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16 votes
Accepted

How to construct a surface of genus $g$ by identifying sides of a $4g$-gon?

Since you asked for an explanation for 2 and 3 (and not 1) I'll assume you understand the basic idea of identifications, and will try to walk you through the process of identifying. Excuse the poor ...
Noah Caplinger's user avatar
12 votes
Accepted

CW complex structure of geometric realization

Here's a sketch of the proof why realizations of simplicial sets are CW complexes. Roughly speaking, a CW complex is a space built by one-by-one gluing in new simplices along their boundaries. If $...
Eric Wofsey's user avatar
10 votes
Accepted

Specific examples of Eilenberg-Maclane spaces?

For $K(G,1)$ spaces there are some geometric methods. The key thing about the $K(G,1)$ property is that a connected CW complex possesses that property if and only if the universal covering space of ...
Lee Mosher's user avatar
  • 122k
10 votes
Accepted

What manifolds $M$ have a $CW-$structure so that the $n-$skeleton, $M_n$, is a manifold for all $n$ aswell?

I believe you gave a homological classification. Let's work with $\mathbb{Z}/2$ coefficients; then Poincare duality tells us that if our lowest homology is in degree $k$, to obtain a manifold from $M^...
Connor Malin's user avatar
  • 11.7k
9 votes
Accepted

An infinite dimensional CW complex always has infinitely many non-trivial homology groups?

A counterexample is $S^\infty$, the union of $S^n$ for all $n\geq 0$. This is a complex with two cells in each dimension, so it is infinite dimensional, but it is contractible.
Matt Samuel's user avatar
  • 58.3k
9 votes
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Is a closed embedding of CW-complexes a cofibration?

This is only a partial answer: If $X$ is a locally finite CW-complex and $A \subset X$ is a closed subspace which is also a CW-complex (but not necessarily a subcomplex), then $i : A \hookrightarrow ...
Paul Frost's user avatar
  • 77.4k
9 votes

Does every finite $CW$ complex have the homotopy type of a smooth manifold? How about infinite $CW$ complexes?

For finite complexes - yes, for infinite complexes in general - no. Why yes: First, each finite CW complex $W$ is homotopy-equivalent to a finite simplicial complex $C$. (This should be in Hatcher's ...
Moishe Kohan's user avatar
  • 99.2k
8 votes

How do you prove a CW complex is locally path connected

You can use the following two general topology facts: A disjoint union of locally path-connected spaces is locally path-connected. A quotient of a locally path-connected space is locally path-...
Jeremy Brazas's user avatar
8 votes
Accepted

Is every space homology equivalent to an Eilenberg–MacLane space?

No. For instance, take $X=S^2$ and suppose $f:X\to K(\pi,n)$ is a homology equivalence. Since $f$ is non-nullhomotopic, $\pi_2(K(\pi,n))\neq 0$, so $n=2$. But then $X$ and $K(\pi,n)$ are simply ...
Eric Wofsey's user avatar
8 votes
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Definition of CW complex: weak topology vs quotient

The two topologies are exactly the same, as you can easily prove by induction on $n$. Indeed, supposing the two topologies are the same on the $(n-1)$-skeleton $X^{n-1}$, the quotient topology on $X^...
Eric Wofsey's user avatar
8 votes

Compact subset of CW complex

There are some important theorems about compact sets that can simplify your work: If $A$ is closed, $A\subseteq B$, and $B$ is compact, then $A$ is compact If $A$ is compact and $f:A\to B$ is ...
Makenzie's user avatar
  • 288
8 votes

CW structure of the universal cover

When you have a covering map $p:\tilde{X}\to X$ with a CW structure on $X$, you can use the homotopy lifting property to lift each cell of $X$ to a collection of cells of $\tilde{X}$, since you can ...
Kyle Miller's user avatar
  • 19.5k
8 votes
Accepted

Wedge sum of spheres is the quotient $X^n/X^{n-1}$

I think you only need to write another pushout square saying that the $n$-sphere is the quotient of the $n$-ball by the $(n-1)$-sphere : $$\begin{CD} S^{n-1} @>>> \{*\}\\ @VVV @VVV\\ D^n @&...
J. Darné's user avatar
  • 1,255
8 votes

Wedge sum of spheres is the quotient $X^n/X^{n-1}$

Alternate answer : $X_n/X_{n-1}$ is a CW-complex (quotient of a CW-complex by a subcomplex) with one 0-cell and only n-cells, and this must be a wedge of n-spheres.
J. Darné's user avatar
  • 1,255
8 votes
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Attaching 1-cells to CW-complex affects homotopy groups?

This is not true. Consider attaching a one-cell to $S^2$ via a constant attaching map. The resulting space is $S^2\vee S^1$. The universal cover of $S^2\vee S^1$ is $\mathbb{R}$ with an $S^2$ wedged ...
Michael Albanese's user avatar
8 votes
Accepted

Confusion about topology on CW complex: weak or final?

We have inclusions $i_k : X_k \to X$. The skeleta $X_i$ have topologies, and with respect to these topologies each $X_i$ is a subspace of $X_{i+1}$. The CW-complex $X$ is then endowed with the final ...
Paul Frost's user avatar
  • 77.4k
8 votes
Accepted

Showing Hawaiian earrings are not CW complexes

Assume $X$ is a CW-complex. As you say, it must be a finite CW-complex because $X$ is compact. Let $e_1, \ldots, e_n$ be its open cells. Note that all $e_i$ are path connected. If a space $Z$ is the ...
Paul Frost's user avatar
  • 77.4k
8 votes
Accepted

Fundamental group of CW complexes

You can't attach just a single 3-cell to fill in the torus, since the inside of a torus is not homeomorphic to a 3-dimensional ball. Instead you should first fill in the "inside" of the ...
Sergey Guminov's user avatar
8 votes
Accepted

Want a simple construction of a relative CW complex (something like a torus) analogous to regular CW Complex constructions.

First let me correct your intuition of a CW complex, and then I'll explain how to extend the corrected intuition to a relative CW complex. For a CW complex $X$ you start from nothing. Next, you add ...
Lee Mosher's user avatar
  • 122k
7 votes
Accepted

Fundamental group of a CW complex only depends on its $2$-skeleton

First of all, $\pi_n(X)$ is the set of pointed homotopy classes of maps; the basepoint matters. There's also a problem in your argument. When you replace $H : S^n\times [0, 1] \to X$ by the cellular ...
Michael Albanese's user avatar
7 votes

integral cohomology ring of real projective space

The ring structure of the integral cohomology can be determined by the ring structure of the mod 2 cohomology. Indeed, the reduction mod $2$ map $H^*(\mathbb{R}P^\infty;\mathbb{Z})\to H^*(\mathbb{R}P^...
Eric Wofsey's user avatar
7 votes

Universal covers of CW complexes and $n$-skeletons

This is not true. Consider $X = S^2$ with a single $0$-cell, a single $1$-cell and two $2$-cells (the hemispheres). As $X$ is simply connected, it is its own universal cover, so $U = X$. However, $X^1 ...
Michael Albanese's user avatar
7 votes
Accepted

Finding a CW decomposition of $X$ having a prescribed point $x\in X$ as a 0-cell.

Here's one simple way to do it. The proof is by induction on the dimension of the cell containing $x$. Let $e$ be the cell containing $x$, let $n$ be the dimension of $e$, and let $\chi : B^n \to X$ ...
Lee Mosher's user avatar
  • 122k
7 votes
Accepted

Isomorphism that is not a bijection in underlying sets?

The condition for an arrow $g:y\to x$ in a category $\mathbf C$ to be the inverse of $f:x\to y$ is equational: $fg=\mathrm{id}_y$ and $gf=\mathrm{id}_x$. Hence it will be preserved by any functor, by ...
Pece's user avatar
  • 11.6k
7 votes
Accepted

Cell decomposition for $\mathbb{C}P^n$ that has $\mathbb{R}P^n$ as a subcomplex?

I believe the following works. Consider a cell decomposition made of subsets of $\mathbb{CP}^n$ of the form $$\{ (x_1+s_1 y_1 i : x_2+s_2 y_2 i : \ldots : x_k+s_k y_k i : 1 : 0 : \ldots : 0) \mid x_i ...
pregunton's user avatar
  • 5,811
7 votes
Accepted

Covering $\Bbb RP^\text{odd}\longrightarrow X$, what can be said about $X$?

If $n = 1$, then $\mathbb{RP}^1 = S^1$ which only covers itself. If $n > 1$, the manifold $\mathbb{RP}^{2n-1}$ covers infinitely many manifolds which are pairwise non-homotopy equivalent. To see ...
Michael Albanese's user avatar
7 votes

Showing Hawaiian earrings are not CW complexes

Here's another way to show it. As you say, if $X$ were a CW-complex, then it would have to be a finite CW-complex since it is compact. Since every finite CW-complex has finitely generated homology, ...
Eric Wofsey's user avatar
7 votes
Accepted

Gluing of Möbius strips

I am not sure why the comment is upvoted: your meaning is clear, take $M_1 = \cdots = M_n = M$ all copies of the Mobius band; there are inclusion maps $f_i: S^1 \to M_i$ for all $i$ parameterizing the ...
j.q's user avatar
  • 86
7 votes
Accepted

Examples of isomorphic group presentations with homotopy NONequivalent complexes

Take $$G=\langle x,y\mid x^2=y^3\rangle.$$ This is the standard presentation for the fundamental group of the trefoil knot. Of note is the fact that this group is Hopfian and non polycyclic-by-finite. ...
Tyrone's user avatar
  • 16.3k
7 votes
Accepted

On a cw complex structure on the $2$- sphere

$\bullet$ For $v=1$ we have $f=1+e$. So, consider a wedge of $e$-many circles on $\Bbb S^2$ with wedge point as $v$, and then fill up the complement of this wedge by $(1+e)$ many $2$-cells. $\bullet$ ...
Sumanta's user avatar
  • 9,614

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