Is every homology map induced by a chain map
If I have a chain map $f:Cto D$, I know that there is an induced map $f_ast: H(C)to H(D)$ on homology.
Is the other way around also true: If I start with a map $g:H(C)to H(D)$ with $C$ and $D$ chain complexes, can I always find a chain map $Cto D$ that induces $g$? I am interested in that question in the special case of homology/chain complexes over a (finite) field, so the question should boil down to linear algebra, but I fail to see an argument in either direction.
If the answer to my question is negative, does it change if $g$ is an isomorphism?
homological-algebra
asked Dec 2 '18 at 13:56
mike
61
add a comment |
If I have a chain map $f:Cto D$, I know that there is an induced map $f_ast: H(C)to H(D)$ on homology.
Is the other way around also true: If I start with a map $g:H(C)to H(D)$ with $C$ and $D$ chain complexes, can I always find a chain map $Cto D$ that induces $g$? I am interested in that question in the special case of homology/chain complexes over a (finite) field, so the question should boil down to linear algebra, but I fail to see an argument in either direction.
If the answer to my question is negative, does it change if $g$ is an isomorphism?
homological-algebra
asked Dec 2 '18 at 13:56
mike
61
add a comment |
If I have a chain map $f:Cto D$, I know that there is an induced map $f_ast: H(C)to H(D)$ on homology.
Is the other way around also true: If I start with a map $g:H(C)to H(D)$ with $C$ and $D$ chain complexes, can I always find a chain map $Cto D$ that induces $g$? I am interested in that question in the special case of homology/chain complexes over a (finite) field, so the question should boil down to linear algebra, but I fail to see an argument in either direction.
If the answer to my question is negative, does it change if $g$ is an isomorphism?
homological-algebra
asked Dec 2 '18 at 13:56
mike
61
If I have a chain map $f:Cto D$, I know that there is an induced map $f_ast: H(C)to H(D)$ on homology.
Is the other way around also true: If I start with a map $g:H(C)to H(D)$ with $C$ and $D$ chain complexes, can I always find a chain map $Cto D$ that induces $g$? I am interested in that question in the special case of homology/chain complexes over a (finite) field, so the question should boil down to linear algebra, but I fail to see an argument in either direction.
If the answer to my question is negative, does it change if $g$ is an isomorphism?
homological-algebra
homological-algebra
asked Dec 2 '18 at 13:56
mike
61
asked Dec 2 '18 at 13:56
mike
61
asked Dec 2 '18 at 13:56
mike
61
asked Dec 2 '18 at 13:56
mike
61
asked Dec 2 '18 at 13:56
mike
61
61
add a comment |
add a comment |
1 Answer
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First, without your requirement to work over a field, this is false. The chain complex $0 to mathbb{F}_2 to 0$ has the same homology as the chain complex $0 to mathbb{Z} xrightarrow{times 2} mathbb{Z} to 0$, but there is no map from the first to the second inducing this.
Once you work over a field, the answer is yes. The following is inspired by Tyler Lawson's answer to https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence.
If you have a chain complex $... to V_n to V_{n-1} to ...$ of vector spaces over a field, then for each $n$ you have a surjective vector space map $V_n to B_{n-1}$, where $B_{n-1}$ is the boundaries, the image of the boundary map. Since these are vector spaces, we can choose a vector space complement (equivalently, a splitting of this map), which is necessarily isomorphic to $Z_n$, the $n$-cycles. That is, $V_n cong Z_n oplus B_{n-1}$. Similarly, since $H_n = Z_n/B_n$, then $Z_n$ is isomorphic to $H_n oplus B_n$. So putting these together, we have $V_n cong (B_n oplus H_n) oplus B_{n-1}$. The boundary map
$$
B_n oplus H_n oplus B_{n-1} to B_{n-1} oplus H_{n-1} oplus B_{n-2}
$$
sends the summand $B_{n-1}$ by the identity to the summand $B_{n-1}$, and is zero on the other summands.
Now the question is, if you have two chain complexes, say $... to V_n to ...$ and $... to V'_n to ...$, and if you have maps $H_n to H'_n$ for each $n$, is there a chain map that induces it? Yes, just decompose $V_n$ and $V'_n$ as above, send the $H_n$ summand to $H'_n$ using the desired map, and send the other summands $B_n$ and $B_{n-1}$ to zero.
answered Dec 3 '18 at 0:34
John Palmieri
69339
add a comment |
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First, without your requirement to work over a field, this is false. The chain complex $0 to mathbb{F}_2 to 0$ has the same homology as the chain complex $0 to mathbb{Z} xrightarrow{times 2} mathbb{Z} to 0$, but there is no map from the first to the second inducing this.
Once you work over a field, the answer is yes. The following is inspired by Tyler Lawson's answer to https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence.
If you have a chain complex $... to V_n to V_{n-1} to ...$ of vector spaces over a field, then for each $n$ you have a surjective vector space map $V_n to B_{n-1}$, where $B_{n-1}$ is the boundaries, the image of the boundary map. Since these are vector spaces, we can choose a vector space complement (equivalently, a splitting of this map), which is necessarily isomorphic to $Z_n$, the $n$-cycles. That is, $V_n cong Z_n oplus B_{n-1}$. Similarly, since $H_n = Z_n/B_n$, then $Z_n$ is isomorphic to $H_n oplus B_n$. So putting these together, we have $V_n cong (B_n oplus H_n) oplus B_{n-1}$. The boundary map
$$
B_n oplus H_n oplus B_{n-1} to B_{n-1} oplus H_{n-1} oplus B_{n-2}
$$
sends the summand $B_{n-1}$ by the identity to the summand $B_{n-1}$, and is zero on the other summands.
Now the question is, if you have two chain complexes, say $... to V_n to ...$ and $... to V'_n to ...$, and if you have maps $H_n to H'_n$ for each $n$, is there a chain map that induces it? Yes, just decompose $V_n$ and $V'_n$ as above, send the $H_n$ summand to $H'_n$ using the desired map, and send the other summands $B_n$ and $B_{n-1}$ to zero.
answered Dec 3 '18 at 0:34
John Palmieri
69339
add a comment |
First, without your requirement to work over a field, this is false. The chain complex $0 to mathbb{F}_2 to 0$ has the same homology as the chain complex $0 to mathbb{Z} xrightarrow{times 2} mathbb{Z} to 0$, but there is no map from the first to the second inducing this.
Once you work over a field, the answer is yes. The following is inspired by Tyler Lawson's answer to https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence.
If you have a chain complex $... to V_n to V_{n-1} to ...$ of vector spaces over a field, then for each $n$ you have a surjective vector space map $V_n to B_{n-1}$, where $B_{n-1}$ is the boundaries, the image of the boundary map. Since these are vector spaces, we can choose a vector space complement (equivalently, a splitting of this map), which is necessarily isomorphic to $Z_n$, the $n$-cycles. That is, $V_n cong Z_n oplus B_{n-1}$. Similarly, since $H_n = Z_n/B_n$, then $Z_n$ is isomorphic to $H_n oplus B_n$. So putting these together, we have $V_n cong (B_n oplus H_n) oplus B_{n-1}$. The boundary map
$$
B_n oplus H_n oplus B_{n-1} to B_{n-1} oplus H_{n-1} oplus B_{n-2}
$$
sends the summand $B_{n-1}$ by the identity to the summand $B_{n-1}$, and is zero on the other summands.
Now the question is, if you have two chain complexes, say $... to V_n to ...$ and $... to V'_n to ...$, and if you have maps $H_n to H'_n$ for each $n$, is there a chain map that induces it? Yes, just decompose $V_n$ and $V'_n$ as above, send the $H_n$ summand to $H'_n$ using the desired map, and send the other summands $B_n$ and $B_{n-1}$ to zero.
answered Dec 3 '18 at 0:34
John Palmieri
69339
add a comment |
First, without your requirement to work over a field, this is false. The chain complex $0 to mathbb{F}_2 to 0$ has the same homology as the chain complex $0 to mathbb{Z} xrightarrow{times 2} mathbb{Z} to 0$, but there is no map from the first to the second inducing this.
Once you work over a field, the answer is yes. The following is inspired by Tyler Lawson's answer to https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence.
If you have a chain complex $... to V_n to V_{n-1} to ...$ of vector spaces over a field, then for each $n$ you have a surjective vector space map $V_n to B_{n-1}$, where $B_{n-1}$ is the boundaries, the image of the boundary map. Since these are vector spaces, we can choose a vector space complement (equivalently, a splitting of this map), which is necessarily isomorphic to $Z_n$, the $n$-cycles. That is, $V_n cong Z_n oplus B_{n-1}$. Similarly, since $H_n = Z_n/B_n$, then $Z_n$ is isomorphic to $H_n oplus B_n$. So putting these together, we have $V_n cong (B_n oplus H_n) oplus B_{n-1}$. The boundary map
$$
B_n oplus H_n oplus B_{n-1} to B_{n-1} oplus H_{n-1} oplus B_{n-2}
$$
sends the summand $B_{n-1}$ by the identity to the summand $B_{n-1}$, and is zero on the other summands.
Now the question is, if you have two chain complexes, say $... to V_n to ...$ and $... to V'_n to ...$, and if you have maps $H_n to H'_n$ for each $n$, is there a chain map that induces it? Yes, just decompose $V_n$ and $V'_n$ as above, send the $H_n$ summand to $H'_n$ using the desired map, and send the other summands $B_n$ and $B_{n-1}$ to zero.
answered Dec 3 '18 at 0:34
John Palmieri
69339
First, without your requirement to work over a field, this is false. The chain complex $0 to mathbb{F}_2 to 0$ has the same homology as the chain complex $0 to mathbb{Z} xrightarrow{times 2} mathbb{Z} to 0$, but there is no map from the first to the second inducing this.
Once you work over a field, the answer is yes. The following is inspired by Tyler Lawson's answer to https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence.
If you have a chain complex $... to V_n to V_{n-1} to ...$ of vector spaces over a field, then for each $n$ you have a surjective vector space map $V_n to B_{n-1}$, where $B_{n-1}$ is the boundaries, the image of the boundary map. Since these are vector spaces, we can choose a vector space complement (equivalently, a splitting of this map), which is necessarily isomorphic to $Z_n$, the $n$-cycles. That is, $V_n cong Z_n oplus B_{n-1}$. Similarly, since $H_n = Z_n/B_n$, then $Z_n$ is isomorphic to $H_n oplus B_n$. So putting these together, we have $V_n cong (B_n oplus H_n) oplus B_{n-1}$. The boundary map
$$
B_n oplus H_n oplus B_{n-1} to B_{n-1} oplus H_{n-1} oplus B_{n-2}
$$
sends the summand $B_{n-1}$ by the identity to the summand $B_{n-1}$, and is zero on the other summands.
Now the question is, if you have two chain complexes, say $... to V_n to ...$ and $... to V'_n to ...$, and if you have maps $H_n to H'_n$ for each $n$, is there a chain map that induces it? Yes, just decompose $V_n$ and $V'_n$ as above, send the $H_n$ summand to $H'_n$ using the desired map, and send the other summands $B_n$ and $B_{n-1}$ to zero.
answered Dec 3 '18 at 0:34
John Palmieri
69339
answered Dec 3 '18 at 0:34
John Palmieri
69339
answered Dec 3 '18 at 0:34
John Palmieri
69339
answered Dec 3 '18 at 0:34
John Palmieri
69339
69339
add a comment |
add a comment |
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