Mixing
Help with homology axioms
$begingroup$
I am currently interested in the calculation $H_1(S^2-I)$ where $I$ is an interval embedded in the unit sphere.
The answer should be 0, and Hatcher proves it in his book in the sections "Classical Computations" at the end of chapter two. The main idea is to bisect $I$ into two intervals $I_+$ and $I_-$ and use mayer vietoris for the sets $A=S^2-I_+$ and $B=S^2-I_-$. As part of this sequence there will be a map $H_1(S^2-I)to H_1(A)oplus H_1(B)$ induced by the inclusion. Now the idea is to say that if $xin H_1(S^2-I)$ is not zero, then a representative of x is not a boundary in either $A$ or $B$ and then repeat the argument, culminating in the fact that we will get a nested sequence of intervals intersecting in one point.
See proposition 2B.1 in Hatcher's "algebraic topology"
My question is - how can one translate this proof to axiomatic homology theory (without using singular homology). I had heard that something like a "colimit axiom" can help with this, but I can't find a suitable reference.
Thank you very much in advance!
geometry algebraic-topology homology-cohomology
edited Jan 21 at 18:27
Paul Frost
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
$endgroup$
add a comment |
$begingroup$
I am currently interested in the calculation $H_1(S^2-I)$ where $I$ is an interval embedded in the unit sphere.
The answer should be 0, and Hatcher proves it in his book in the sections "Classical Computations" at the end of chapter two. The main idea is to bisect $I$ into two intervals $I_+$ and $I_-$ and use mayer vietoris for the sets $A=S^2-I_+$ and $B=S^2-I_-$. As part of this sequence there will be a map $H_1(S^2-I)to H_1(A)oplus H_1(B)$ induced by the inclusion. Now the idea is to say that if $xin H_1(S^2-I)$ is not zero, then a representative of x is not a boundary in either $A$ or $B$ and then repeat the argument, culminating in the fact that we will get a nested sequence of intervals intersecting in one point.
See proposition 2B.1 in Hatcher's "algebraic topology"
My question is - how can one translate this proof to axiomatic homology theory (without using singular homology). I had heard that something like a "colimit axiom" can help with this, but I can't find a suitable reference.
Thank you very much in advance!
geometry algebraic-topology homology-cohomology
edited Jan 21 at 18:27
Paul Frost
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
$endgroup$
add a comment |
$begingroup$
I am currently interested in the calculation $H_1(S^2-I)$ where $I$ is an interval embedded in the unit sphere.
The answer should be 0, and Hatcher proves it in his book in the sections "Classical Computations" at the end of chapter two. The main idea is to bisect $I$ into two intervals $I_+$ and $I_-$ and use mayer vietoris for the sets $A=S^2-I_+$ and $B=S^2-I_-$. As part of this sequence there will be a map $H_1(S^2-I)to H_1(A)oplus H_1(B)$ induced by the inclusion. Now the idea is to say that if $xin H_1(S^2-I)$ is not zero, then a representative of x is not a boundary in either $A$ or $B$ and then repeat the argument, culminating in the fact that we will get a nested sequence of intervals intersecting in one point.
See proposition 2B.1 in Hatcher's "algebraic topology"
My question is - how can one translate this proof to axiomatic homology theory (without using singular homology). I had heard that something like a "colimit axiom" can help with this, but I can't find a suitable reference.
Thank you very much in advance!
geometry algebraic-topology homology-cohomology
edited Jan 21 at 18:27
Paul Frost
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
$endgroup$
I am currently interested in the calculation $H_1(S^2-I)$ where $I$ is an interval embedded in the unit sphere.
The answer should be 0, and Hatcher proves it in his book in the sections "Classical Computations" at the end of chapter two. The main idea is to bisect $I$ into two intervals $I_+$ and $I_-$ and use mayer vietoris for the sets $A=S^2-I_+$ and $B=S^2-I_-$. As part of this sequence there will be a map $H_1(S^2-I)to H_1(A)oplus H_1(B)$ induced by the inclusion. Now the idea is to say that if $xin H_1(S^2-I)$ is not zero, then a representative of x is not a boundary in either $A$ or $B$ and then repeat the argument, culminating in the fact that we will get a nested sequence of intervals intersecting in one point.
See proposition 2B.1 in Hatcher's "algebraic topology"
My question is - how can one translate this proof to axiomatic homology theory (without using singular homology). I had heard that something like a "colimit axiom" can help with this, but I can't find a suitable reference.
Thank you very much in advance!
geometry algebraic-topology homology-cohomology
geometry algebraic-topology homology-cohomology
edited Jan 21 at 18:27
Paul Frost
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
edited Jan 21 at 18:27
Paul Frost
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
edited Jan 21 at 18:27
Paul Frost
11.4k3934
edited Jan 21 at 18:27
Paul Frost
11.4k3934
edited Jan 21 at 18:27
Paul Frost
11.4k3934
11.4k3934
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
asked Jan 21 at 12:49
Benny ZackBenny Zack
365213
365213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It sounds to me like you are saying: given an arc $I$ with $H_1(S^2 - I) neq 0$, you may find an infinite sequence of arcs $I_k subset I_{k+1} subset cdots $ with $I = I_0$, and $bigcap I_k = {pt}$, with a class $x_0 in H_1(S^2 - I_0)$ so that $H_1(S^2 - I_k) to H_1(S^2 - I_{k+1})$ sends $x_k$ injectively to some class $x_{k+1}$.
A fact about singular homology is that if you have an increasing sequence $U_i$ of open sets of a space $X$, so that $overline U_i subset U_{i+1}$, then $H_*(bigcup U_i) = text{colim} H_*(U_i)$; see prop 3.33 of Hatcher. This is certainly true in our situation above, and identifies $$H_1 Bbb R^2 = H_1(S^2 setminus {pt}) = text{colim} H_1(S^2 - I_k),$$
but we have seen above that the class $x_0$ does not die in any $H_1(S^2 - I_k)$, and hence does not die in the colimit.
Now you'd like to understand how to discuss the direct limit axiom; see this excellent MO thread. In particular, you simply need to know that your homology theory preserves infinite wedge products.
$endgroup$
add a comment |
$begingroup$
This is not really an answer to your question concerning the axioms of a homology theory, but it shows that $H_1(S^2 setminus I) = 0$ for any ordinary homology theory.
The reason is that $S^2 setminus I$ is contractible. This follows from Lukas Geyer's answer to Connected Partitions of Spheres. We conclude that $S^2 setminus I$ is simply connected. Now apply the Riemann mapping theorem.
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
add a comment |
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2 Answers
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$begingroup$
It sounds to me like you are saying: given an arc $I$ with $H_1(S^2 - I) neq 0$, you may find an infinite sequence of arcs $I_k subset I_{k+1} subset cdots $ with $I = I_0$, and $bigcap I_k = {pt}$, with a class $x_0 in H_1(S^2 - I_0)$ so that $H_1(S^2 - I_k) to H_1(S^2 - I_{k+1})$ sends $x_k$ injectively to some class $x_{k+1}$.
A fact about singular homology is that if you have an increasing sequence $U_i$ of open sets of a space $X$, so that $overline U_i subset U_{i+1}$, then $H_*(bigcup U_i) = text{colim} H_*(U_i)$; see prop 3.33 of Hatcher. This is certainly true in our situation above, and identifies $$H_1 Bbb R^2 = H_1(S^2 setminus {pt}) = text{colim} H_1(S^2 - I_k),$$
but we have seen above that the class $x_0$ does not die in any $H_1(S^2 - I_k)$, and hence does not die in the colimit.
Now you'd like to understand how to discuss the direct limit axiom; see this excellent MO thread. In particular, you simply need to know that your homology theory preserves infinite wedge products.
$endgroup$
add a comment |
$begingroup$
It sounds to me like you are saying: given an arc $I$ with $H_1(S^2 - I) neq 0$, you may find an infinite sequence of arcs $I_k subset I_{k+1} subset cdots $ with $I = I_0$, and $bigcap I_k = {pt}$, with a class $x_0 in H_1(S^2 - I_0)$ so that $H_1(S^2 - I_k) to H_1(S^2 - I_{k+1})$ sends $x_k$ injectively to some class $x_{k+1}$.
A fact about singular homology is that if you have an increasing sequence $U_i$ of open sets of a space $X$, so that $overline U_i subset U_{i+1}$, then $H_*(bigcup U_i) = text{colim} H_*(U_i)$; see prop 3.33 of Hatcher. This is certainly true in our situation above, and identifies $$H_1 Bbb R^2 = H_1(S^2 setminus {pt}) = text{colim} H_1(S^2 - I_k),$$
but we have seen above that the class $x_0$ does not die in any $H_1(S^2 - I_k)$, and hence does not die in the colimit.
Now you'd like to understand how to discuss the direct limit axiom; see this excellent MO thread. In particular, you simply need to know that your homology theory preserves infinite wedge products.
$endgroup$
add a comment |
$begingroup$
It sounds to me like you are saying: given an arc $I$ with $H_1(S^2 - I) neq 0$, you may find an infinite sequence of arcs $I_k subset I_{k+1} subset cdots $ with $I = I_0$, and $bigcap I_k = {pt}$, with a class $x_0 in H_1(S^2 - I_0)$ so that $H_1(S^2 - I_k) to H_1(S^2 - I_{k+1})$ sends $x_k$ injectively to some class $x_{k+1}$.
A fact about singular homology is that if you have an increasing sequence $U_i$ of open sets of a space $X$, so that $overline U_i subset U_{i+1}$, then $H_*(bigcup U_i) = text{colim} H_*(U_i)$; see prop 3.33 of Hatcher. This is certainly true in our situation above, and identifies $$H_1 Bbb R^2 = H_1(S^2 setminus {pt}) = text{colim} H_1(S^2 - I_k),$$
but we have seen above that the class $x_0$ does not die in any $H_1(S^2 - I_k)$, and hence does not die in the colimit.
Now you'd like to understand how to discuss the direct limit axiom; see this excellent MO thread. In particular, you simply need to know that your homology theory preserves infinite wedge products.
$endgroup$
It sounds to me like you are saying: given an arc $I$ with $H_1(S^2 - I) neq 0$, you may find an infinite sequence of arcs $I_k subset I_{k+1} subset cdots $ with $I = I_0$, and $bigcap I_k = {pt}$, with a class $x_0 in H_1(S^2 - I_0)$ so that $H_1(S^2 - I_k) to H_1(S^2 - I_{k+1})$ sends $x_k$ injectively to some class $x_{k+1}$.
A fact about singular homology is that if you have an increasing sequence $U_i$ of open sets of a space $X$, so that $overline U_i subset U_{i+1}$, then $H_*(bigcup U_i) = text{colim} H_*(U_i)$; see prop 3.33 of Hatcher. This is certainly true in our situation above, and identifies $$H_1 Bbb R^2 = H_1(S^2 setminus {pt}) = text{colim} H_1(S^2 - I_k),$$
but we have seen above that the class $x_0$ does not die in any $H_1(S^2 - I_k)$, and hence does not die in the colimit.
Now you'd like to understand how to discuss the direct limit axiom; see this excellent MO thread. In particular, you simply need to know that your homology theory preserves infinite wedge products.
answered Jan 21 at 15:14
user98602
add a comment |
add a comment |
$begingroup$
This is not really an answer to your question concerning the axioms of a homology theory, but it shows that $H_1(S^2 setminus I) = 0$ for any ordinary homology theory.
The reason is that $S^2 setminus I$ is contractible. This follows from Lukas Geyer's answer to Connected Partitions of Spheres. We conclude that $S^2 setminus I$ is simply connected. Now apply the Riemann mapping theorem.
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
add a comment |
$begingroup$
This is not really an answer to your question concerning the axioms of a homology theory, but it shows that $H_1(S^2 setminus I) = 0$ for any ordinary homology theory.
The reason is that $S^2 setminus I$ is contractible. This follows from Lukas Geyer's answer to Connected Partitions of Spheres. We conclude that $S^2 setminus I$ is simply connected. Now apply the Riemann mapping theorem.
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
add a comment |
$begingroup$
This is not really an answer to your question concerning the axioms of a homology theory, but it shows that $H_1(S^2 setminus I) = 0$ for any ordinary homology theory.
The reason is that $S^2 setminus I$ is contractible. This follows from Lukas Geyer's answer to Connected Partitions of Spheres. We conclude that $S^2 setminus I$ is simply connected. Now apply the Riemann mapping theorem.
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
$endgroup$
This is not really an answer to your question concerning the axioms of a homology theory, but it shows that $H_1(S^2 setminus I) = 0$ for any ordinary homology theory.
The reason is that $S^2 setminus I$ is contractible. This follows from Lukas Geyer's answer to Connected Partitions of Spheres. We conclude that $S^2 setminus I$ is simply connected. Now apply the Riemann mapping theorem.
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
edited Jan 21 at 23:47
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
answered Jan 21 at 22:17
Paul FrostPaul Frost
11.4k3934
11.4k3934
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
add a comment |
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
Now apply the Hurewicz theorem for $n=1$, $H_1 = pi_1^{text{ab}}$! That is a lot less effort than Riemann mapping :)
$endgroup$
– user98602
Jan 22 at 14:07
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
@MikeMiller You are right if we work with singular homology. For an arbitrary homology theory satisfying the Eilenberg-Steenrod axioms I am not sure whether $H_1 = pi_1^{ab}$ is true. Probably we need again some addtional axioms. Anyway, my argument strongly depends on the space $S^2$. It cannot be generalized to $S^n$ for $n > 2$ where we can find wild arcs whose complements are not simply connected. Hatcher's and your arguments work for all $n$.
$endgroup$
– Paul Frost
Jan 22 at 15:11
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
Since we are considering open subsets of Euclidean space, they may be triangulated, and all imaginable homology theories agree, given the infinite wedge axiom. But you are right, and I hadn't considered that. The standard proof is essentially a composite isomorphism $pi_1^{ab} to Omega_1^{or} to H_1$, and for the last map to be clear one needs geometric representatives.
$endgroup$
– user98602
Jan 22 at 16:28
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
$begingroup$
@MikeMiller So the (infinite) wedge axiom is in fact the key since it assures uniqueness on the class of all CW-complexes.
$endgroup$
– Paul Frost
Jan 22 at 16:40
add a comment |
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