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Intuition for local degree formula for singular homology map between n-spheres
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In Hatcher's Algebraic Topology, he gives a formula for computing the degree of a map $f_*: H_n(S^n) to H_n(S^n)$ in case some point $y$ in $S^n$ has preimage consisting only of finitely many points $x_i$. Then $deg f = sum_i deg f mid_x_i$. 
He uses the above diagram to illustrate the proof.
Let $U_i$ be disjoint neighbourhoods of the $x_i$. Then by excision, he writes, the central term $H_n(S^n, S^n - f^-1(y))$ is the direct sum of the groups $H_n(U_i, U_i - x_i) cong mathbb Z$.
Now I think this is what he means:
The interior of $bigcup_i U_i ,cup (S^n - f^-1(y))$ covers $S^n$ so we can apply the excision theorem. Since the $U_i$ are all disjoint, we get an isomorphism $bigoplus_i H_n(U_i, U_i - x_i) cong H_n(S^n, S^n - f^-1(y))$.
I would like to know whether I'm on the right track here.
Apart from this, I would like to gain some geometric intuition as to why the $n$-th homology group of the $n$-sphere relative to the $n$-sphere with $k$ points removed is $bigoplus_i=1^k mathbb Z$.
Since $(S^n, S^n - f^-1(y))$ is not what Hatcher calls a good pair we are technically not allowed to think in terms of $S^n/(S^n - f^-1(y))$. I've tried that anyway but it didn't help.
Considering the special case of $H_1(S^1 / S^1 - x_1, x_2, x_3)$, we see that the elements of this homology group are generated by paths in $S^1$ with endpoints not in $x_1, x_2, x_3$. I can't see intuitively how this gives $H_1 cong mathbb Z oplus mathbb Z oplus mathbb Z$ and I would appreciate any help.
algebraic-topology homology-cohomology homology-sphere
asked Jul 27 at 10:20
Lukas Kofler
5811518
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In Hatcher's Algebraic Topology, he gives a formula for computing the degree of a map $f_*: H_n(S^n) to H_n(S^n)$ in case some point $y$ in $S^n$ has preimage consisting only of finitely many points $x_i$. Then $deg f = sum_i deg f mid_x_i$.
He uses the above diagram to illustrate the proof.
Let $U_i$ be disjoint neighbourhoods of the $x_i$. Then by excision, he writes, the central term $H_n(S^n, S^n - f^-1(y))$ is the direct sum of the groups $H_n(U_i, U_i - x_i) cong mathbb Z$.
Now I think this is what he means:
The interior of $bigcup_i U_i ,cup (S^n - f^-1(y))$ covers $S^n$ so we can apply the excision theorem. Since the $U_i$ are all disjoint, we get an isomorphism $bigoplus_i H_n(U_i, U_i - x_i) cong H_n(S^n, S^n - f^-1(y))$.
I would like to know whether I'm on the right track here.
Apart from this, I would like to gain some geometric intuition as to why the $n$-th homology group of the $n$-sphere relative to the $n$-sphere with $k$ points removed is $bigoplus_i=1^k mathbb Z$.
Since $(S^n, S^n - f^-1(y))$ is not what Hatcher calls a good pair we are technically not allowed to think in terms of $S^n/(S^n - f^-1(y))$. I've tried that anyway but it didn't help.
Considering the special case of $H_1(S^1 / S^1 - x_1, x_2, x_3)$, we see that the elements of this homology group are generated by paths in $S^1$ with endpoints not in $x_1, x_2, x_3$. I can't see intuitively how this gives $H_1 cong mathbb Z oplus mathbb Z oplus mathbb Z$ and I would appreciate any help.
algebraic-topology homology-cohomology homology-sphere
asked Jul 27 at 10:20
Lukas Kofler
5811518
1
One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31
add a comment |Â
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0
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up vote
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down vote
favorite
In Hatcher's Algebraic Topology, he gives a formula for computing the degree of a map $f_*: H_n(S^n) to H_n(S^n)$ in case some point $y$ in $S^n$ has preimage consisting only of finitely many points $x_i$. Then $deg f = sum_i deg f mid_x_i$.
He uses the above diagram to illustrate the proof.
Let $U_i$ be disjoint neighbourhoods of the $x_i$. Then by excision, he writes, the central term $H_n(S^n, S^n - f^-1(y))$ is the direct sum of the groups $H_n(U_i, U_i - x_i) cong mathbb Z$.
Now I think this is what he means:
The interior of $bigcup_i U_i ,cup (S^n - f^-1(y))$ covers $S^n$ so we can apply the excision theorem. Since the $U_i$ are all disjoint, we get an isomorphism $bigoplus_i H_n(U_i, U_i - x_i) cong H_n(S^n, S^n - f^-1(y))$.
I would like to know whether I'm on the right track here.
Apart from this, I would like to gain some geometric intuition as to why the $n$-th homology group of the $n$-sphere relative to the $n$-sphere with $k$ points removed is $bigoplus_i=1^k mathbb Z$.
Since $(S^n, S^n - f^-1(y))$ is not what Hatcher calls a good pair we are technically not allowed to think in terms of $S^n/(S^n - f^-1(y))$. I've tried that anyway but it didn't help.
Considering the special case of $H_1(S^1 / S^1 - x_1, x_2, x_3)$, we see that the elements of this homology group are generated by paths in $S^1$ with endpoints not in $x_1, x_2, x_3$. I can't see intuitively how this gives $H_1 cong mathbb Z oplus mathbb Z oplus mathbb Z$ and I would appreciate any help.
algebraic-topology homology-cohomology homology-sphere
asked Jul 27 at 10:20
Lukas Kofler
5811518
In Hatcher's Algebraic Topology, he gives a formula for computing the degree of a map $f_*: H_n(S^n) to H_n(S^n)$ in case some point $y$ in $S^n$ has preimage consisting only of finitely many points $x_i$. Then $deg f = sum_i deg f mid_x_i$.
He uses the above diagram to illustrate the proof.
Let $U_i$ be disjoint neighbourhoods of the $x_i$. Then by excision, he writes, the central term $H_n(S^n, S^n - f^-1(y))$ is the direct sum of the groups $H_n(U_i, U_i - x_i) cong mathbb Z$.
Now I think this is what he means:
The interior of $bigcup_i U_i ,cup (S^n - f^-1(y))$ covers $S^n$ so we can apply the excision theorem. Since the $U_i$ are all disjoint, we get an isomorphism $bigoplus_i H_n(U_i, U_i - x_i) cong H_n(S^n, S^n - f^-1(y))$.
I would like to know whether I'm on the right track here.
Apart from this, I would like to gain some geometric intuition as to why the $n$-th homology group of the $n$-sphere relative to the $n$-sphere with $k$ points removed is $bigoplus_i=1^k mathbb Z$.
Since $(S^n, S^n - f^-1(y))$ is not what Hatcher calls a good pair we are technically not allowed to think in terms of $S^n/(S^n - f^-1(y))$. I've tried that anyway but it didn't help.
Considering the special case of $H_1(S^1 / S^1 - x_1, x_2, x_3)$, we see that the elements of this homology group are generated by paths in $S^1$ with endpoints not in $x_1, x_2, x_3$. I can't see intuitively how this gives $H_1 cong mathbb Z oplus mathbb Z oplus mathbb Z$ and I would appreciate any help.
algebraic-topology homology-cohomology homology-sphere
asked Jul 27 at 10:20
Lukas Kofler
5811518
asked Jul 27 at 10:20
Lukas Kofler
5811518
asked Jul 27 at 10:20
Lukas Kofler
5811518
asked Jul 27 at 10:20
Lukas Kofler
5811518
5811518
1
One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31
add a comment |Â
1
One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31
1
1
One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31
One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31
add a comment |Â
1 Answer
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You excise the set $B = S^n - bigcup_i=1^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^-1(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^-1(y)) approx H_n(bigcup_i=1^k U_i, bigcup_i=1^k (U_i - x_i)) = H_n(bigcup_i=1^k( U_i, U_i - x_i)) approx bigoplus_i=1^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^-1(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^-1(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
answered Jul 27 at 13:53
Paul Frost
3,613420
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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up vote
1
down vote
You excise the set $B = S^n - bigcup_i=1^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^-1(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^-1(y)) approx H_n(bigcup_i=1^k U_i, bigcup_i=1^k (U_i - x_i)) = H_n(bigcup_i=1^k( U_i, U_i - x_i)) approx bigoplus_i=1^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^-1(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^-1(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
answered Jul 27 at 13:53
Paul Frost
3,613420
add a comment |Â
up vote
1
down vote
You excise the set $B = S^n - bigcup_i=1^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^-1(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^-1(y)) approx H_n(bigcup_i=1^k U_i, bigcup_i=1^k (U_i - x_i)) = H_n(bigcup_i=1^k( U_i, U_i - x_i)) approx bigoplus_i=1^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^-1(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^-1(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
answered Jul 27 at 13:53
Paul Frost
3,613420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You excise the set $B = S^n - bigcup_i=1^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^-1(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^-1(y)) approx H_n(bigcup_i=1^k U_i, bigcup_i=1^k (U_i - x_i)) = H_n(bigcup_i=1^k( U_i, U_i - x_i)) approx bigoplus_i=1^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^-1(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^-1(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
answered Jul 27 at 13:53
Paul Frost
3,613420
You excise the set $B = S^n - bigcup_i=1^k U_i$. This is possible because $B$ is closed and contained in $S^n - f^-1(y)$ which is open. Therefore
$$H_n(S^n,S^n - f^-1(y)) approx H_n(bigcup_i=1^k U_i, bigcup_i=1^k (U_i - x_i)) = H_n(bigcup_i=1^k( U_i, U_i - x_i)) approx bigoplus_i=1^k H_n(U_i, U_i - x_i) .$$
An element of $H_n(S^n,S^n - f^-1(y))$ is a homology class of a chain of singular $n$-simplices in $S^n$ whose boundary is contained in $S^n - f^-1(y)$. The most elementary such chains are single $n$-simplices in $S^n$ which contain exactly one $x_i$ in its interior. It can be shown that they generate the complete relative homology group. I am not sure whether this fact is intuitively clear, but the purpose of homology theory is to give a formal framework.
answered Jul 27 at 13:53
Paul Frost
3,613420
answered Jul 27 at 13:53
Paul Frost
3,613420
answered Jul 27 at 13:53
Paul Frost
3,613420
answered Jul 27 at 13:53
Paul Frost
3,613420
3,613420
add a comment |Â
add a comment |Â
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One way to understand the relative homology $H_*(S^n,S^n-f^-1(y))$ is by Alexander duality, as discussed by Hatcher around Th. 3.44 and Co, 3.45.
– Tyrone
Jul 27 at 12:31