Relative homology: the splitting $C_*(X,A) rightarrow C_*(X)$
$begingroup$
Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:
The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.
By splitting, does it mean $C_*(X) cong C_*(A) oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?
algebraic-topology
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
$endgroup$
add a comment |
$begingroup$
Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:
The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.
By splitting, does it mean $C_*(X) cong C_*(A) oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?
algebraic-topology
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
$endgroup$
1
$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
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^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39
add a comment |
$begingroup$
Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:
The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.
By splitting, does it mean $C_*(X) cong C_*(A) oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?
algebraic-topology
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
$endgroup$
Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:
The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.
By splitting, does it mean $C_*(X) cong C_*(A) oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?
algebraic-topology
algebraic-topology
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
asked Dec 10 '18 at 1:04
YuliyaYuliya
132
132
1
$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
$begingroup$
^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39
add a comment |
1
$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
$begingroup$
^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39
1
1
$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
$begingroup$
^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39
add a comment |
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$begingroup$
I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) cong C_n(A) oplus C_n(X, A)$ in $textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes.
$endgroup$
– Robert Cardona
Dec 10 '18 at 1:14
$begingroup$
^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology.
$endgroup$
– Randall
Dec 10 '18 at 1:43
$begingroup$
If it split in the category of chain complexes then you would have $H_*(X) = H_*(A) oplus H_*(X,A)$. You can check in examples that this often is not true.
$endgroup$
– Mike Miller
Dec 10 '18 at 3:36
$begingroup$
The splitting, for Bredon, is the collection of maps $j: C_n(X,A) to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) to H_n(X)$, and in general $H_n(X) not cong H_n(A) oplus H_n(X,A)$.
$endgroup$
– John Palmieri
Dec 10 '18 at 4:39