Difference between revisions of "Notes:CW-Complex"

Latest revision as of 01:00, 23 January 2017

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, $(X,\mathcal{ J })$, and a collection of (pairwise) disjoint open cells, $\{e_\alpha\}_{\alpha\in I}$, with $X\eq\bigcup_{\alpha\in I}e_\alpha$, such that:

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

Hatcher: Algebraic Topology - Appendix

A CW-Complex is constructed as follows:

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

   the cell $e^n_\alpha$ is the homeomorphic image of $D^n_\alpha - \partial D^n_\alpha$ under the quotient map

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

Algebraic Topology: An Intuitive Approach

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

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

  $\vdots$

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

Klein bottle example

I will almost certainly loose my paper notes.

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

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

Now we need an attaching map:

• $h_1:\partial X^{(1)}\rightarrow X^0$

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

• $\partial X^{(1)}\eq\{(a,-1),(a,1),(b,-1),(b,1),(c,-1),(c,1)\}$

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

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

Now we can form an adjunction space:

• $$X^1:\eq\frac{X^0\coprod X^{(1)} }{\langle x\sim h_1(x)\rangle}\eq X^0 \cup_{h_1} X^{(1)}$$
  • It is easy to see that $X^0\coprod X^{(1)}$ "looks like" 3 lines of length $2$ that are disconnected and a point, also disconnected.
  • We then identify the end points of those 3 lines with the point $v$
    • 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?

$2$-cells

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

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

The attaching map:

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

Sphere example

• $$X^0:\eq\coprod_{i\in \{u,v,w\} }i\eq\{(u,u),(v,v),(w,w)\}$$
• $$X^{(1)}:\eq\coprod_{i\in\{a,b,c\} }i\eq\bigcup_{i\in\{a,b,c\} }\left\{(i,p)\ \vert\ p\in\overline{\mathbb{B}^1}\right\}$$

Now we need an attaching map, $h_1:\partial X^{(1)}\rightarrow X^0$ that is continuous, where the boundary is considered with $$X^{(1)}\subseteq\coprod_{i\in\{a,b,c\} }\mathbb{R}$$

• $\partial X^{(1)}\eq\{(a,-1),(a,1),(b,-1),(b,1),(c,-1),(c,1)\}$

From the diagram we define:

• $h_1:(a,-1)\mapsto (w,w)$
• $h_1:(a,1)\mapsto (v,v)$
• $h_1:(b,-1)\mapsto (v,v)$
• $h_1:(b,1)\mapsto (u,u)$
• $h_1:(c,-1)\mapsto (v,v)$
• $h_1:(c,1)\mapsto (v,v)$

Considering $\partial X^{(1)}\subseteq X^{(1)}$ as a subspace and $X^0$ with the discrete topology things look continuous.... I mean the pre-image of $\{(v,v)\}$ say has a few "components" but yeah there's an open set in $X^{(1)}$ which intersected with $\partial X^{(1)}$ is that set surely. Check this later but looking good.

• Define $$X^1:\eq X^0\cup_{h_1} X^{(1)}:\eq\frac{X^0\coprod X^{(1)} }{\langle x\sim h_1(x) \rangle}$$

Notes

I drew some pictures of the triangles, $A$ and $B$ joined up as needed and they do indeed attach to this $1$-skeleton, to form something homeomorphic to the sphere. So looking good so far!

Notes

1. Jump up ↑ $\mathbb{B}^m\eq\text{Int}\left(\overline{\mathbb{B}^m}\right)$
2. Jump up ↑ Into means nothing special, all functions map the domain into the co-domain, it is a common first-year mistake to look at the association of "onto" with "surjection" and associate into with "injection" - I mention this here to record Munkres' exact phrasing

References