derived functor that preserves weak equivalences
$begingroup$
Suppose we have a functor $F:Arightarrow B$ between model categories.
1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
It will be great to see the construction of such derived functor in detail. Thank you.
reference-request model-categories
asked Nov 25 '18 at 18:39
ABCABC
755
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$begingroup$
Suppose we have a functor $F:Arightarrow B$ between model categories.
1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
It will be great to see the construction of such derived functor in detail. Thank you.
reference-request model-categories
asked Nov 25 '18 at 18:39
ABCABC
755
$endgroup$
add a comment |
$begingroup$
Suppose we have a functor $F:Arightarrow B$ between model categories.
1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
It will be great to see the construction of such derived functor in detail. Thank you.
reference-request model-categories
asked Nov 25 '18 at 18:39
ABCABC
755
$endgroup$
Suppose we have a functor $F:Arightarrow B$ between model categories.
1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
It will be great to see the construction of such derived functor in detail. Thank you.
reference-request model-categories
reference-request model-categories
asked Nov 25 '18 at 18:39
ABCABC
755
asked Nov 25 '18 at 18:39
ABCABC
755
asked Nov 25 '18 at 18:39
ABCABC
755
asked Nov 25 '18 at 18:39
ABCABC
755
asked Nov 25 '18 at 18:39
ABCABC
755
755
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer is yes.
Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.
Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.
answered Nov 25 '18 at 19:01
the Lthe L
7551920
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1 Answer
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$begingroup$
The answer is yes.
Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.
Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.
answered Nov 25 '18 at 19:01
the Lthe L
7551920
$endgroup$
add a comment |
$begingroup$
The answer is yes.
Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.
Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.
answered Nov 25 '18 at 19:01
the Lthe L
7551920
$endgroup$
add a comment |
$begingroup$
The answer is yes.
Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.
Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.
answered Nov 25 '18 at 19:01
the Lthe L
7551920
$endgroup$
The answer is yes.
Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.
Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.
answered Nov 25 '18 at 19:01
the Lthe L
7551920
answered Nov 25 '18 at 19:01
the Lthe L
7551920
answered Nov 25 '18 at 19:01
the Lthe L
7551920
answered Nov 25 '18 at 19:01
the Lthe L
7551920
7551920
add a comment |
add a comment |
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