I am looking for an example of a well-pointed space in which no (sufficiently small) neighbourhood of the base-point is contractible. As usual, a well-pointed space is a pointed space in which the inclusion of the base-point is a Hurewicz cofibration.

In the reverse direction, I am looking for conditions under which a well-pointed space has a basis of contractible neighbourhoods around the base-point.

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    $\begingroup$ There may well be examples of well-pointed spaces in which the basepoint does not have arbitrarily small contractible neighborhoods. But I believe that well-pointed spaces are locally contractible at the basepoint in the following weaker sense: For every neighborhood $U$ of the basepoint there is a neighborhood $V\subset U$ of the basepoint such that the inclusion $V\to U$ is based homotopic to the constant map. $\endgroup$ – Tom Goodwillie Feb 3 '12 at 3:31
  • $\begingroup$ It seems like the characterization of Hurewicz cofibrations as NDR-pairs says more or less exactly that some neighborhood of a nondegenerate basepoint can be contracted back to the basepoint. $\endgroup$ – Mike Shulman Feb 3 '12 at 16:54
  • $\begingroup$ @Mike: Unfortunately, the deformation of the neighbourhood to the base-point does not have to remain within the neighbourhood. Thus, the neighbourhood need not be contractible. $\endgroup$ – Ricardo Andrade Feb 3 '12 at 23:25
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    $\begingroup$ @Ricardo: yes, but it may be assumed to remain within a slightly bigger neighborhood (using that the defomation is pointed and continuous). So what Mike said gives a proof of what Tom said. $\endgroup$ – Sergey Melikhov Feb 11 '12 at 17:47

My initial answer was fully ignorant of the literature on the subject; I'm now recalling that it does exist. In 1934, Borsuk and Mazurkiewicz constructed a $2$-dimensional AR (=contractible, locally contractible $2$-dimensional compactum) $X$ that is not a countable union of smaller ARs (Sur les rétractes absolus indécomposables, C.R.. Acad. Sci. Paris 199 (1934), 110-112; see Borsuk's "Theory of retracts", Section VI.4). The proof actually shows that $X$ is not a countable union of contractible proper subsets (open or not). Thus $X$ contains points that do not have a basis of contractible neighborhoods.

A simplified version $Y$ of $X$ (constructed below), which is an AR failing to have a basis of contractible neighborhoods at a certain point $\infty$, is also forward tame at $\infty$ and so answers Ricardo's revised question at the Algebraic Topology Discussion List. $Y$ looks somewhat like a $2$-skeleton of $W^+$, the example in my previous answer, except that $W^+$ is not forward tame. A google search for "Mazurkiewicz singularity" (this is how Borsuk called the kind of ANRs that do not have arbitrarily fine countable covers by ARs) returns a paper by Armentrout which has an example somewhat similar to $W^+$ (also a decomposition space and a manifold factor) though more laborious.

Note also a positive result of Begle proved using Zorn's lemma.

Here's a construction of $X$ (which I hope is easier to read than Borsuk's account) and also of $Y$.

Let $Q=D^2/\{a,b\}$, the quotient of the $2$-disk where two distinct interior points $a,b$ are identified with each other. Let $\partial Q$ be the image of $\partial D^2$ in $Q$, and let $C$ be the image in $Q$ of a path joining $a$ and $b$ in the interior of $D^2$. Let $(Q_1,\partial Q_1)=(D^2,\partial D^2)$, and form $(Q_k,\partial Q_k)$ by attaching $Q_{k-1}$ to $(Q,\partial Q)$ along a homeomorphism $\partial Q_{k-1}\to C$. Then $Q_k$ is contractible, but no proper subset of $Q_k$ containing $\partial Q_k$ is contractible (or even acyclic).

Now $Y$ is just the one point compactification $(Q_\infty)^+$ of the direct limit $Q_\infty$ of the chain of inclusions $Q_1\subset Q_2\subset\dots$.

To see that $Y$ is locally contractible, observe that $D^2\times [0,1]$ collapses onto a copy of $Q_2$ such that $\partial Q_2=\partial D^2\times\{1\}$ and $Q_1=D^2\times\{0\}$. It follows that $D^2\times [0,\infty)$ admits a proper deformation retraction onto a copy of $Q_\infty$. Thus $Q_\infty$ is proper homotopy equivalent to $[0,\infty)$. Then $Q_\infty$ is forward tame in the sense of Quinn (see Definition 7.1, p.75 and Proposition 9.6, p.99 in Hughes-Ranicki), and therefore its one-point compactification is locally contractible.

For the construction of $X$ we will need $k$ copies $C_1^{Q_k},\dots,C_k^{Q_k}$ of $S^1$ embedded in $Q_k$, namely $C_1^{Q_k}=\partial Q_k$, and $C_{i+1}^{Q_k}=C_i^{Q_{k-1}}$. Now $X$ is formed by attaching increasingly smaller copies of $Q_1,Q_8,Q_{64},...$ to the Sierpinsky carpet so that for each $k$ of the form $8^m$, the $k$ circles $C_1^{Q_k},\dots,C_k^{Q_k}$ in $Q_k$ are attached to the boundaries of the disks removed at the $m$th stage of the construction of the Sierpinski carpet, and additionally $(k-1)$ arcs connecting $C_i^{Q_k}$ with $C_{i+1}^{Q_{k+1}}$ are identified with analogous arcs in the Sierpinsky carpet. This can, and must be done so that every neighborhood of every point of the Sierpinsky carpet contains $C_1^{Q_k}=\partial Q_k$ for some $k$ (of the form $8^m$).

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    $\begingroup$ @Sergey: Truly great work! Thank you so much. I am still thinking about the details. In the meantime, just a few remarks. 1) If I understand the example correctly, to make $Q_k$ contractible, it seems that you want to start with $(Q_1,\partial Q_1)=(D^2,\partial D^2)$ (instead of $(Q,\partial Q)$). This change doesn't make much difference other than ensuring that $Q_k$ and $Q_\infty$ are contractible, and either possibility seems to give an answer to my question. $\endgroup$ – Ricardo Andrade Feb 17 '12 at 3:17
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    $\begingroup$ [continuation of comment] 2) The part about collapsing $D^2\times [0,1]$ mapping onto $Q$ does not seem to work unless you replace $Q$ by $Q_2$, and only after implementing the change I mentioned in (1). 3) The numbers seem to be a little off in the last paragraph. Sorry about the nitpicking. I am just trying to make sure I understand the example correctly by working out the kinks. $\endgroup$ – Ricardo Andrade Feb 17 '12 at 3:25
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    $\begingroup$ I do have two vague questions, though. Did the inspiration for $Y$ come only from the example of Borsuk-Mazurkiewicz? Any advice on how to find relevant examples like these in the old literature on geometric topology? Thanks a lot for your amazing answers. I may eventually mark this answer as the accepted one. $\endgroup$ – Ricardo Andrade Feb 17 '12 at 5:06
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    $\begingroup$ Ricardo, thank you for correcting my errors and for your interesting questions! Good starting points are Ferry's notes, Borsuk's "Theory of Retracts", the Daverman-Venema book and Daverman's "Decompositions of Manifolds", also Bing and Moise's books. As for $Y$, I was initially thinking of the one-point compactification of the mapping telescope of some inverse sequense of compact polyhedra. This is a rather large class of spaces for which simple characterizations of forward tameness and local contractibility are known, see theorems 3.12 and 6.12 in arxiv.org/abs/0812.1407 $\endgroup$ – Sergey Melikhov Feb 17 '12 at 10:41

Edit of edit: I'm adding still more details and references.

The one-point compactification $W^+=W\cup\{\infty\}$ of the Whitehead manifold $W$ is locally contractible (in the sense of geometric topology - see Goodwillie's comment above) and hence well-pointed at $\infty$.

This can be seen directly, and also follows from the fact $W^+\times\Bbb R$ is homeomorphic to $S^3\times\Bbb R$ (for this homeomorphism see Chapter 4 "Manifold factors" in Ferry's notes). Perhaps this implication works as an argument for judging what definition of "locally contractible" is the "right" one, so I'm including the details.

Since $\{\infty\}\times\Bbb R\subset W^+\times\Bbb R$ is just a copy of $\Bbb R$ in $S^3\times\Bbb R$ (a wild copy, though - see my comment below), every neighborhood $U$ of $\{\infty\}\times\Bbb R$ in $W^+\times\Bbb R$ contains a smaller neighborhood $V$ such that the inclusion $i:V\to U$ is homotopic to a map $V\to\Bbb\{\infty\}\times\Bbb R$. (Using that $\Bbb R$ is an ANR to get the map, and that $S^3\times\Bbb R$ is locally contractible to get the homotopy.) Since $\Bbb R$ is contractible, $i$ is null-homotopic. Then every neighborhood $U'$ of $\infty$ in $W^+$ contains a smaller neighborhood $V'$ (namely $V'=V\cap W^+\times\{0\}$) that contracts in $U:=U'\times\Bbb R$. This contraction can be projected down to $W^+$.

On the other hand $W^+$ has no basis of contractible neighborhoods at $\infty$.

Indeed $W$ is the union of compact submanifolds $W_1\subset W_2\subset\dots$ of $S^3$, each $W_k$ being the closure of the complement of a solid torus $T_k$. If $U$ is a contractible neighborhood of infinity in $W^+$ contained in $W^+\setminus W_1$, then $U$ contains $W^+\setminus W_k$ for some $k$. Now $W\setminus W_k$ contains a knot $K_k$ isotopic to the core of the solid torus $S^3\setminus W_k$ by an isotopy of that solid torus. Thus $K_k$ is contained in the contractible subset $U$ of $W^+\setminus W_1$.

Consider $f:S^3\to W^+$ that is the identity on $W$ and sends $S^3\setminus W$ to $\infty$ (in fact it is a quotient map). This map has acyclic point inverses (in the sense of Cech cohomology) so $f^{-1}(U)$ is an acyclic subset of $S^3\setminus W_1$ containing $K_k$. (This is using the Vietoris-Begle theorem, which is a special case of the Leray spectral sequence for a continuous map - see for instance Bredon's "Sheaf Theory".) But it is not hard to show that there exists no such acyclic subset (see N. Smythe, $n$-linking and $n$-splitting, Amer. J. Math. 92 (1970), 272-282).

Smythe's proof of the latter assertion is very lovely. He calls compact subsets $A,B\subset S^3$ $n$-split (this is the revised definition, on p.277), if there exists a sequence of compact subpolyhedra $A\subset P_0\subset\dots\subset P_{n+1}\subset S^3\setminus B$ such that each inclusion $P_i\subset P_{i+1}$ is trivial on reduced integral homology groups. Thus “(−1)-split” means “disjoint”, and a two-component link has 0-split components iff it has zero linking number. It is also easy to see that a two-component link has 1-split components iff it is a boundary link (i.e. the components bound disjoint Seifert surfaces in $S^3$). It is easy to see, using the Alexander duality, that the relation of being $n$-split is symmetric.

Returning to the Whitehead manifold, it suffices to show that $K_k$ is not $(k-1)$-split from $W_1$, or equivalently from the core $K_0$ of the solid torus $W_1$. We prove by induction a slightly stronger assertion, that every essential (i.e. not null-homotopic) simple closed curve in the solid torus $T_k$ (=the closure of $S^3\setminus W_k$) is not $(k-1)$-split from $K_0$. Assume that this holds with $k=n$, and let $C$ be an essential simple closed curve in $T_{n+1}$. If $C$ is $n$-split from $K_0$, then $C$ bounds an orientable surface $F$ in $S^3\setminus K_0$ that is $(n-1)$-split from $K_0$. Without loss of generality $D:=F\cap\partial T_n$ is a closed $1$-manifold. By the induction hypothesis, every simple closed curve in $F\cap T_n$ is null-homotopic in $T_n$; in particular, so is each component of $D$.

Let $\tilde T_n$ be the universal cover of $T_n$. Since $C$ lies in $T_{n+1}$, it is null-homotopic in $T_n$, and so lifts to a simple closed curve $C_0$ in $\tilde T_n$. Similarly $F\cap T_n$ lifts to a compact surface in $\tilde T_n$ with boundary $\tilde C_0\cup\tilde D_0$. The translate $C_1=t(C_0)$, where $t$ is a generator of the covering translation group $\Bbb Z$, has linking number zero with every component of $\tilde D_0$, and therefore also with $\tilde C_0$. On the other hand if $[C]=m\in\Bbb Z=\pi_1(T_{n+1})$, then $\tilde C_0$ has linking number $m^2$ with $\tilde C_1$. So $m=0$ and $C$ is null-homotopic in $T_{n+1}$, which is a contradiction.

  • $\begingroup$ This is a very interesting answer. Absolutely not what I had in mind. I am still wading through the details. I do have a question, though. Do you know if the homeomorphism $f:W^+ \times \Bbb{R}\rightarrow S^3\times \Bbb{R}$ can be chosen so that $f(\{\infty\}\times\Bbb{R})=\{p\}\times\Bbb{R}$? $\endgroup$ – Ricardo Andrade Feb 11 '12 at 0:29
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    $\begingroup$ No, $f(\infty\times\Bbb R)$ is a wild curve in $S^3\times\Bbb R$ because for each $k$ there is a loop in $W\setminus W_k$ that is not null-homotopic in $W\setminus W_1$. Hence every punctured neighborhood of $\infty$ in $W^+$ contains a loop that is not null-homotopic in $W\setminus W_1$. This loop, considered as being in $W^+\times 0$, is also not null-homotopic in $W\times\Bbb R\setminus W_1\times\Bbb R$. Thus $W^+\times\Bbb R$ is not "locally 1-co-connected" at $\infty\times\Bbb R$ and hence $f(\infty\times\Bbb R)$ is wild (see pp.31-32 in calvin.edu/~venema/embeddingsbook). $\endgroup$ – Sergey Melikhov Feb 11 '12 at 15:04
  • $\begingroup$ @Sergey: Many thanks for the edit and answer to my comment. Is it then true that the image of $\{\infty\}\times\Bbb{R}$ in $S^3\times\Bbb{R}$ cannot be locally flat at any point? $\endgroup$ – Ricardo Andrade Feb 11 '12 at 22:35
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    $\begingroup$ Yes, for every $t\in\Bbb R$ and every $\epsilon>0$ we can find a loop in the punctured $\epsilon$-neighborhood of $(\infty,t)$ in $W^+\times\\{t\\}$ that is not null-homotopic in $W\times\Bbb R\setminus W_1\times\Bbb R$. So $f(\infty\times\Bbb R)$ is not locally flat at $f(\infty,t)$. $\endgroup$ – Sergey Melikhov Feb 12 '12 at 1:24

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