Finding a mistake using Mayer-VietorisProve that $H_n(A sqcup B) cong H_n(A) oplus H_n(B)$ for all $n in...

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Finding a mistake using Mayer-Vietoris


Prove that $H_n(A sqcup B) cong H_n(A) oplus H_n(B)$ for all $n in mathbb{Z}$.Rigorous application of Mayer-Vietoris to a quotient of $S^{2}times I$Mayer-Vietoris where $Acap B$ bounds $A$ and $B$Calculating the homology groups of a simplicial complex using a Mayer-Vietoris sequence$H_1(S^1)$ with Mayer-Vietoris sequenceMisunderstanding connected with the Mayer-Vietoris sequenceCompute the homology groups using Mayer-Vietoris sequenceComputing the homology of genus $g$ surface, using Mayer-VietorisConfusion in using Mayer-Vietoris theoremHomology of Klein bottle with Mayer-Vietoris













1












$begingroup$


I was computing the homology of $S^3-coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $mathbb{R}^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbb{R}^2$ and finaly the last one ends up with a space homotopic to $S^1$. Therefore, $H_2(S^3-coprod_{i=1}^4 I_i)=0$.



But in other calculations this caused me some problems so I decide to do it formally using Mayer-Vietoris. I had already computed $H_*(S^3-Isqcup I)$, giving me a consistent result with the intuition above, i.e. $H_2(S^3-Isqcup I)=mathbb{Z}$.



Now I decompose $S^3=(S^3-coprod_{i=1}^2 I_i)cup (S^3-coprod_{i=3}^4 I_i)$. From Mayer-Vietoris there is a short exact sequence



$$0to H_3(S^3)to H_2(S^3-coprod_{i=1}^4 I_i)to H_2(S^3-I_1sqcup I_2)oplus H_2(S^3-I_3sqcup I_4)to 0$$



From my calculations this would be



$0tomathbb{Z}to H_2(S^3-coprod_{i=1}^4 I_i)to mathbb{Z}oplusmathbb{Z}to 0$



But then $H_2(S^3-coprod_{i=1}^4 I_i)neq 0$, which is inconsistent with my first reasoning. Where is the mistake?






general-topology proof-verification algebraic-topology homology-cohomology homological-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
    $endgroup$
    – Servaes
    55 mins ago












  • $begingroup$
    @Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
    $endgroup$
    – Javi
    53 mins ago












  • $begingroup$
    @Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
    $endgroup$
    – Javi
    50 mins ago
















1












$begingroup$


I was computing the homology of $S^3-coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $mathbb{R}^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbb{R}^2$ and finaly the last one ends up with a space homotopic to $S^1$. Therefore, $H_2(S^3-coprod_{i=1}^4 I_i)=0$.



But in other calculations this caused me some problems so I decide to do it formally using Mayer-Vietoris. I had already computed $H_*(S^3-Isqcup I)$, giving me a consistent result with the intuition above, i.e. $H_2(S^3-Isqcup I)=mathbb{Z}$.



Now I decompose $S^3=(S^3-coprod_{i=1}^2 I_i)cup (S^3-coprod_{i=3}^4 I_i)$. From Mayer-Vietoris there is a short exact sequence



$$0to H_3(S^3)to H_2(S^3-coprod_{i=1}^4 I_i)to H_2(S^3-I_1sqcup I_2)oplus H_2(S^3-I_3sqcup I_4)to 0$$



From my calculations this would be



$0tomathbb{Z}to H_2(S^3-coprod_{i=1}^4 I_i)to mathbb{Z}oplusmathbb{Z}to 0$



But then $H_2(S^3-coprod_{i=1}^4 I_i)neq 0$, which is inconsistent with my first reasoning. Where is the mistake?






general-topology proof-verification algebraic-topology homology-cohomology homological-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
    $endgroup$
    – Servaes
    55 mins ago












  • $begingroup$
    @Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
    $endgroup$
    – Javi
    53 mins ago












  • $begingroup$
    @Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
    $endgroup$
    – Javi
    50 mins ago














1












1








1


0



$begingroup$


I was computing the homology of $S^3-coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $mathbb{R}^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbb{R}^2$ and finaly the last one ends up with a space homotopic to $S^1$. Therefore, $H_2(S^3-coprod_{i=1}^4 I_i)=0$.



But in other calculations this caused me some problems so I decide to do it formally using Mayer-Vietoris. I had already computed $H_*(S^3-Isqcup I)$, giving me a consistent result with the intuition above, i.e. $H_2(S^3-Isqcup I)=mathbb{Z}$.



Now I decompose $S^3=(S^3-coprod_{i=1}^2 I_i)cup (S^3-coprod_{i=3}^4 I_i)$. From Mayer-Vietoris there is a short exact sequence



$$0to H_3(S^3)to H_2(S^3-coprod_{i=1}^4 I_i)to H_2(S^3-I_1sqcup I_2)oplus H_2(S^3-I_3sqcup I_4)to 0$$



From my calculations this would be



$0tomathbb{Z}to H_2(S^3-coprod_{i=1}^4 I_i)to mathbb{Z}oplusmathbb{Z}to 0$



But then $H_2(S^3-coprod_{i=1}^4 I_i)neq 0$, which is inconsistent with my first reasoning. Where is the mistake?






general-topology proof-verification algebraic-topology homology-cohomology homological-algebra







share|cite|improve this question











$endgroup$




I was computing the homology of $S^3-coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $mathbb{R}^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbb{R}^2$ and finaly the last one ends up with a space homotopic to $S^1$. Therefore, $H_2(S^3-coprod_{i=1}^4 I_i)=0$.



But in other calculations this caused me some problems so I decide to do it formally using Mayer-Vietoris. I had already computed $H_*(S^3-Isqcup I)$, giving me a consistent result with the intuition above, i.e. $H_2(S^3-Isqcup I)=mathbb{Z}$.



Now I decompose $S^3=(S^3-coprod_{i=1}^2 I_i)cup (S^3-coprod_{i=3}^4 I_i)$. From Mayer-Vietoris there is a short exact sequence



$$0to H_3(S^3)to H_2(S^3-coprod_{i=1}^4 I_i)to H_2(S^3-I_1sqcup I_2)oplus H_2(S^3-I_3sqcup I_4)to 0$$



From my calculations this would be



$0tomathbb{Z}to H_2(S^3-coprod_{i=1}^4 I_i)to mathbb{Z}oplusmathbb{Z}to 0$



But then $H_2(S^3-coprod_{i=1}^4 I_i)neq 0$, which is inconsistent with my first reasoning. Where is the mistake?






general-topology proof-verification algebraic-topology homology-cohomology homological-algebra




general-topology proof-verification algebraic-topology homology-cohomology homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 52 mins ago







Javi

















asked 1 hour ago









JaviJavi

2,7572827




2,7572827








  • 2




    $begingroup$
    How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
    $endgroup$
    – Servaes
    55 mins ago












  • $begingroup$
    @Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
    $endgroup$
    – Javi
    53 mins ago












  • $begingroup$
    @Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
    $endgroup$
    – Javi
    50 mins ago














  • 2




    $begingroup$
    How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
    $endgroup$
    – Servaes
    55 mins ago












  • $begingroup$
    @Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
    $endgroup$
    – Javi
    53 mins ago












  • $begingroup$
    @Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
    $endgroup$
    – Javi
    50 mins ago








2




2




$begingroup$
How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
$endgroup$
– Servaes
55 mins ago






$begingroup$
How exactly is $coprod_{i=1}^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
$endgroup$
– Servaes
55 mins ago














$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
$endgroup$
– Javi
53 mins ago






$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbb{R}^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetilde{H}_i(S^n-h(D^k))=0$ for all $i$ and every embedding $h$ of $D^k$ into $S^n$, so I am using that implicitly for $n=3,k=1$.
$endgroup$
– Javi
53 mins ago














$begingroup$
@Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
$endgroup$
– Javi
50 mins ago




$begingroup$
@Servaes My original problem incluided intervals rather than points, but the result should be the same, and I would obtain the same mistake, so I guess I'm doing something wrong somewhere.
$endgroup$
– Javi
50 mins ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopic to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh that's it, I was so dumb! Thank you
    $endgroup$
    – Javi
    37 mins ago






  • 2




    $begingroup$
    @Javi, happens to all of us.
    $endgroup$
    – Carsten S
    36 mins ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopic to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh that's it, I was so dumb! Thank you
    $endgroup$
    – Javi
    37 mins ago






  • 2




    $begingroup$
    @Javi, happens to all of us.
    $endgroup$
    – Carsten S
    36 mins ago
















5












$begingroup$

By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopic to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh that's it, I was so dumb! Thank you
    $endgroup$
    – Javi
    37 mins ago






  • 2




    $begingroup$
    @Javi, happens to all of us.
    $endgroup$
    – Carsten S
    36 mins ago














5












5








5





$begingroup$

By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopic to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.






share|cite|improve this answer











$endgroup$



By your argument at the beginning, removing four points from a two-sphere should yield something which is homotopy equivalent to a zero-sphere. This is obviously not true. Your error is that after getting a space homotopic to $S^2$, you proceed as if it is actually homeomorphic to $S^2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 mins ago

























answered 39 mins ago









Carsten SCarsten S

7,02311334




7,02311334












  • $begingroup$
    Oh that's it, I was so dumb! Thank you
    $endgroup$
    – Javi
    37 mins ago






  • 2




    $begingroup$
    @Javi, happens to all of us.
    $endgroup$
    – Carsten S
    36 mins ago


















  • $begingroup$
    Oh that's it, I was so dumb! Thank you
    $endgroup$
    – Javi
    37 mins ago






  • 2




    $begingroup$
    @Javi, happens to all of us.
    $endgroup$
    – Carsten S
    36 mins ago
















$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
37 mins ago




$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
37 mins ago




2




2




$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
36 mins ago




$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
36 mins ago


















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