Mixing
Hatcher Problem 2.1.16 (b)
I am trying to do the stated problem in Hatcher:
Show $H_1(X,A) = 0$ iff $H_1(A) to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.
Now I have reduced the problem to showing that $i_ast : H_0(A) to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$ldots to H_1(X) to H_1(X,A) to H_0(A) to H_0(X) to H_0(X,A) to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_ast$ is not injective. Then there is a $tau in C_0(A)$ such that $[tau circ i] = 0$ but $[tau] neq 0$. That is to say, $tau circ i = partial(sigma)$ for some $sigma in C_1(X)$ but $tau$ is not the boundary of any $sigma'in C_1(A)$. However I'm confused because to me the only way for $tau circ i$ to be the boundary of a singular $1$ - simplex $sigma$ in $X$ is if $sigma$ is a loop. What's wrong here?
Thanks.
algebraic-topology
add a comment |
I am trying to do the stated problem in Hatcher:
Show $H_1(X,A) = 0$ iff $H_1(A) to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.
Now I have reduced the problem to showing that $i_ast : H_0(A) to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$ldots to H_1(X) to H_1(X,A) to H_0(A) to H_0(X) to H_0(X,A) to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_ast$ is not injective. Then there is a $tau in C_0(A)$ such that $[tau circ i] = 0$ but $[tau] neq 0$. That is to say, $tau circ i = partial(sigma)$ for some $sigma in C_1(X)$ but $tau$ is not the boundary of any $sigma'in C_1(A)$. However I'm confused because to me the only way for $tau circ i$ to be the boundary of a singular $1$ - simplex $sigma$ in $X$ is if $sigma$ is a loop. What's wrong here?
Thanks.
algebraic-topology
Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40
add a comment |
I am trying to do the stated problem in Hatcher:
Show $H_1(X,A) = 0$ iff $H_1(A) to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.
Now I have reduced the problem to showing that $i_ast : H_0(A) to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$ldots to H_1(X) to H_1(X,A) to H_0(A) to H_0(X) to H_0(X,A) to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_ast$ is not injective. Then there is a $tau in C_0(A)$ such that $[tau circ i] = 0$ but $[tau] neq 0$. That is to say, $tau circ i = partial(sigma)$ for some $sigma in C_1(X)$ but $tau$ is not the boundary of any $sigma'in C_1(A)$. However I'm confused because to me the only way for $tau circ i$ to be the boundary of a singular $1$ - simplex $sigma$ in $X$ is if $sigma$ is a loop. What's wrong here?
Thanks.
algebraic-topology
I am trying to do the stated problem in Hatcher:
Show $H_1(X,A) = 0$ iff $H_1(A) to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.
Now I have reduced the problem to showing that $i_ast : H_0(A) to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$ldots to H_1(X) to H_1(X,A) to H_0(A) to H_0(X) to H_0(X,A) to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_ast$ is not injective. Then there is a $tau in C_0(A)$ such that $[tau circ i] = 0$ but $[tau] neq 0$. That is to say, $tau circ i = partial(sigma)$ for some $sigma in C_1(X)$ but $tau$ is not the boundary of any $sigma'in C_1(A)$. However I'm confused because to me the only way for $tau circ i$ to be the boundary of a singular $1$ - simplex $sigma$ in $X$ is if $sigma$ is a loop. What's wrong here?
Thanks.
algebraic-topology
algebraic-topology
edited Nov 11 '12 at 23:40
asked Nov 10 '12 at 6:28
user38268
Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40
add a comment |
Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40
Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40
add a comment |
1 Answer
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So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) to H_1(X)$ is surjective and that $H_0(A) to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) neq 0$, so let $sigma neq 0 in H_1(X,A)$. Then you have two possibilities:
- Either $partial sigma = 0 in H_0(A)$. Then by exactness, there exists $gamma in H_1(X)$ such that $j_*gamma = sigma$. But then $i_* : H_0(A) to H_0(X)$ cannot be surjective: if you have $gamma = i_*alpha$, then $j_*gamma = j_*i_*alpha = 0$ by exactness.
- Or $partial sigma$ is nonzero. But then $i_*(partialsigma) = 0$ by exactness, so $i_* : H_0(A) to H_0(X)$ is not injective.
So all that's left to prove is that $i_* H_0(A) to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b in A$ in the two distinct path components. Then $[a]-[b] in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) to H_0(X)$ is not injective. Let $0 neq alpha in H_0(A)$ be such that $i_*alpha = 0$. Write $alpha = sum_{k in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(alpha) = sum_{k in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $alpha neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
add a comment |
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So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) to H_1(X)$ is surjective and that $H_0(A) to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) neq 0$, so let $sigma neq 0 in H_1(X,A)$. Then you have two possibilities:
- Either $partial sigma = 0 in H_0(A)$. Then by exactness, there exists $gamma in H_1(X)$ such that $j_*gamma = sigma$. But then $i_* : H_0(A) to H_0(X)$ cannot be surjective: if you have $gamma = i_*alpha$, then $j_*gamma = j_*i_*alpha = 0$ by exactness.
- Or $partial sigma$ is nonzero. But then $i_*(partialsigma) = 0$ by exactness, so $i_* : H_0(A) to H_0(X)$ is not injective.
So all that's left to prove is that $i_* H_0(A) to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b in A$ in the two distinct path components. Then $[a]-[b] in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) to H_0(X)$ is not injective. Let $0 neq alpha in H_0(A)$ be such that $i_*alpha = 0$. Write $alpha = sum_{k in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(alpha) = sum_{k in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $alpha neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
add a comment |
So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) to H_1(X)$ is surjective and that $H_0(A) to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) neq 0$, so let $sigma neq 0 in H_1(X,A)$. Then you have two possibilities:
- Either $partial sigma = 0 in H_0(A)$. Then by exactness, there exists $gamma in H_1(X)$ such that $j_*gamma = sigma$. But then $i_* : H_0(A) to H_0(X)$ cannot be surjective: if you have $gamma = i_*alpha$, then $j_*gamma = j_*i_*alpha = 0$ by exactness.
- Or $partial sigma$ is nonzero. But then $i_*(partialsigma) = 0$ by exactness, so $i_* : H_0(A) to H_0(X)$ is not injective.
So all that's left to prove is that $i_* H_0(A) to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b in A$ in the two distinct path components. Then $[a]-[b] in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) to H_0(X)$ is not injective. Let $0 neq alpha in H_0(A)$ be such that $i_*alpha = 0$. Write $alpha = sum_{k in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(alpha) = sum_{k in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $alpha neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
add a comment |
So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) to H_1(X)$ is surjective and that $H_0(A) to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) neq 0$, so let $sigma neq 0 in H_1(X,A)$. Then you have two possibilities:
- Either $partial sigma = 0 in H_0(A)$. Then by exactness, there exists $gamma in H_1(X)$ such that $j_*gamma = sigma$. But then $i_* : H_0(A) to H_0(X)$ cannot be surjective: if you have $gamma = i_*alpha$, then $j_*gamma = j_*i_*alpha = 0$ by exactness.
- Or $partial sigma$ is nonzero. But then $i_*(partialsigma) = 0$ by exactness, so $i_* : H_0(A) to H_0(X)$ is not injective.
So all that's left to prove is that $i_* H_0(A) to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b in A$ in the two distinct path components. Then $[a]-[b] in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) to H_0(X)$ is not injective. Let $0 neq alpha in H_0(A)$ be such that $i_*alpha = 0$. Write $alpha = sum_{k in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(alpha) = sum_{k in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $alpha neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) to H_1(X)$ is surjective and that $H_0(A) to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) neq 0$, so let $sigma neq 0 in H_1(X,A)$. Then you have two possibilities:
- Either $partial sigma = 0 in H_0(A)$. Then by exactness, there exists $gamma in H_1(X)$ such that $j_*gamma = sigma$. But then $i_* : H_0(A) to H_0(X)$ cannot be surjective: if you have $gamma = i_*alpha$, then $j_*gamma = j_*i_*alpha = 0$ by exactness.
- Or $partial sigma$ is nonzero. But then $i_*(partialsigma) = 0$ by exactness, so $i_* : H_0(A) to H_0(X)$ is not injective.
So all that's left to prove is that $i_* H_0(A) to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b in A$ in the two distinct path components. Then $[a]-[b] in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) to H_0(X)$ is not injective. Let $0 neq alpha in H_0(A)$ be such that $i_*alpha = 0$. Write $alpha = sum_{k in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(alpha) = sum_{k in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $alpha neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
answered Nov 20 '18 at 8:18
Najib Idrissi
40.9k470138
40.9k470138
add a comment |
add a comment |
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Why do you assume $sigma'in H_1(A)$ in contrast to $sigmain C_1(X)$? Homology classes are always represented by cycles, so this would mean that $partial(sigma')=0$. If it's just a typo, maybe you could correct it along with the map in the quote which is $H_1(A)to H_1(X)$.
– Stefan Hamcke
Nov 10 '12 at 17:41
@StefanH. I have corrected the first point in the comment. I don't understand which map you're talking about in the second.
– user38268
Nov 10 '12 at 22:40
Why must $sigma$ be a loop? Are you assuming that $tau$ is a single vertex rather than an arbitrary 0-chain in $A$?
– John Palmieri
Nov 10 '12 at 23:49
@BenjaLim. I was talking about the map $H_1(A)to H_1(X)$ in the quotation of the problem, where you switched $A$ and $X$.
– Stefan Hamcke
Nov 11 '12 at 13:06
@StefanH. I have corrected it now. Thanks.
– user38268
Nov 11 '12 at 23:40