A pair of spaces equivalent to a pair of CW-complexes
$begingroup$
Suppose that $X$ is a CW-complex and $Y$ a CW-subcomplex of $X$. Let $A$ be a closed subspace of $Z$ such that
$Z-A$ is homeomorhic to $X-Y$ and
$Z/A$ homeomorphic to $X/Y$ and- The closure of $Z-A$ in $Z$ is $Z$ it self.
Does it follow that the embedding $Arightarrow Z$
is a cofibration ?
reference-request at.algebraic-topology gn.general-topology cw-complexes
asked May 28 at 14:24
cellularcellular
835 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose that $X$ is a CW-complex and $Y$ a CW-subcomplex of $X$. Let $A$ be a closed subspace of $Z$ such that
$Z-A$ is homeomorhic to $X-Y$ and
$Z/A$ homeomorphic to $X/Y$ and- The closure of $Z-A$ in $Z$ is $Z$ it self.
Does it follow that the embedding $Arightarrow Z$
is a cofibration ?
reference-request at.algebraic-topology gn.general-topology cw-complexes
asked May 28 at 14:24
cellularcellular
835 bronze badges
$endgroup$
$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56
add a comment
|
$begingroup$
Suppose that $X$ is a CW-complex and $Y$ a CW-subcomplex of $X$. Let $A$ be a closed subspace of $Z$ such that
$Z-A$ is homeomorhic to $X-Y$ and
$Z/A$ homeomorphic to $X/Y$ and- The closure of $Z-A$ in $Z$ is $Z$ it self.
Does it follow that the embedding $Arightarrow Z$
is a cofibration ?
reference-request at.algebraic-topology gn.general-topology cw-complexes
asked May 28 at 14:24
cellularcellular
835 bronze badges
$endgroup$
Suppose that $X$ is a CW-complex and $Y$ a CW-subcomplex of $X$. Let $A$ be a closed subspace of $Z$ such that
$Z-A$ is homeomorhic to $X-Y$ and
$Z/A$ homeomorphic to $X/Y$ and- The closure of $Z-A$ in $Z$ is $Z$ it self.
Does it follow that the embedding $Arightarrow Z$
is a cofibration ?
reference-request at.algebraic-topology gn.general-topology cw-complexes
reference-request at.algebraic-topology gn.general-topology cw-complexes
asked May 28 at 14:24
cellularcellular
835 bronze badges
asked May 28 at 14:24
cellularcellular
835 bronze badges
edited May 28 at 14:47
cellular
asked May 28 at 14:24
cellularcellular
835 bronze badges
asked May 28 at 14:24
cellularcellular
835 bronze badges
asked May 28 at 14:24
cellularcellular
835 bronze badges
835 bronze badges
$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56
add a comment
|
$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56
$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56
add a comment
|
1 Answer
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If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.
Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $Ucong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.
$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{itheta}mapsto re^{i(theta+frac{1}{1-r})}$$Take the real open interval with endpoints $pm 1$ and hit it with this map.
Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $Ato Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $Vsubset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Zto W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $pin U$ yields a point $i(p)in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
$endgroup$
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
add a comment
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$begingroup$
If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.
Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $Ucong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.
$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{itheta}mapsto re^{i(theta+frac{1}{1-r})}$$Take the real open interval with endpoints $pm 1$ and hit it with this map.
Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $Ato Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $Vsubset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Zto W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $pin U$ yields a point $i(p)in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
$endgroup$
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
add a comment
|
$begingroup$
If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.
Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $Ucong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.
$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{itheta}mapsto re^{i(theta+frac{1}{1-r})}$$Take the real open interval with endpoints $pm 1$ and hit it with this map.
Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $Ato Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $Vsubset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Zto W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $pin U$ yields a point $i(p)in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
$endgroup$
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
add a comment
|
$begingroup$
If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.
Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $Ucong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.
$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{itheta}mapsto re^{i(theta+frac{1}{1-r})}$$Take the real open interval with endpoints $pm 1$ and hit it with this map.
Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $Ato Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $Vsubset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Zto W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $pin U$ yields a point $i(p)in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
$endgroup$
If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.
Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $Ucong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.
$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{itheta}mapsto re^{i(theta+frac{1}{1-r})}$$Take the real open interval with endpoints $pm 1$ and hit it with this map.
Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $Ato Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $Vsubset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Zto W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $pin U$ yields a point $i(p)in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
answered May 28 at 17:59
Tom GoodwillieTom Goodwillie
41.4k3 gold badges114 silver badges203 bronze badges
41.4k3 gold badges114 silver badges203 bronze badges
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
add a comment
|
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
$begingroup$
More simply, $Ato Z$ is not a cofibration because here is a trivial fibration for which a section defined on $A$ does not extend to $Z$: the "identity" map from the disjoint union of $A$ and $U$ to $Z$.
$endgroup$
– Tom Goodwillie
May 28 at 21:26
add a comment
|
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$begingroup$
You should clarify what you mean by "cofibration". The term has different meanings in different contexts. I am guessing you mean Quillen's model theory on Top, so a map that has the same left lifting properties as the inclusions $S^{n-1}to D^n$,
$endgroup$
– Tom Goodwillie
May 28 at 17:37
$begingroup$
@TomGoodwillie Yes, exactly.
$endgroup$
– cellular
May 28 at 17:56