76

Take the composition of a degree one map $f:T^3\to S^3$ with the Hopf map $g:S^3\to S^2$, where $T^3$ is the 3-torus. This composition is trivial on homotopy groups since $T^3$ is aspherical and $\pi_1S^2=0$. It is trivial on $H_i$ for $i>0$ since this is true for $g$. If $gf$ were nullhomotopic we could lift a nullhomotopy to a homotopy of $f$ to a map ...

18

I presume by "acyclic" you are referring to homology with $\mathbb{Z}$ coefficients. There are many such examples.
For instance, you can take two elements $u,v$ in the free group $F_2$ of rank 2 that satisfy the $C'(1/6)$ small-cancellation condition, and also such that $u,v$ together generate the abelianisation $\mathbb{Z}^2$. Explicit examples are easy ...

17

A mapping space $\mathrm{Map}(X,Y)$ between two finite CW-complexes never admits a cell structure if both $X$ and $Y$ are positive-dimensional. If you use the compact-open topology, this essentially follows from this answer: the mapping space is a complete metric space and thus satisfies the Baire category theorem, but if every component of $Y$ is positive-...

answered Dec 25 '13 at 19:13

Eric Wofsey

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16

The Higman group with presentation
$$\langle{a,b,c,d}\mid{aba^{-1}b^{-2}},~bcb^{-1}c^{-2},~cdc^{-1}d^{-2},~
dad^{-1}a^{-2}\rangle$$
is perfect, and the 2-complex associated to this presentation
has Euler characteristic 0. Hence this complex is acyclic.
It is in fact aspherical, but it may be simpler to observe that
Higman's group is also an iterated ...

16

Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_{s} \rangle$ with the same number of generators and relations.
Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_1(X) = G$ and the cellular chain ...

15

Pick a torus, and add two discs along a meridian and a longitude. You get a 2-complex homotopic to a sphere that does not contain a sphere. This generalises easily to any genus by picking a genus-$g$ surface.
More generally, a finite 2-complex contains finitely many surfaces, and there are some moves (like the Matveev - Piergallini move) that preserve the (...

14

This is true for finite $\pi_1$ and false for infinite $\pi_1$: Let $\widetilde{X}$ denote the universal cover of $X$, then $\Omega\widetilde{X}$ is the unit connected component of $\Omega X$, and $\Omega X = \coprod_{\pi_1(X)} \Omega\widetilde{X}$. So if $\pi_1$ is infinite, then certainly $H_0(\Omega X)$ is not finitely generated as others have noted in ...

14

Let $X$ be a finite complex. Then the functor
$$\lim_X:\operatorname{Fun}(X,\operatorname{Sp})\to \operatorname{Sp}$$
sending a local system of spectra $E$ to its limit preserves all colimits. Indeed it preserves all finite colimits by stability, and it preserves all filtered colimits by the finiteness of $X$. Therefore it is of the form
$$\lim_X E \cong \...

13

For our CW-complex I'm going to take $X = \Bbb{CP}^\infty$ (as a based space), whose skeleta are $\Bbb{CP}^n$. The cohomology theory will be more difficult to construct.
For any $k \geq 1$, let $E_k$ be the spectrum which is the homotopy fiber of the map
$$
Sq^{2^k} \cdots Sq^8 Sq^4 Sq^2: \Sigma^2 H\Bbb{Z}/2 \to \Sigma^{2^{k+1}} H\Bbb{Z}/2
$$
so that, for ...

answered Feb 10 '14 at 6:10

Tyler Lawson

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13

In the simply connected case, the answer is yes.
In the general case, the theory was worked out in complete detail by Wall in the paper:
Wall, C. T. C. Finiteness conditions for CW-complexes. Ann. of Math. (2) 81 1965 56–69.
See "Theorem E."
Here is a description of the result:
In the non-simply connected case, to get that your complex $X$ is ...

13

The authors of this book are attempting to use CW structures to justify certain cohomology isomorphisms, but this seems to be the wrong approach since some of their claims about CW structures are just not true. For example, they say a vector bundle over a CW complex base space has a CW structure such that the complement of the zero section is a subcomplex, ...

11

I think an answer to a question close to yours can be found in [1] and in the survey in the first section of [2].
There some necessary and sufficient conditions are given for a simplicial complex to be a topological manifold. In particular you can look at Theorem 3.5 of [1] or Theorem 1.12 of [2]. (They are basically the if and only if version of Adam's ...

11

I do not claim that I know what Adams meant; for all I know his definition of "product" may include products which are not bilinear. Take the following with a grain of salt.
What I suspect is the following. Adams says, just before your quote:
"In order to construct such products, G.W. Whitehead introduced the notion of a 'pairing of spectra'. A pairing of ...

11

As requested I am writing this as an answer.
No there are spaces with vanished homology which are not homotopy equivalent to finite CW-complexes.
For example if $G$ is an acyclic group, then the classifying space will be a space with vanishing homology. It is not contractible as $\pi_1 BG = G$.
Now a finite CW-complex will always have $\pi_1$ a finitely ...

answered Apr 26 '17 at 22:20

Chris Schommer-Pries

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11

First note that Hatcher's exercise says "where direct limits mean mapping telescopes", so he is defining $\underset{\rightarrow}{\lim}$ to mean the telescope. I disapprove of that quite strongly. The telescope is the same as the homotopy colimit, and the standard notation for that is $\underset{\rightarrow}{\text{holim}}$. Anyway, if we write $T$ for the ...

answered Oct 13 '18 at 10:55

Neil Strickland

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10

Let me start with the fact that, in one sense, it's true that Type 1 complexes are all that are "needed." That's true in the sense that complexes built from Type 2 and 3 cells have the $G$-homotopy type of $G$-CW complexes (meaning ones built from Type 1 cells). I'm not sure of a reference, but the proof is relatively straightforward: Cells of Types 2 and 3 ...

10

For any finite CW-complex $X$ and any basepoint $x \in X$, the fundamental group $\pi_1(X,x)$ is finitely presented. (This is a consequence of the Seifert-van Kampen theorem.) In particular, the group itself is a quotient of a finitely generated free group, and hence must be a countable set.
However, if $G$ is a Lie group of positive dimension, then the ...

answered Mar 14 '20 at 19:44

Tyler Lawson

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9

It seems to me that the stronger statement is true. If $f$ is only an isomorphism on homotopy in degrees $* \leq n$ then the homotopy fibre $F$ is $(n-1)$-connected and its Hurewicz map is an isomorphism in degree $n$. Considering the Serre spectral sequence of the fibration seqeuence $F \to \widetilde{X} \to \widetilde{Y}$, and using the fact that it ...

answered Jul 5 '15 at 19:43

Oscar Randal-Williams

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9

Take $X$ to be the Moore space for the group $\mathbb Z[\frac1p]$ in dimension $n$ (realized by a telescope of $n$-spheres mapping to each other vie the times-$p$ map), and take $h^*$ to be ordinary cohomology with coefficients in $\mathbb Z$.
The universal coefficient theorem gives you a sort exact sequence
$$
0\longrightarrow \mathrm{Ext}(H_n(X,\mathbb Z),...

answered Feb 10 '14 at 5:04

André Henriques

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9

Take $X=S^3$. Then no closed manifold of dimension at least 6 has the same homotopy type.

8

This text gives some details explaining that even with triangulations this may not work: a construction of a non-combinatorial triangulation (as double suspension of a homology sphere) and further references.

8

You can build them using a homology decomposition and read off the number of cells in each dimension from the homology groups.
For any simply-connected space, this will give you a construction by iterated cofiber sequences $M_n \to X(n-1) \to X(n)$ where $M_n$ is a Moore space and $X = \mathrm{colim} X(n)$.
If you construct $M_n$ efficiently and give $X(n)$...

8

Yes, there is in fact a handle decomposition with one 0-handle, two 1-handles, and a single 2-handle. In general, for a k-bridge knot, you'll get k 1-handles and $k-1$ 2-handles. The way that I learned this was to think of the knot (this works for any knot) as a solid wire that you are holding in a bathtub as the water fills in. At first you see a 0-handle, ...

8

I think isos up to dimension $n$ is enough to deduce $f_*:[A,X]\to[A,Y]$ onto when $dim(A)\leq n$. This gives a right inverse to $f$ which also induces isos up to $n$ and hence has a right inverse. So $f$ has a homotopy inverse.

8

If the Morse function $f$ is perfect, then, for any choice of metric $g$, the attaching maps cannot be homeomorphism. Indeed if the Morse function was perfect, then the boundary operator of the associated Morse-Smale complex is trivial. If the attaching map was a homeomorphism, then the boundary operator cannot be zero. This can be seen from Thm. 4....

answered Feb 23 '16 at 22:17

Liviu Nicolaescu

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8

For the smooth case:
Via Morse theory the claim is equivalent to having a Morse function with only one maximum, or only one minimum, or to have a handle decomposition with only one 0-handle.
Assume you have several 0-handles. By connectedness they have to be connected via 1-handles. Smale's handle cancellation says that a k-handle can be canceled against a ...

8

In general, no. Assuming, say, that the structure maps $SE_n\to E_{n+1}$ are inclusions, the correct statement is
$$
E^n(X) \cong [X,\lim_k \Omega^k E_{n+k}],
$$
and if $X$ is not compact that is not the same as $\lim_k [X,\Omega^k E_{n+k}] = \lim_k[S^kX,E_{n+k}]$. Peter May drilled this into me when I was a student (and after!). The general statement (for $...

8

Here is a picture from Topology and Groupoids
It is meant to show in (i) the Cayley graph of the presentation $\mathcal P$ of $G=S3$, $\{x,y:x^3,y^2,xyxy\}$. The Cayley graph is the $1$-skeleton of the universal cover of the cell complex $K(\mathcal P)$ of the presentation. The picture (ii) is a chosen tree in the Cayley graph. The diagram shows 3 ...

8

Here are some details which are related to Tyler's comment.
I recommend looking at the paper "Induced Fibrations and Cofibrations" by Tudor Ganea (1967). For connected based spaces $X$ and $Y$, there is a fibration up to homotopy
$$
\Sigma (\Omega X) \wedge (\Omega Y) \to X\vee Y \to X\times Y
$$
where the first map in the display is a kind of ...

8

More generally, suppose $n \le m$ are non-negative integers,
$X$ is a CW complex of dimension $\le n$, $M$ is a non-empty, closed $m$-manifold,
and $X$ and $M$ have the same homotopy type.
It is well known that a non-empty closed $m$-manifold has non-trivial mod 2 homology in degree $m$, whereas a CW complex of dimension $\le n$ has no homology above ...

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