More on flimsy spaces
$begingroup$
I've recently encountered this very nice question on flimsy spaces and have come up with the following generalised version of the question, one which isn't also answered (in an obvious fashion) by one of the answers posted there:
Definition: Let $n$ and $k$ be two non-negative integers. A $k$-connected topological space is called $(n,k)$-flimsy if the space, obtained from $X$ by removing any choice of $n-1$ distinct points inside $X$, is $k$-conntected, and the space, obtained from $X$ by removing any choice of $n$ distinct points, is not $k$-connected.
It is evident from the definition that there always exists $(1,k)$ and $(2,k)$-flimsy spaces for each $k in mathbb N_{0}$ (Take $mathbb R^{k+1}$ and $S^{k+1}$). One could probably also come up with more exotic (non-CW) examples.
Question: Is it true that no $(3,k)$-flimsy space exists ?
The original question is concerned with the case $k = 0$. I don't see how the point-set theoretical methods employed in the proof can be extended to higher $k$. Any help is appreciated.
general-topology algebraic-topology homotopy-theory
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
$endgroup$
add a comment |
$begingroup$
I've recently encountered this very nice question on flimsy spaces and have come up with the following generalised version of the question, one which isn't also answered (in an obvious fashion) by one of the answers posted there:
Definition: Let $n$ and $k$ be two non-negative integers. A $k$-connected topological space is called $(n,k)$-flimsy if the space, obtained from $X$ by removing any choice of $n-1$ distinct points inside $X$, is $k$-conntected, and the space, obtained from $X$ by removing any choice of $n$ distinct points, is not $k$-connected.
It is evident from the definition that there always exists $(1,k)$ and $(2,k)$-flimsy spaces for each $k in mathbb N_{0}$ (Take $mathbb R^{k+1}$ and $S^{k+1}$). One could probably also come up with more exotic (non-CW) examples.
Question: Is it true that no $(3,k)$-flimsy space exists ?
The original question is concerned with the case $k = 0$. I don't see how the point-set theoretical methods employed in the proof can be extended to higher $k$. Any help is appreciated.
general-topology algebraic-topology homotopy-theory
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
$endgroup$
2
$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
1
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02
add a comment |
$begingroup$
I've recently encountered this very nice question on flimsy spaces and have come up with the following generalised version of the question, one which isn't also answered (in an obvious fashion) by one of the answers posted there:
Definition: Let $n$ and $k$ be two non-negative integers. A $k$-connected topological space is called $(n,k)$-flimsy if the space, obtained from $X$ by removing any choice of $n-1$ distinct points inside $X$, is $k$-conntected, and the space, obtained from $X$ by removing any choice of $n$ distinct points, is not $k$-connected.
It is evident from the definition that there always exists $(1,k)$ and $(2,k)$-flimsy spaces for each $k in mathbb N_{0}$ (Take $mathbb R^{k+1}$ and $S^{k+1}$). One could probably also come up with more exotic (non-CW) examples.
Question: Is it true that no $(3,k)$-flimsy space exists ?
The original question is concerned with the case $k = 0$. I don't see how the point-set theoretical methods employed in the proof can be extended to higher $k$. Any help is appreciated.
general-topology algebraic-topology homotopy-theory
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
$endgroup$
I've recently encountered this very nice question on flimsy spaces and have come up with the following generalised version of the question, one which isn't also answered (in an obvious fashion) by one of the answers posted there:
Definition: Let $n$ and $k$ be two non-negative integers. A $k$-connected topological space is called $(n,k)$-flimsy if the space, obtained from $X$ by removing any choice of $n-1$ distinct points inside $X$, is $k$-conntected, and the space, obtained from $X$ by removing any choice of $n$ distinct points, is not $k$-connected.
It is evident from the definition that there always exists $(1,k)$ and $(2,k)$-flimsy spaces for each $k in mathbb N_{0}$ (Take $mathbb R^{k+1}$ and $S^{k+1}$). One could probably also come up with more exotic (non-CW) examples.
Question: Is it true that no $(3,k)$-flimsy space exists ?
The original question is concerned with the case $k = 0$. I don't see how the point-set theoretical methods employed in the proof can be extended to higher $k$. Any help is appreciated.
general-topology algebraic-topology homotopy-theory
general-topology algebraic-topology homotopy-theory
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
asked Dec 17 '18 at 19:57
Berni WatermanBerni Waterman
958713
958713
2
$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
1
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02
add a comment |
2
$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
1
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02
2
2
$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
1
1
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02
add a comment |
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$begingroup$
If $M$ is a noncompact manifold, then $M setminus {pt} simeq M vee S^{dim M - 1}$. In particular, if $M$ is any manifold, then $pi_k(M setminus {p_1, p_2, cdots, p_n}) = pi_k(M setminus{p_1}) oplus (pi_k(S^{dim M - 1}))^{n-1}.$ In particular, if a manifold minus $n$ points fails to be $k$-connected, the same is true for the manifold minus two points. So there are no $(n,k)$-flimsy manifolds for $n > 2$, and a $(2,k)$-flimsy manifold must be compact.
$endgroup$
– Mike Miller
Dec 17 '18 at 20:14
$begingroup$
Could you elaborate a little on the homotopy equivalence of your first sentence ? I have to admit that I don't quite see it.
$endgroup$
– Berni Waterman
Dec 18 '18 at 10:56
1
$begingroup$
The picture is: Triangulate M. It is known (and easy to see when M is the interior of a compact manifold) that M deformation retracts onto a subcomplex of 1 dimension less. The thing to do is to argue that you can also deformation retract onto something of the form $M' cup sigma$, where $M'$ is that subcomplex and $sigma$ is a simplex which may be collapsed onto its boundary faces; one may argue this is homotopy equivalent relative to $sigma$ to $M' vee sigma$. Now, puncturing inside $sigma$ leaves you with an extra sphere.
$endgroup$
– Mike Miller
Dec 19 '18 at 9:02