Mixing
Chain complex homotopy equivalent to its homology
$begingroup$
Let $(C_*,d)$ be a chain complex of vector spaces over a field $F$. Can always be constructed a homotopy equivalence $C_* rightarrow H(C_*)$ ? (Here $H(C_*)$ is seen as a chain complex with $d=0$)
There is a split exact sequence $0 rightarrow B_* rightarrow Z_* rightarrow H_*(C) rightarrow 0$ so we have a map $H_*(C) rightarrow Z_* subseteq C_*$ which induces a quasi-isomorphism. Is this map the homotopy equivalence between $C_*$ and its homology?
homology-cohomology homological-algebra
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
$endgroup$
add a comment |
$begingroup$
Let $(C_*,d)$ be a chain complex of vector spaces over a field $F$. Can always be constructed a homotopy equivalence $C_* rightarrow H(C_*)$ ? (Here $H(C_*)$ is seen as a chain complex with $d=0$)
There is a split exact sequence $0 rightarrow B_* rightarrow Z_* rightarrow H_*(C) rightarrow 0$ so we have a map $H_*(C) rightarrow Z_* subseteq C_*$ which induces a quasi-isomorphism. Is this map the homotopy equivalence between $C_*$ and its homology?
homology-cohomology homological-algebra
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
$endgroup$
$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53
add a comment |
$begingroup$
Let $(C_*,d)$ be a chain complex of vector spaces over a field $F$. Can always be constructed a homotopy equivalence $C_* rightarrow H(C_*)$ ? (Here $H(C_*)$ is seen as a chain complex with $d=0$)
There is a split exact sequence $0 rightarrow B_* rightarrow Z_* rightarrow H_*(C) rightarrow 0$ so we have a map $H_*(C) rightarrow Z_* subseteq C_*$ which induces a quasi-isomorphism. Is this map the homotopy equivalence between $C_*$ and its homology?
homology-cohomology homological-algebra
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
$endgroup$
Let $(C_*,d)$ be a chain complex of vector spaces over a field $F$. Can always be constructed a homotopy equivalence $C_* rightarrow H(C_*)$ ? (Here $H(C_*)$ is seen as a chain complex with $d=0$)
There is a split exact sequence $0 rightarrow B_* rightarrow Z_* rightarrow H_*(C) rightarrow 0$ so we have a map $H_*(C) rightarrow Z_* subseteq C_*$ which induces a quasi-isomorphism. Is this map the homotopy equivalence between $C_*$ and its homology?
homology-cohomology homological-algebra
homology-cohomology homological-algebra
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
edited Jan 17 at 17:53
C. Zhihao
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
asked Jan 17 at 17:37
C. ZhihaoC. Zhihao
489214
489214
$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53
add a comment |
$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53
$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53
add a comment |
1 Answer
1
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$begingroup$
In the case of a field, this is true. We can write $Z = Boplus i(H)$ for $i : Hto Z$ a section of the projection and $C = Loplus Boplus i(H)$. Define $j : Hto C$ using $i$, and define $q : Cto H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : Cto C$ by picking a section $s : Bto C$ of $d : Cto B$ and sending
$bin B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
- If $lin L$, then $(hd+dh)l = l = l-iql = (1-iq)(l)$,
- If $bin B$, then $(hd+dh)b = dsb = b = (1-iq)(b)$,
- If $zin i(H)$, then $(hd+dh)z = 0 = (1-iq)(z)$,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
$endgroup$
add a comment |
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$begingroup$
In the case of a field, this is true. We can write $Z = Boplus i(H)$ for $i : Hto Z$ a section of the projection and $C = Loplus Boplus i(H)$. Define $j : Hto C$ using $i$, and define $q : Cto H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : Cto C$ by picking a section $s : Bto C$ of $d : Cto B$ and sending
$bin B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
- If $lin L$, then $(hd+dh)l = l = l-iql = (1-iq)(l)$,
- If $bin B$, then $(hd+dh)b = dsb = b = (1-iq)(b)$,
- If $zin i(H)$, then $(hd+dh)z = 0 = (1-iq)(z)$,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
$endgroup$
add a comment |
$begingroup$
In the case of a field, this is true. We can write $Z = Boplus i(H)$ for $i : Hto Z$ a section of the projection and $C = Loplus Boplus i(H)$. Define $j : Hto C$ using $i$, and define $q : Cto H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : Cto C$ by picking a section $s : Bto C$ of $d : Cto B$ and sending
$bin B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
- If $lin L$, then $(hd+dh)l = l = l-iql = (1-iq)(l)$,
- If $bin B$, then $(hd+dh)b = dsb = b = (1-iq)(b)$,
- If $zin i(H)$, then $(hd+dh)z = 0 = (1-iq)(z)$,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
$endgroup$
add a comment |
$begingroup$
In the case of a field, this is true. We can write $Z = Boplus i(H)$ for $i : Hto Z$ a section of the projection and $C = Loplus Boplus i(H)$. Define $j : Hto C$ using $i$, and define $q : Cto H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : Cto C$ by picking a section $s : Bto C$ of $d : Cto B$ and sending
$bin B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
- If $lin L$, then $(hd+dh)l = l = l-iql = (1-iq)(l)$,
- If $bin B$, then $(hd+dh)b = dsb = b = (1-iq)(b)$,
- If $zin i(H)$, then $(hd+dh)z = 0 = (1-iq)(z)$,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
$endgroup$
In the case of a field, this is true. We can write $Z = Boplus i(H)$ for $i : Hto Z$ a section of the projection and $C = Loplus Boplus i(H)$. Define $j : Hto C$ using $i$, and define $q : Cto H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : Cto C$ by picking a section $s : Bto C$ of $d : Cto B$ and sending
$bin B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
- If $lin L$, then $(hd+dh)l = l = l-iql = (1-iq)(l)$,
- If $bin B$, then $(hd+dh)b = dsb = b = (1-iq)(b)$,
- If $zin i(H)$, then $(hd+dh)z = 0 = (1-iq)(z)$,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
answered Jan 17 at 17:53
Pedro Tamaroff♦Pedro Tamaroff
97k10153297
97k10153297
add a comment |
add a comment |
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$begingroup$
Presumably, where $H_*(C)$ has a zero $d$ as a chain complex?
$endgroup$
– Thomas Andrews
Jan 17 at 17:51
$begingroup$
Yes $H(C_*)$ carries the zero differential. I will point it out explicitly.
$endgroup$
– C. Zhihao
Jan 17 at 17:53