Mixing
What is an example of an Eilenberg-MacLane space that is not homotopy equivalent to a CW complex?
$begingroup$
The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.
algebraic-topology
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
$endgroup$
add a comment |
$begingroup$
The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.
algebraic-topology
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
$endgroup$
add a comment |
$begingroup$
The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.
algebraic-topology
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
$endgroup$
The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.
algebraic-topology
algebraic-topology
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
asked Jan 12 at 21:50
Connor MalinConnor Malin
758
758
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.
Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
$endgroup$
add a comment |
$begingroup$
For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
$endgroup$
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
add a comment |
$begingroup$
I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
$endgroup$
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.
Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
$endgroup$
add a comment |
$begingroup$
The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.
Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
$endgroup$
add a comment |
$begingroup$
The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.
Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
$endgroup$
The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.
Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
answered Jan 13 at 12:33
Paul FrostPaul Frost
10.7k3933
10.7k3933
add a comment |
add a comment |
$begingroup$
For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
$endgroup$
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
add a comment |
$begingroup$
For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
$endgroup$
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
add a comment |
$begingroup$
For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
$endgroup$
For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
answered Jan 13 at 11:41
TyroneTyrone
4,65511225
4,65511225
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
add a comment |
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant.
$endgroup$
– Mike Miller
Jan 13 at 15:39
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
$begingroup$
@MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1simeq K(mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $Xrightarrow S^1$.
$endgroup$
– Tyrone
Jan 13 at 17:29
1
1
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer.
$endgroup$
– Mike Miller
Jan 13 at 17:32
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
$begingroup$
@MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $pi_0Y=|Y|$ and $pi_1Y=0$, so since $pi_1Xcongmathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;).
$endgroup$
– Tyrone
Jan 13 at 17:42
add a comment |
$begingroup$
I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
$endgroup$
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
add a comment |
$begingroup$
I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
$endgroup$
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
add a comment |
$begingroup$
I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
$endgroup$
I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
answered Jan 12 at 22:39
Connor MalinConnor Malin
758
758
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
add a comment |
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
2
2
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
$begingroup$
Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample!
$endgroup$
– Mike Miller
Jan 12 at 22:43
add a comment |
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