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Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.
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Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.
This was my (failed) attempt to solve this problem.
Suppose that $X simeq c$. Pick a point $p in X$. We want to show that there exists a retract $r : X top$ such that for the map $i_p : p to X$ defined by $i_p (p) = p$ we have $i_p circ r simeq textId_X$.
Let $r : X to p$ be defined by $r(x) = p$ for all $x in X$. We claim that this is the required retraction. Then $i_p circ r =r$. So the problem boils down to showing that $r simeq textId_X$.
Now since $X simeq c$, there exist continuous maps $f : X to c$ and $g : c to X$ such that $f circ g simeq textId_c$ and $g circ f simeq textId_X$. Now note that there can only be one such map $f$, the map defined by $f(x) = c$ for all $x in X$ (which is trivially continuous).
At this point I figured that if I showed that $r simeq g circ f$, then I'd arrive at $r simeq textId_X$. But for any $x in X$ we have $(gcirc f) (x) = g(f(x)) = g(c)$ and since we don't know exactly what function $g$ is, $g(c)$ could be anything.
Now I'd have to define an explicit homotopy betwen $r$ and $g circ f$ but I'm not sure how to do that. Intuitively I need to find some way to continuously 'move' $g(c)$ to $p$ in $X$ via this homotopy, and usually if $X$ is convex then this can be done using a straight line homotopy. I'm not sure how to construct any analogous homotopy in this case.
How can I prove the above?
general-topology algebraic-topology
asked Jul 27 at 6:10
Perturbative
3,47911039
add a comment |Â
up vote
3
down vote
favorite
Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.
This was my (failed) attempt to solve this problem.
Suppose that $X simeq c$. Pick a point $p in X$. We want to show that there exists a retract $r : X top$ such that for the map $i_p : p to X$ defined by $i_p (p) = p$ we have $i_p circ r simeq textId_X$.
Let $r : X to p$ be defined by $r(x) = p$ for all $x in X$. We claim that this is the required retraction. Then $i_p circ r =r$. So the problem boils down to showing that $r simeq textId_X$.
Now since $X simeq c$, there exist continuous maps $f : X to c$ and $g : c to X$ such that $f circ g simeq textId_c$ and $g circ f simeq textId_X$. Now note that there can only be one such map $f$, the map defined by $f(x) = c$ for all $x in X$ (which is trivially continuous).
At this point I figured that if I showed that $r simeq g circ f$, then I'd arrive at $r simeq textId_X$. But for any $x in X$ we have $(gcirc f) (x) = g(f(x)) = g(c)$ and since we don't know exactly what function $g$ is, $g(c)$ could be anything.
Now I'd have to define an explicit homotopy betwen $r$ and $g circ f$ but I'm not sure how to do that. Intuitively I need to find some way to continuously 'move' $g(c)$ to $p$ in $X$ via this homotopy, and usually if $X$ is convex then this can be done using a straight line homotopy. I'm not sure how to construct any analogous homotopy in this case.
How can I prove the above?
general-topology algebraic-topology
asked Jul 27 at 6:10
Perturbative
3,47911039
Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
2
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.
This was my (failed) attempt to solve this problem.
Suppose that $X simeq c$. Pick a point $p in X$. We want to show that there exists a retract $r : X top$ such that for the map $i_p : p to X$ defined by $i_p (p) = p$ we have $i_p circ r simeq textId_X$.
Let $r : X to p$ be defined by $r(x) = p$ for all $x in X$. We claim that this is the required retraction. Then $i_p circ r =r$. So the problem boils down to showing that $r simeq textId_X$.
Now since $X simeq c$, there exist continuous maps $f : X to c$ and $g : c to X$ such that $f circ g simeq textId_c$ and $g circ f simeq textId_X$. Now note that there can only be one such map $f$, the map defined by $f(x) = c$ for all $x in X$ (which is trivially continuous).
At this point I figured that if I showed that $r simeq g circ f$, then I'd arrive at $r simeq textId_X$. But for any $x in X$ we have $(gcirc f) (x) = g(f(x)) = g(c)$ and since we don't know exactly what function $g$ is, $g(c)$ could be anything.
Now I'd have to define an explicit homotopy betwen $r$ and $g circ f$ but I'm not sure how to do that. Intuitively I need to find some way to continuously 'move' $g(c)$ to $p$ in $X$ via this homotopy, and usually if $X$ is convex then this can be done using a straight line homotopy. I'm not sure how to construct any analogous homotopy in this case.
How can I prove the above?
general-topology algebraic-topology
asked Jul 27 at 6:10
Perturbative
3,47911039
Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.
This was my (failed) attempt to solve this problem.
Suppose that $X simeq c$. Pick a point $p in X$. We want to show that there exists a retract $r : X top$ such that for the map $i_p : p to X$ defined by $i_p (p) = p$ we have $i_p circ r simeq textId_X$.
Let $r : X to p$ be defined by $r(x) = p$ for all $x in X$. We claim that this is the required retraction. Then $i_p circ r =r$. So the problem boils down to showing that $r simeq textId_X$.
Now since $X simeq c$, there exist continuous maps $f : X to c$ and $g : c to X$ such that $f circ g simeq textId_c$ and $g circ f simeq textId_X$. Now note that there can only be one such map $f$, the map defined by $f(x) = c$ for all $x in X$ (which is trivially continuous).
At this point I figured that if I showed that $r simeq g circ f$, then I'd arrive at $r simeq textId_X$. But for any $x in X$ we have $(gcirc f) (x) = g(f(x)) = g(c)$ and since we don't know exactly what function $g$ is, $g(c)$ could be anything.
Now I'd have to define an explicit homotopy betwen $r$ and $g circ f$ but I'm not sure how to do that. Intuitively I need to find some way to continuously 'move' $g(c)$ to $p$ in $X$ via this homotopy, and usually if $X$ is convex then this can be done using a straight line homotopy. I'm not sure how to construct any analogous homotopy in this case.
How can I prove the above?
general-topology algebraic-topology
asked Jul 27 at 6:10
Perturbative
3,47911039
asked Jul 27 at 6:10
Perturbative
3,47911039
asked Jul 27 at 6:10
Perturbative
3,47911039
asked Jul 27 at 6:10
Perturbative
3,47911039
3,47911039
Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
2
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42
add a comment |Â
Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
2
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42
Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
2
2
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42
add a comment |Â
2 Answers
2
active
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votes
up vote
1
down vote
accepted
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.
Suppose $X simeq *$ . Let $f : X to *$ be the homotopy equivalence and $g : * to X$, defined as $g(*) = p$ for some $p in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g circ f : X to X$ is a constant map homotopic to $textId_X$. Let $H : X times I to X$ be the homotopy, with $H(x,0) = textId_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x in X$, $H(x,cdot) : I to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $iota_q circ r simeq textId_X$, where $r : X to q$ is a constant map (retraction) and $iota_q : q to X$ is the inclusion map. Let $k := iota_q circ r$.
As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.
Note that $h = g circ f$ and $k = iota_q circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $alpha : I to X$ from $alpha(0) = p$ to $alpha(1) =q$. Define a map $F : X times I to X$ by $F(x,t) = alpha(t)$. This means that $F(x,0) = alpha(0) = p = h(x)$, and $F(x,1) = alpha(1) = q = k(x)$ for any $x in X$. So $h simeq iota_q circ r$, and $h simeq textId_X$ by hypothesis. Therefore $iota_q circ r simeq textId_X$.
answered Jul 27 at 10:33
Sou
2,7042820
add a comment |Â
up vote
1
down vote
Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = p $ with $p in X$. By (1) there is a unique map $r : X to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P to X$. It is characaterized by $i circ r simeq Id_X$ ($r circ i = Id_P$ is trivial). Let $H : X times I to X$ be a homotopy such that $H_0 = Id_X , H_1 = i circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P to X$ (which is given by $j(p) in X$) is a homotopy inverse for $r$. This comes from the fact that $j circ r = Id_X circ j circ r simeq i circ r circ j circ r = i circ r simeq Id_X$.
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) in P times I$. In fact, a strong deformation retraction does not always exist.
answered Jul 27 at 9:23
Paul Frost
3,611420
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.
Suppose $X simeq *$ . Let $f : X to *$ be the homotopy equivalence and $g : * to X$, defined as $g(*) = p$ for some $p in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g circ f : X to X$ is a constant map homotopic to $textId_X$. Let $H : X times I to X$ be the homotopy, with $H(x,0) = textId_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x in X$, $H(x,cdot) : I to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $iota_q circ r simeq textId_X$, where $r : X to q$ is a constant map (retraction) and $iota_q : q to X$ is the inclusion map. Let $k := iota_q circ r$.
As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.
Note that $h = g circ f$ and $k = iota_q circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $alpha : I to X$ from $alpha(0) = p$ to $alpha(1) =q$. Define a map $F : X times I to X$ by $F(x,t) = alpha(t)$. This means that $F(x,0) = alpha(0) = p = h(x)$, and $F(x,1) = alpha(1) = q = k(x)$ for any $x in X$. So $h simeq iota_q circ r$, and $h simeq textId_X$ by hypothesis. Therefore $iota_q circ r simeq textId_X$.
answered Jul 27 at 10:33
Sou
2,7042820
add a comment |Â
up vote
1
down vote
accepted
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.
Suppose $X simeq *$ . Let $f : X to *$ be the homotopy equivalence and $g : * to X$, defined as $g(*) = p$ for some $p in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g circ f : X to X$ is a constant map homotopic to $textId_X$. Let $H : X times I to X$ be the homotopy, with $H(x,0) = textId_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x in X$, $H(x,cdot) : I to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $iota_q circ r simeq textId_X$, where $r : X to q$ is a constant map (retraction) and $iota_q : q to X$ is the inclusion map. Let $k := iota_q circ r$.
As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.
Note that $h = g circ f$ and $k = iota_q circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $alpha : I to X$ from $alpha(0) = p$ to $alpha(1) =q$. Define a map $F : X times I to X$ by $F(x,t) = alpha(t)$. This means that $F(x,0) = alpha(0) = p = h(x)$, and $F(x,1) = alpha(1) = q = k(x)$ for any $x in X$. So $h simeq iota_q circ r$, and $h simeq textId_X$ by hypothesis. Therefore $iota_q circ r simeq textId_X$.
answered Jul 27 at 10:33
Sou
2,7042820
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.
Suppose $X simeq *$ . Let $f : X to *$ be the homotopy equivalence and $g : * to X$, defined as $g(*) = p$ for some $p in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g circ f : X to X$ is a constant map homotopic to $textId_X$. Let $H : X times I to X$ be the homotopy, with $H(x,0) = textId_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x in X$, $H(x,cdot) : I to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $iota_q circ r simeq textId_X$, where $r : X to q$ is a constant map (retraction) and $iota_q : q to X$ is the inclusion map. Let $k := iota_q circ r$.
As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.
Note that $h = g circ f$ and $k = iota_q circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $alpha : I to X$ from $alpha(0) = p$ to $alpha(1) =q$. Define a map $F : X times I to X$ by $F(x,t) = alpha(t)$. This means that $F(x,0) = alpha(0) = p = h(x)$, and $F(x,1) = alpha(1) = q = k(x)$ for any $x in X$. So $h simeq iota_q circ r$, and $h simeq textId_X$ by hypothesis. Therefore $iota_q circ r simeq textId_X$.
answered Jul 27 at 10:33
Sou
2,7042820
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.
Suppose $X simeq *$ . Let $f : X to *$ be the homotopy equivalence and $g : * to X$, defined as $g(*) = p$ for some $p in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g circ f : X to X$ is a constant map homotopic to $textId_X$. Let $H : X times I to X$ be the homotopy, with $H(x,0) = textId_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x in X$, $H(x,cdot) : I to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $iota_q circ r simeq textId_X$, where $r : X to q$ is a constant map (retraction) and $iota_q : q to X$ is the inclusion map. Let $k := iota_q circ r$.
As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.
Note that $h = g circ f$ and $k = iota_q circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $alpha : I to X$ from $alpha(0) = p$ to $alpha(1) =q$. Define a map $F : X times I to X$ by $F(x,t) = alpha(t)$. This means that $F(x,0) = alpha(0) = p = h(x)$, and $F(x,1) = alpha(1) = q = k(x)$ for any $x in X$. So $h simeq iota_q circ r$, and $h simeq textId_X$ by hypothesis. Therefore $iota_q circ r simeq textId_X$.
answered Jul 27 at 10:33
Sou
2,7042820
edited Jul 27 at 19:34
answered Jul 27 at 10:33
Sou
2,7042820
answered Jul 27 at 10:33
Sou
2,7042820
answered Jul 27 at 10:33
Sou
2,7042820
2,7042820
add a comment |Â
add a comment |Â
up vote
1
down vote
Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = p $ with $p in X$. By (1) there is a unique map $r : X to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P to X$. It is characaterized by $i circ r simeq Id_X$ ($r circ i = Id_P$ is trivial). Let $H : X times I to X$ be a homotopy such that $H_0 = Id_X , H_1 = i circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P to X$ (which is given by $j(p) in X$) is a homotopy inverse for $r$. This comes from the fact that $j circ r = Id_X circ j circ r simeq i circ r circ j circ r = i circ r simeq Id_X$.
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) in P times I$. In fact, a strong deformation retraction does not always exist.
answered Jul 27 at 9:23
Paul Frost
3,611420
add a comment |Â
up vote
1
down vote
Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = p $ with $p in X$. By (1) there is a unique map $r : X to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P to X$. It is characaterized by $i circ r simeq Id_X$ ($r circ i = Id_P$ is trivial). Let $H : X times I to X$ be a homotopy such that $H_0 = Id_X , H_1 = i circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P to X$ (which is given by $j(p) in X$) is a homotopy inverse for $r$. This comes from the fact that $j circ r = Id_X circ j circ r simeq i circ r circ j circ r = i circ r simeq Id_X$.
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) in P times I$. In fact, a strong deformation retraction does not always exist.
answered Jul 27 at 9:23
Paul Frost
3,611420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = p $ with $p in X$. By (1) there is a unique map $r : X to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P to X$. It is characaterized by $i circ r simeq Id_X$ ($r circ i = Id_P$ is trivial). Let $H : X times I to X$ be a homotopy such that $H_0 = Id_X , H_1 = i circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P to X$ (which is given by $j(p) in X$) is a homotopy inverse for $r$. This comes from the fact that $j circ r = Id_X circ j circ r simeq i circ r circ j circ r = i circ r simeq Id_X$.
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) in P times I$. In fact, a strong deformation retraction does not always exist.
answered Jul 27 at 9:23
Paul Frost
3,611420
Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = p $ with $p in X$. By (1) there is a unique map $r : X to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P to X$. It is characaterized by $i circ r simeq Id_X$ ($r circ i = Id_P$ is trivial). Let $H : X times I to X$ be a homotopy such that $H_0 = Id_X , H_1 = i circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P to X$ (which is given by $j(p) in X$) is a homotopy inverse for $r$. This comes from the fact that $j circ r = Id_X circ j circ r simeq i circ r circ j circ r = i circ r simeq Id_X$.
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) in P times I$. In fact, a strong deformation retraction does not always exist.
answered Jul 27 at 9:23
Paul Frost
3,611420
answered Jul 27 at 9:23
Paul Frost
3,611420
answered Jul 27 at 9:23
Paul Frost
3,611420
answered Jul 27 at 9:23
Paul Frost
3,611420
3,611420
add a comment |Â
add a comment |Â
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Maybe the answer on this question is helpful.
– drhab
Jul 27 at 6:43
2
depends whether you consider a "strong DR" (the deformation is required to be stationary on the point) or not. Hatcher defines a DR with this additional constraint and then the statement is simply false, see counter-examples on page 18, pi.math.cornell.edu/~hatcher/AT/AT.pdf
– Peter Franek
Jul 27 at 8:00
@PeterFranek I think in this case, OP means DR with no additional constraint.
– Sou
Jul 27 at 16:42