Commutativity up to homotopy implies strict commutativity, for lifting problems
Clash Royale CLAN TAG#URR8PPP
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Suppose we have a commutative diagram
$requireAMScd$
beginCD
A @>>> X \
@VVV & @VVV \
W @>>> Y\
endCD
where the map $Arightarrow W$ is a cofibration and the map $Xrightarrow Y$ is a fibration. Suppose also that there exists a map $Wrightarrow X$ that makes the diagram commute up to homotopy.
Is then true that we can find a map $Wrightarrow X$ that makes the diagram strictly commute?
at.algebraic-topology homotopy-theory
edited 12 hours ago
David White
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
5
down vote
favorite
Suppose we have a commutative diagram
$requireAMScd$
beginCD
A @>>> X \
@VVV & @VVV \
W @>>> Y\
endCD
where the map $Arightarrow W$ is a cofibration and the map $Xrightarrow Y$ is a fibration. Suppose also that there exists a map $Wrightarrow X$ that makes the diagram commute up to homotopy.
Is then true that we can find a map $Wrightarrow X$ that makes the diagram strictly commute?
at.algebraic-topology homotopy-theory
edited 12 hours ago
David White
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
1
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose we have a commutative diagram
$requireAMScd$
beginCD
A @>>> X \
@VVV & @VVV \
W @>>> Y\
endCD
where the map $Arightarrow W$ is a cofibration and the map $Xrightarrow Y$ is a fibration. Suppose also that there exists a map $Wrightarrow X$ that makes the diagram commute up to homotopy.
Is then true that we can find a map $Wrightarrow X$ that makes the diagram strictly commute?
at.algebraic-topology homotopy-theory
edited 12 hours ago
David White
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose we have a commutative diagram
$requireAMScd$
beginCD
A @>>> X \
@VVV & @VVV \
W @>>> Y\
endCD
where the map $Arightarrow W$ is a cofibration and the map $Xrightarrow Y$ is a fibration. Suppose also that there exists a map $Wrightarrow X$ that makes the diagram commute up to homotopy.
Is then true that we can find a map $Wrightarrow X$ that makes the diagram strictly commute?
at.algebraic-topology homotopy-theory
at.algebraic-topology homotopy-theory
edited 12 hours ago
David White
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 hours ago
David White
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 hours ago
David White
10.9k45998
edited 12 hours ago
David White
10.9k45998
edited 12 hours ago
David White
10.9k45998
10.9k45998
asked 13 hours ago
Diego95
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
Diego95
461
asked 13 hours ago
Diego95
461
461
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Diego95 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
1
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago
add a comment |Â
2
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
1
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago
2
2
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
1
1
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = A to W$. Then, because $f:Xto Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.
answered 12 hours ago
David White
10.9k45998
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = A to W$. Then, because $f:Xto Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.
answered 12 hours ago
David White
10.9k45998
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
add a comment |Â
up vote
6
down vote
I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = A to W$. Then, because $f:Xto Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.
answered 12 hours ago
David White
10.9k45998
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = A to W$. Then, because $f:Xto Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.
answered 12 hours ago
David White
10.9k45998
I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = A to W$. Then, because $f:Xto Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.
answered 12 hours ago
David White
10.9k45998
answered 12 hours ago
David White
10.9k45998
answered 12 hours ago
David White
10.9k45998
answered 12 hours ago
David White
10.9k45998
10.9k45998
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
add a comment |Â
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
Let call $g:Wrightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this?
– Diego95
8 hours ago
add a comment |Â
Diego95 is a new contributor. Be nice, and check out our Code of Conduct.
Diego95 is a new contributor. Be nice, and check out our Code of Conduct.
Diego95 is a new contributor. Be nice, and check out our Code of Conduct.
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2
What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibrationâ€Â/“cofibration†in $Top$, or some well-behaved subcategory thereof?
– Peter LeFanu Lumsdaine
12 hours ago
1
I'm working in Top category, more precisely $(W, A) $ is a CW-pair, but I'm not sure whether this hypothesis is necessary.
– Diego95
12 hours ago
I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$.
– dhy
9 hours ago