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Why are some maps cofibrations
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I am currently reading a textbook in which it is claimed that both the following maps are (Hurewitz) cofibrations $Avee Arightarrow Awedge (I_+)$ (the first $A$ is mapped to $Atimes0$ and the second to $Atimes1$) and $Arightarrow CA$, where $CA$ is the reduced cone on $A$, and $A$ is a non-degenerately based space.
I have thought about showing that these subspaces are NDR using the DR of $Awedge (I_+)$ to $A$, but my attempts have not been successful so far.
algebraic-topology
asked Aug 2 at 6:48
user09127
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I am currently reading a textbook in which it is claimed that both the following maps are (Hurewitz) cofibrations $Avee Arightarrow Awedge (I_+)$ (the first $A$ is mapped to $Atimes0$ and the second to $Atimes1$) and $Arightarrow CA$, where $CA$ is the reduced cone on $A$, and $A$ is a non-degenerately based space.
I have thought about showing that these subspaces are NDR using the DR of $Awedge (I_+)$ to $A$, but my attempts have not been successful so far.
algebraic-topology
asked Aug 2 at 6:48
user09127
31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am currently reading a textbook in which it is claimed that both the following maps are (Hurewitz) cofibrations $Avee Arightarrow Awedge (I_+)$ (the first $A$ is mapped to $Atimes0$ and the second to $Atimes1$) and $Arightarrow CA$, where $CA$ is the reduced cone on $A$, and $A$ is a non-degenerately based space.
I have thought about showing that these subspaces are NDR using the DR of $Awedge (I_+)$ to $A$, but my attempts have not been successful so far.
algebraic-topology
asked Aug 2 at 6:48
user09127
31
I am currently reading a textbook in which it is claimed that both the following maps are (Hurewitz) cofibrations $Avee Arightarrow Awedge (I_+)$ (the first $A$ is mapped to $Atimes0$ and the second to $Atimes1$) and $Arightarrow CA$, where $CA$ is the reduced cone on $A$, and $A$ is a non-degenerately based space.
I have thought about showing that these subspaces are NDR using the DR of $Awedge (I_+)$ to $A$, but my attempts have not been successful so far.
algebraic-topology
asked Aug 2 at 6:48
user09127
31
asked Aug 2 at 6:48
user09127
31
asked Aug 2 at 6:48
user09127
31
asked Aug 2 at 6:48
user09127
31
31
add a comment |Â
add a comment |Â
1 Answer
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You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A subset X$. The definition of NDR pairs can be generalized to arbitrary $A subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A to X$, $f : A to B$, $i' : B to Y$ and $f' : X to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X times Y, X times B cup A times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, a_0 )$ is an NDR pair.
$A vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is
$$A vee A = (A times 0, 1 cup a_0 times I) / a_0 times I .$$
Let $p : A times 0, 1 cup a_0 times I to A vee A$ denote the quotient map.
$A wedge I_+$ is nothing else than $A times I / a_0 times I$. Let $p' : A times I to A times I / a_0 times I$ denote the quotient map. Its restriction $p''$ to $A times 0, 1 cup a_0 times I$ induces a unique map $i' : A vee A to A times I / a_0 times I$ such that $i' circ p = p'' = p' circ i$ where $i : A times 0, 1 cup a_0 times I to A times I$ denotes the inclusion map. Obviously $A_k = p(A times k ) subset A vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A times k ) subset A times I / a_0 times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A times I, A times 0, 1 cup a_0 times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A times I / a_0 times I) / A'_1 .$$
Let $r' : A times I / a_0 times I to CA$ denote the quotient map. The map $i'' : A to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A times 0, 1 cup a_0 times I to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A vee A to A$ such that $R = r circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.
answered Aug 2 at 14:53
Paul Frost
3,483420
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
add a comment |Â
1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A subset X$. The definition of NDR pairs can be generalized to arbitrary $A subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A to X$, $f : A to B$, $i' : B to Y$ and $f' : X to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X times Y, X times B cup A times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, a_0 )$ is an NDR pair.
$A vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is
$$A vee A = (A times 0, 1 cup a_0 times I) / a_0 times I .$$
Let $p : A times 0, 1 cup a_0 times I to A vee A$ denote the quotient map.
$A wedge I_+$ is nothing else than $A times I / a_0 times I$. Let $p' : A times I to A times I / a_0 times I$ denote the quotient map. Its restriction $p''$ to $A times 0, 1 cup a_0 times I$ induces a unique map $i' : A vee A to A times I / a_0 times I$ such that $i' circ p = p'' = p' circ i$ where $i : A times 0, 1 cup a_0 times I to A times I$ denotes the inclusion map. Obviously $A_k = p(A times k ) subset A vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A times k ) subset A times I / a_0 times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A times I, A times 0, 1 cup a_0 times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A times I / a_0 times I) / A'_1 .$$
Let $r' : A times I / a_0 times I to CA$ denote the quotient map. The map $i'' : A to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A times 0, 1 cup a_0 times I to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A vee A to A$ such that $R = r circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.
answered Aug 2 at 14:53
Paul Frost
3,483420
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
add a comment |Â
up vote
0
down vote
accepted
You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A subset X$. The definition of NDR pairs can be generalized to arbitrary $A subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A to X$, $f : A to B$, $i' : B to Y$ and $f' : X to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X times Y, X times B cup A times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, a_0 )$ is an NDR pair.
$A vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is
$$A vee A = (A times 0, 1 cup a_0 times I) / a_0 times I .$$
Let $p : A times 0, 1 cup a_0 times I to A vee A$ denote the quotient map.
$A wedge I_+$ is nothing else than $A times I / a_0 times I$. Let $p' : A times I to A times I / a_0 times I$ denote the quotient map. Its restriction $p''$ to $A times 0, 1 cup a_0 times I$ induces a unique map $i' : A vee A to A times I / a_0 times I$ such that $i' circ p = p'' = p' circ i$ where $i : A times 0, 1 cup a_0 times I to A times I$ denotes the inclusion map. Obviously $A_k = p(A times k ) subset A vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A times k ) subset A times I / a_0 times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A times I, A times 0, 1 cup a_0 times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A times I / a_0 times I) / A'_1 .$$
Let $r' : A times I / a_0 times I to CA$ denote the quotient map. The map $i'' : A to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A times 0, 1 cup a_0 times I to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A vee A to A$ such that $R = r circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.
answered Aug 2 at 14:53
Paul Frost
3,483420
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A subset X$. The definition of NDR pairs can be generalized to arbitrary $A subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A to X$, $f : A to B$, $i' : B to Y$ and $f' : X to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X times Y, X times B cup A times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, a_0 )$ is an NDR pair.
$A vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is
$$A vee A = (A times 0, 1 cup a_0 times I) / a_0 times I .$$
Let $p : A times 0, 1 cup a_0 times I to A vee A$ denote the quotient map.
$A wedge I_+$ is nothing else than $A times I / a_0 times I$. Let $p' : A times I to A times I / a_0 times I$ denote the quotient map. Its restriction $p''$ to $A times 0, 1 cup a_0 times I$ induces a unique map $i' : A vee A to A times I / a_0 times I$ such that $i' circ p = p'' = p' circ i$ where $i : A times 0, 1 cup a_0 times I to A times I$ denotes the inclusion map. Obviously $A_k = p(A times k ) subset A vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A times k ) subset A times I / a_0 times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A times I, A times 0, 1 cup a_0 times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A times I / a_0 times I) / A'_1 .$$
Let $r' : A times I / a_0 times I to CA$ denote the quotient map. The map $i'' : A to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A times 0, 1 cup a_0 times I to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A vee A to A$ such that $R = r circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.
answered Aug 2 at 14:53
Paul Frost
3,483420
You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A subset X$. The definition of NDR pairs can be generalized to arbitrary $A subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A to X$, $f : A to B$, $i' : B to Y$ and $f' : X to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X times Y, X times B cup A times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, a_0 )$ is an NDR pair.
$A vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is
$$A vee A = (A times 0, 1 cup a_0 times I) / a_0 times I .$$
Let $p : A times 0, 1 cup a_0 times I to A vee A$ denote the quotient map.
$A wedge I_+$ is nothing else than $A times I / a_0 times I$. Let $p' : A times I to A times I / a_0 times I$ denote the quotient map. Its restriction $p''$ to $A times 0, 1 cup a_0 times I$ induces a unique map $i' : A vee A to A times I / a_0 times I$ such that $i' circ p = p'' = p' circ i$ where $i : A times 0, 1 cup a_0 times I to A times I$ denotes the inclusion map. Obviously $A_k = p(A times k ) subset A vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A times k ) subset A times I / a_0 times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A times I, A times 0, 1 cup a_0 times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A times I / a_0 times I) / A'_1 .$$
Let $r' : A times I / a_0 times I to CA$ denote the quotient map. The map $i'' : A to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A times 0, 1 cup a_0 times I to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A vee A to A$ such that $R = r circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.
answered Aug 2 at 14:53
Paul Frost
3,483420
edited Aug 2 at 23:20
answered Aug 2 at 14:53
Paul Frost
3,483420
answered Aug 2 at 14:53
Paul Frost
3,483420
answered Aug 2 at 14:53
Paul Frost
3,483420
3,483420
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
add a comment |Â
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
Thanks a lot! Btw, the book I'm reading works in CGWH, so cofibrations are always closed inclusions.
– user09127
yesterday
add a comment |Â
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