Finding a mistake using Mayer-VietorisRigorous application of Mayer-Vietoris to a quotient of $S^{2}times...
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Finding a mistake using Mayer-Vietoris
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 sequenceMayer-Vietoris in reduced homology for a torus.$H_1(S^1)$ with Mayer-Vietoris sequenceCompute the homology groups using Mayer-Vietoris sequenceComputing the homology of genus $g$ surface, using Mayer-VietorisBoundary Map in Mayer Vietoris and Homology of Knot ComplementConfusion in using Mayer-Vietoris theoremHomology of Klein bottle with Mayer-Vietoris
$begingroup$
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
asked 16 hours ago
JaviJavi
2,7972827
$endgroup$
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 $mathbb{R}^3$, 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
asked 16 hours ago
JaviJavi
2,7972827
$endgroup$
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
16 hours 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
16 hours 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
16 hours ago
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 $mathbb{R}^3$, 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
asked 16 hours ago
JaviJavi
2,7972827
$endgroup$
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
asked 16 hours ago
JaviJavi
2,7972827
asked 16 hours ago
JaviJavi
2,7972827
edited 16 hours ago
Javi
asked 16 hours ago
JaviJavi
2,7972827
asked 16 hours ago
JaviJavi
2,7972827
asked 16 hours ago
JaviJavi
2,7972827
2,7972827
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
16 hours 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
16 hours 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
16 hours ago
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
16 hours 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
16 hours 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
16 hours 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
16 hours 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
16 hours 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
16 hours 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
16 hours 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
16 hours 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
16 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 15 hours ago
Carsten SCarsten S
7,19811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
add a comment |
Your Answer
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1 Answer
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votes
$begingroup$
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 15 hours ago
Carsten SCarsten S
7,19811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
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 15 hours ago
Carsten SCarsten S
7,19811436
$endgroup$
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
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 15 hours ago
Carsten SCarsten S
7,19811436
$endgroup$
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 15 hours ago
Carsten SCarsten S
7,19811436
edited 15 hours ago
answered 15 hours ago
Carsten SCarsten S
7,19811436
answered 15 hours ago
Carsten SCarsten S
7,19811436
answered 15 hours ago
Carsten SCarsten S
7,19811436
7,19811436
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
add a comment |
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
1
1
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
$begingroup$
Oh that's it, I was so dumb! Thank you
$endgroup$
– Javi
15 hours ago
11
11
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
$begingroup$
@Javi, happens to all of us.
$endgroup$
– Carsten S
15 hours ago
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
16 hours 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
16 hours 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
16 hours ago