Notes:CW-Complex

Overview

I get CW-Complexes in terms of what they are but no so much in terms of a formal definition. This page details my research.

Munkres: Elements of Algebraic Topology

A CW-Complex is a topological space, [ilmath](X,\mathcal{ J })[/ilmath], and a collection of (pairwise) disjoint open cells, [ilmath]\{e_\alpha\}_{\alpha\in I} [/ilmath], with [ilmath]X\eq\bigcup_{\alpha\in I}e_\alpha[/ilmath], such that:

1. [ilmath](X,\mathcal{ J })[/ilmath] is a Hausdorff space
2. For each open [ilmath]m[/ilmath]-cell, [ilmath]e_\alpha[/ilmath], there exists a continuous map, [ilmath]f_\alpha:\overline{\mathbb{B}^m}\rightarrow X[/ilmath] such that:
   1. [ilmath]f_\alpha[/ilmath] maps [ilmath]\mathbb{B}^m[/ilmath] homeomorphically onto [ilmath]e_\alpha[/ilmath] and
   2. [ilmath]f_\alpha\left(\partial\left(\overline{\mathbb{B}^m}\right)\right)[/ilmath] "into" a finite union of open cells, each of dimension (strictly) less than [ilmath]m[/ilmath]
3. A set [ilmath]A\in\mathcal{P}(X)[/ilmath] is closed in [ilmath](X,\mathcal{ J })[/ilmath] if and only if [ilmath]\forall\alpha\in I[A\cap\overline{e_\alpha}\text{ is closed in }\overline{e_\alpha}][/ilmath]

Hatcher: Algebraic Topology - Appendix

A CW-Complex is constructed as follows:

1. Start with [ilmath]X^0[/ilmath], the [ilmath]0[/ilmath]-cells of [ilmath]X[/ilmath]
2. Inductively, form the [ilmath]n[/ilmath]-skeleton, [ilmath]X^n[/ilmath], from [ilmath]X^{n-1} [/ilmath] by attaching [ilmath]n[/ilmath]-cells, [ilmath]e^n_\alpha[/ilmath] via maps, [ilmath]\varphi_\alpha:\mathbb{S}^{n-1}\rightarrow X^{n-1} [/ilmath].
   • This means that [ilmath]X^n[/ilmath] is the quotient space of [ilmath]X^{n-1}\coprod_\alpha D_\alpha^n[/ilmath] under the identifications:
     • [ilmath]x\sim \varphi_\alpha(x) [/ilmath] for [ilmath]x\in \partial D^n_\alpha[/ilmath]

   the cell [ilmath]e^n_\alpha[/ilmath] is the homeomorphic image of [ilmath]D^n_\alpha - \partial D^n_\alpha[/ilmath] under the quotient map

3. [ilmath]X\eq\bigcup_{n\in\mathbb{N} }X^n[/ilmath] with the weak topology.
   • A set [ilmath]A\in\mathcal{P}(X)[/ilmath] is open if and only if [ilmath]\forall n\in\mathbb{N}[A\cap X^n\text{ is open in }X^n][/ilmath]

Algebraic Topology: An Intuitive Approach

We build an "attaching space" called a (finite) cell complex inductively from the following recipe:

• Ingredients:
  • [ilmath]k_0[/ilmath] closed [ilmath]0[/ilmath]-cells, [ilmath]\bar{e}_1^0,\ldots,\bar{e}_{k_0}^0[/ilmath]
  • [ilmath]k_1[/ilmath] closed [ilmath]1[/ilmath]-cells, [ilmath]\bar{e}_1^1,\ldots,\bar{e}_{k_1}^1[/ilmath]

  [ilmath]\vdots[/ilmath]

  • [ilmath]k_n[/ilmath] closed [ilmath]n[/ilmath]-cells, [ilmath]\bar{e}_1^n,\ldots,\bar{e}_{k_n}^n[/ilmath]
• Construction:
  • [ilmath]X^0:\eq\coprod_{i\eq 1}^{k_0}\bar{e}_i^0[/ilmath]
  • Set [ilmath]X^{(1)}:\eq\coprod_{i\eq 1}^{k_1}\bar{e}_i^1[/ilmath]
  • Define [ilmath]\partial X^{(1)}:\eq\coprod_{i\eq 1}^{k_1}\partial\bar{e}_i^1[/ilmath] (where we consider each [ilmath]\bar{e}^1_i[/ilmath] as a subspace of [ilmath]\mathbb{R} [/ilmath]
    • We could consider [ilmath]X^{(1)} [/ilmath] as a subset of [ilmath]\coprod_{i\eq 1}^{k_1}\mathbb{R} [/ilmath] for boundary purposes.
  • We must now construct an attaching map: [ilmath]h_1:\partial X^{(1)}\rightarrow X^0[/ilmath] to attach [ilmath]X^{(1)} [/ilmath] to [ilmath]X^0[/ilmath]
  • Define: [math]X^1:\eq X^0\cup_{h_1}X^{(1)} :\eq\frac{X^0\coprod X^{(1)} }{\langle x\sim h_1(x)\rangle} [/math]
  • Set [ilmath]X^{(2)}:\eq\coprod_{i\eq 1}^{k_2}\bar{e}_i^2[/ilmath]
  • Specify an attaching map, [ilmath]h_2:\partial X^{(2)}\rightarrow X^1 [/ilmath]
  • And so on until we obtain [ilmath]X^n[/ilmath], then let [ilmath]X:\eq X^n[/ilmath] - this final product is an [ilmath]n[/ilmath]-dimensional cell complex.
    • For each [ilmath]q\in\{0,\ldots,n\} [/ilmath] we call [ilmath]X^q[/ilmath] a [ilmath]q[/ilmath]-skeleton of [ilmath]X[/ilmath].
    • For a cell complex [ilmath]X[/ilmath] we get 3 maps:
      1. For each [ilmath]q[/ilmath]-cell, [ilmath]e^q_j[/ilmath] we have the canonical inclusion map: [ilmath]i_{q,j}:\bar{e}^q_j\rightarrow X^{(q)} [/ilmath]
      2. The canonical quotient map: [ilmath]\pi:X^{(q)}\rightarrow X^q[/ilmath] Caveat:what on earth.... - oh okay, might be canonical injection followed by projection of the quotient
      3. The inclusion map [ilmath]i:X^q\rightarrow X[/ilmath]
    • The composition of these maps: [ilmath]\phi^q_j:\eq i\circ\pi\circ i_{q,j}:\bar{e}^q_j\rightarrow X[/ilmath]
      • Called the characteristic map of the [ilmath]e^q_j[/ilmath] cell.
        • The restriction of the characteristic map to the boundary, [ilmath]\partial\bar{e}^q_j[/ilmath] should agree with the restriction of the attaching map [ilmath]h_q:\partial X^{(q)}\rightarrow X^{q-1} [/ilmath] to [ilmath]\partial\bar{e}^q_j[/ilmath]

Klein bottle example

[ilmath]\xymatrix{ v \bullet \ar@{<-}@<.65ex>[d]_a \ar@{<-}[rr]^b & & \bullet v \ar@{<-}@<-.65ex>[d]^a \ar[dll]_c \\ v \bullet \ar[rr]_b & & \bullet v}[/ilmath] With [ilmath]2[/ilmath]-cells [ilmath]A[/ilmath] and [ilmath]B[/ilmath]: [ilmath]A[/ilmath] oriented [ilmath](-a)+(-c)+b[/ilmath] and [ilmath]B[/ilmath] oriented [ilmath]c+b+a[/ilmath]
A CW-complex for the Klein bottle

I will almost certainly loose my paper notes.

• [ilmath]X^0:\eq\{(v,v)\} [/ilmath]
• [math]X^{(1)}:\eq\coprod_{i\in\{a,b,c\} }\overline{\mathbb{B}^1}\eq\bigcup_{j\in\{a,b,c\} }\big\{(j,p)\ \vert\ p\in \overline{\mathbb{B}^1}\big\} \eq\{\underbrace{(a,-1),\ldots,(a,1)}_{a},\underbrace{(b,-1),\ldots,(b,1)}_{b},\underbrace{(c,-1),\ldots,(c,1)}_c\} [/math]

At this point [ilmath]X^0[/ilmath] "looks like" a point and [ilmath]X^{(1)} [/ilmath] "looks like" 3 separate straight lines.

Now we need an attaching map:

• [ilmath]h_1:\partial X^{(1)}\rightarrow X^0[/ilmath]

The boundary is with [ilmath]X^{(1)} [/ilmath] considered as a subset of [ilmath]\coprod_{i\in\{a,b,c\} }\mathbb{R} [/ilmath], so in this case:

• [ilmath]\partial X^{(1)}\eq\{(a,-1),(a,1),(b,-1),(b,1),(c,-1),(c,1)\} [/ilmath]

Of course [ilmath]h_1[/ilmath] maps every point in the boundary to [ilmath](v,v)[/ilmath] - the only vertex there is.

Note that [ilmath]h_1[/ilmath] is continuous, as [ilmath]h_1^{-1}(\emptyset)\eq\emptyset[/ilmath] and [ilmath]h_1^{-1}(\{(v,v)\})\eq\partial X^{(1)} [/ilmath] (we consider the codomain with the subspace topology, [ilmath]X^0[/ilmath] really can only have the trivial topology as a topology.

Now we can form an adjunction space:

• [math]X^1:\eq\frac{X^0\coprod X^{(1)} }{\langle x\sim h_1(x)\rangle}\eq X^0 \cup_{h_1} X^{(1)} [/math]
  • It is easy to see that [ilmath]X^0\coprod X^{(1)} [/ilmath] "looks like" 3 lines of length [ilmath]2[/ilmath] that are disconnected and a point, also disconnected.
  • We then identify the end points of those 3 lines with the point [ilmath]v[/ilmath]
    • Caveat:I think there are a few ways to do this ultimately the space "looks like" a point with 3 loops coming off it. Like a clover shape. But how do we preserve orientation? Does it matter? What do the different directions of each loop (and as the image of which of the 3 lines) correspond to?

[ilmath]2[/ilmath]-cells

This is slightly trickier. Note: it doesn't matter if we consider a [ilmath]\overline{\mathbb{B}^2} [/ilmath] as a "disk" or a "square", as these are homeomorphic.

• [math]X^{(2)}:\eq A\coprod B[/math] which is the set that contains [ilmath](i,(x,y))[/ilmath] given [ilmath]i\eq A[/ilmath] or [ilmath]i\eq B[/ilmath] and [ilmath](x,y)\in\overline{\mathbb{B}^2} [/ilmath].

The attaching map:

• [ilmath]h_2:\partial X^{(2)}\rightarrow X^1[/ilmath] - where we consider [ilmath]\partial X^{(2)} [/ilmath] as a subset of [ilmath]\mathbb{R}^2\coprod\mathbb{R}^2[/ilmath], meaning:
  • [ilmath]\partial X^{(2)}\eq\left\{(i,(x,y))\ \vert\ i\in\{A,B\}\wedge (x,y)\in\mathbb{S}^1\right\} [/ilmath] - [ilmath]\mathbb{S}^1[/ilmath] is a circle centred at the origin of radius 1.