Proving that $X = A cup B$ is a disconnection then $H_n(X) cong H_n(A) oplus H_n(B)$ using excision












2












$begingroup$



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







share|cite|improve this question









$endgroup$








  • 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
















2












$begingroup$



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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$



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







share|cite|improve this question









$endgroup$





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






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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














  • 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










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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.






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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.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 22:52

























        answered Dec 5 '18 at 22:40









        Paul FrostPaul Frost

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        11.4k3934






























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