Finding a mistake using Mayer-Vietoris
Clash Royale CLAN TAG#URR8PPP
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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 $mathbbR^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbbR^2$ 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. $H_2(S^3-Isqcup I)=mathbbZ$.
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
$0tomathbbZto H_2(S^3-coprod_i=1^4 I_i)to mathbbZoplusmathbbZto 0$
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
asked Feb 28 at 14:35
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
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 $mathbbR^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbbR^2$ 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. $H_2(S^3-Isqcup I)=mathbbZ$.
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
$0tomathbbZto H_2(S^3-coprod_i=1^4 I_i)to mathbbZoplusmathbbZto 0$
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
asked Feb 28 at 14:35
JaviJavi
3,0212832
$endgroup$
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
Feb 28 at 14:42
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@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
Feb 28 at 14:47
add a comment |
$begingroup$
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 $mathbbR^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbbR^2$ 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. $H_2(S^3-Isqcup I)=mathbbZ$.
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
$0tomathbbZto H_2(S^3-coprod_i=1^4 I_i)to mathbbZoplusmathbbZto 0$
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
asked Feb 28 at 14:35
JaviJavi
3,0212832
$endgroup$
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 $mathbbR^3$, removing another one gives an espace homotopic to $S^2$, removing the third one results in something homotopic to $mathbbR^2$ 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. $H_2(S^3-Isqcup I)=mathbbZ$.
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
$0tomathbbZto H_2(S^3-coprod_i=1^4 I_i)to mathbbZoplusmathbbZto 0$
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
asked Feb 28 at 14:35
JaviJavi
3,0212832
asked Feb 28 at 14:35
JaviJavi
3,0212832
edited Feb 28 at 14:45
Javi
asked Feb 28 at 14:35
JaviJavi
3,0212832
asked Feb 28 at 14:35
JaviJavi
3,0212832
asked Feb 28 at 14:35
JaviJavi
3,0212832
3,0212832
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
Feb 28 at 14:42
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@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
Feb 28 at 14:47
add a comment |
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
Feb 28 at 14:42
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@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
Feb 28 at 14:47
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
Feb 28 at 14:42
$begingroup$
How exactly is $coprod_i=1^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
$endgroup$
– Servaes
Feb 28 at 14:42
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@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
Feb 28 at 14:47
$begingroup$
@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
Feb 28 at 14:47
add a comment |
1 Answer
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votes
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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 $S^2$.
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
add a comment |
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$begingroup$
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 $S^2$.
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
add a comment |
$begingroup$
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 $S^2$.
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
add a comment |
$begingroup$
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 $S^2$.
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
$endgroup$
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 $S^2$.
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
edited Feb 28 at 15:25
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
answered Feb 28 at 14:58
Carsten SCarsten S
7,28811436
7,28811436
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
add a comment |
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
1
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
Feb 28 at 15:00
12
12
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
Feb 28 at 15:01
add a comment |
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$begingroup$
How exactly is $coprod_i=1^4 I_i$ a subset of $S^3$? And why not just remove four distinct points?
$endgroup$
– Servaes
Feb 28 at 14:42
$begingroup$
@Servaes I'm considering an embedding, the most simple possible, for example an isometry onto its image on $mathbbR^3$ and then adding the point of infinity. Anyways, there is a theorem saying that $widetildeH_i(S^n-h(D^k))=0$ 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
Feb 28 at 14:44
$begingroup$
@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
Feb 28 at 14:47