorem to get the desired result, however I have no way of knowing if $f$ is injective and if $g$ is surjective.
I'm not sure how to proceed and prove this using excision alone.
algebraic-topology
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
$endgroup$
If $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$
If I use the Mayer-Vietoris sequence I can prove this easily, however the book I'm reading hinted at using excision to prove this.
So I tried to prove this using excision but I ended up getting stuck, this is what I've done so far.
Since $B subseteq X$ we obtain a long exact sequence
$$ dots to H_n(B) to H_n(X) to H_n(X, B) to H_{n-1}(B) to dots$$
and observe that because $X = A cup B$ by Excision we get $H_n(A) cong H_n(X, B)$ and so we obtain the following long exact sequence
$$ dots to H_n(B) xrightarrow{f} H_n(X) xrightarrow{g} H_n(A) to H_{n-1}(B) to dots$$
This is where I get stuck, because if $f$ was injective and $g$ was surjective I could use exactness and the first isomorphism theorem to get the desired result, however I have no way of knowing if $f$ is injective and if $g$ is surjective.
I'm not sure how to proceed and prove this using excision alone.
algebraic-topology
algebraic-topology
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
asked Dec 5 '18 at 20:12
PerturbativePerturbative
4,40121553
4,40121553
1
$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
1
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
1
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39
add a comment |
1
$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
1
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
1
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39
1
1
$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
1
1
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
1
1
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39
add a comment |
1 Answer
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Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) cong H_n(A) oplus H_n(B)$.
Let $i : A to X$ denote the inclusion map. Choose $a in A$ and define $r : X to A$ by $r(x) = x$ for $x in A$ and $r(x) = a$ for $x in B$. Then $ri = id$, hence $i_* : H_n(A) to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences
$$0 to H_n(A) stackrel{i_*}{rightarrow} H_n(X) stackrel{j_*}{rightarrow} H_n(X,A) to 0 .$$
We conclude
$$H_n(X) approx H_n(A) oplus H_n(X,A) .$$
Excision gives us an isomorphism $H_n(B) = H_n(B,emptyset) to H_n(X,A)$ which completes the proof.
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
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add a comment |
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$begingroup$
Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) cong H_n(A) oplus H_n(B)$.
Let $i : A to X$ denote the inclusion map. Choose $a in A$ and define $r : X to A$ by $r(x) = x$ for $x in A$ and $r(x) = a$ for $x in B$. Then $ri = id$, hence $i_* : H_n(A) to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences
$$0 to H_n(A) stackrel{i_*}{rightarrow} H_n(X) stackrel{j_*}{rightarrow} H_n(X,A) to 0 .$$
We conclude
$$H_n(X) approx H_n(A) oplus H_n(X,A) .$$
Excision gives us an isomorphism $H_n(B) = H_n(B,emptyset) to H_n(X,A)$ which completes the proof.
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
$endgroup$
add a comment |
$begingroup$
Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) cong H_n(A) oplus H_n(B)$.
Let $i : A to X$ denote the inclusion map. Choose $a in A$ and define $r : X to A$ by $r(x) = x$ for $x in A$ and $r(x) = a$ for $x in B$. Then $ri = id$, hence $i_* : H_n(A) to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences
$$0 to H_n(A) stackrel{i_*}{rightarrow} H_n(X) stackrel{j_*}{rightarrow} H_n(X,A) to 0 .$$
We conclude
$$H_n(X) approx H_n(A) oplus H_n(X,A) .$$
Excision gives us an isomorphism $H_n(B) = H_n(B,emptyset) to H_n(X,A)$ which completes the proof.
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
$endgroup$
add a comment |
$begingroup$
Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) cong H_n(A) oplus H_n(B)$.
Let $i : A to X$ denote the inclusion map. Choose $a in A$ and define $r : X to A$ by $r(x) = x$ for $x in A$ and $r(x) = a$ for $x in B$. Then $ri = id$, hence $i_* : H_n(A) to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences
$$0 to H_n(A) stackrel{i_*}{rightarrow} H_n(X) stackrel{j_*}{rightarrow} H_n(X,A) to 0 .$$
We conclude
$$H_n(X) approx H_n(A) oplus H_n(X,A) .$$
Excision gives us an isomorphism $H_n(B) = H_n(B,emptyset) to H_n(X,A)$ which completes the proof.
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
$endgroup$
Let $H_*$ be any homology theory satisfying the Eilenberg-Steenrod axioms with the possible exception of the dimension axiom. We shall prove that $H_n(X) cong H_n(A) oplus H_n(B)$.
Let $i : A to X$ denote the inclusion map. Choose $a in A$ and define $r : X to A$ by $r(x) = x$ for $x in A$ and $r(x) = a$ for $x in B$. Then $ri = id$, hence $i_* : H_n(A) to H_n(X)$ is a monomorphism with left inverse $r_*$. This shows that the long exact sequence of the pair $(X,A)$ collapses to split short exact sequences
$$0 to H_n(A) stackrel{i_*}{rightarrow} H_n(X) stackrel{j_*}{rightarrow} H_n(X,A) to 0 .$$
We conclude
$$H_n(X) approx H_n(A) oplus H_n(X,A) .$$
Excision gives us an isomorphism $H_n(B) = H_n(B,emptyset) to H_n(X,A)$ which completes the proof.
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
edited Dec 5 '18 at 22:52
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
answered Dec 5 '18 at 22:40
Paul FrostPaul Frost
11.4k3934
11.4k3934
add a comment |
add a comment |
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$begingroup$
You could always argue by hand that $f$ and $g$ are injective and surjective chain $sigma$ in $X$. Then you can write $sigma=sigma_A+sigma_B$ where $sigma_A$ is the sum of simplices lying in $A$ and so forth. It is pretty clear that $partial sigma_B=0$. So your original cycle is actually trivial just in $A$. However this line of argument can be readily adapted to not even use excision. The chain complex $C_n(X)$ itself breaks up as a direct sum $C_n(A)oplus C_n(B)$.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:53
1
$begingroup$
I think any proof that doesn't just check this at the chain level is silly.
$endgroup$
– user98602
Dec 5 '18 at 21:39
1
$begingroup$
Note that if $X$ is the disjoint union of $Asqcup B$, then both inclusions admit retractions and therefore the inclusion has to induce an injective map on homology and the retraction, which induces the map from $H_*(X,B)to H_*(A)$ given via excision has to be surjective and therefore the long exact sequence breaks down to a short exact sequence, which admits a splitting.
$endgroup$
– ThorbenK
Dec 5 '18 at 22:39