$K_star$ and $L_star$ are homotopic as chain complex. Is it true that...
$begingroup$
Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.
$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.
abstract-algebra homological-algebra
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
$endgroup$
add a comment |
$begingroup$
Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.
$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.
abstract-algebra homological-algebra
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
$endgroup$
$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
3
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
1
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35
add a comment |
$begingroup$
Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.
$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.
abstract-algebra homological-algebra
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
$endgroup$
Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.
$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.
abstract-algebra homological-algebra
abstract-algebra homological-algebra
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
edited Dec 19 '18 at 19:33
Eric Wofsey
192k14217351
192k14217351
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
asked Dec 19 '18 at 19:17
user45765user45765
2,6882724
2,6882724
$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
3
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
1
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35
add a comment |
$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
3
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
1
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35
$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
3
3
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
1
1
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35
add a comment |
1 Answer
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oldest
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$begingroup$
It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:
$H_*(underline{Hom}(L,K)) = 0$,
$H_0(underline{Hom}(L,K)) = 0$,
$L$ is contractible, i.e. $text{id}_L simeq 0$.
In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.
answered Dec 21 '18 at 11:44
BenBen
4,318617
$endgroup$
add a comment |
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1 Answer
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$begingroup$
It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:
$H_*(underline{Hom}(L,K)) = 0$,
$H_0(underline{Hom}(L,K)) = 0$,
$L$ is contractible, i.e. $text{id}_L simeq 0$.
In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.
answered Dec 21 '18 at 11:44
BenBen
4,318617
$endgroup$
add a comment |
$begingroup$
It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:
$H_*(underline{Hom}(L,K)) = 0$,
$H_0(underline{Hom}(L,K)) = 0$,
$L$ is contractible, i.e. $text{id}_L simeq 0$.
In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.
answered Dec 21 '18 at 11:44
BenBen
4,318617
$endgroup$
add a comment |
$begingroup$
It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:
$H_*(underline{Hom}(L,K)) = 0$,
$H_0(underline{Hom}(L,K)) = 0$,
$L$ is contractible, i.e. $text{id}_L simeq 0$.
In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.
answered Dec 21 '18 at 11:44
BenBen
4,318617
$endgroup$
It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:
$H_*(underline{Hom}(L,K)) = 0$,
$H_0(underline{Hom}(L,K)) = 0$,
$L$ is contractible, i.e. $text{id}_L simeq 0$.
In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.
answered Dec 21 '18 at 11:44
BenBen
4,318617
answered Dec 21 '18 at 11:44
BenBen
4,318617
answered Dec 21 '18 at 11:44
BenBen
4,318617
answered Dec 21 '18 at 11:44
BenBen
4,318617
4,318617
add a comment |
add a comment |
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$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33
3
$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33
1
$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35
$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35