Homotopy type and CW complexes

Let $X$ be a topological space and $f : S^{n-1} \to X$ a continuous function. Let $X_f = X \cup_f D^n$ be the quotient space formed from the disjoint union by identifying $x \in S^{n-1}$ with $f(x) \in X$.

Suppose that $h : X \to Y$ is a homotopy equivalence with homotopy inverse $g : Y \to X$; that is, $gh$ is homotopic to $1_X$ and $hg$ is homotopic to $1_Y$. Name the homotopies $H : X \times I \to X$, $H_t(x) = H(x,t)$, $H_0(x) = gh(x)$, $H_1(x) = x$, and $G : Y \times I \to Y$, $G_t(y) = G(y,t)$, $G_0(y) = hg(y)$, $G_1(y) = y$. Let $X_f = X \cup_f D^n$ (notation of Exercise 5.3.1) for a fixed attaching map $f : S^{n-1} \to X$, and similarly $Y_{hf}$, $X_{ghf}$.

Figure for Exercise 5.3.2.

The dunce hat is the space obtained from a single triangle (a $2$-simplex) by identifying all three of its edges together in the same direction (so that, reading around the boundary of the triangle, all three edges are labeled with the same letter $a$ traversed the same way).

A subcomplex $T$ of a connected $1$-skeleton $X^1$ which contains all of the vertices and deformation-retracts to any one of its $0$-cells (through a process of collapsing $1$-cells one edge at a time) is called a maximal tree in $X^1$. Such a maximal tree can always be shown to exist by induction on the number of $0$-cells: if there is only one $0$-cell, it is itself the maximal tree; otherwise, choosing a $0$-cell $e^0$ and a maximal subcomplex $K$ of $X^1$ not containing $e^0$ (with, say, $p$ path components, each of which has a maximal tree connecting its $0$-cells by the inductive hypothesis), adding a $1$-cell from $e^0$ to each path component of $K$ gives a maximal tree in $X^1$.

Use Exercise 5.3.2 and a maximal tree $T$ in $X$ to show that there is a homotopy-equivalent CW complex $Y$ with a single $0$-cell.

Consider the CW complex whose $1$-skeleton $X^1$ is pictured in Figure 5.10. Here $A, B, C$ denote $0$-cells and $a,b,c,d,e,f$ are $1$-cells. Suppose that there are two $2$-cells attached to $X^1$, which can be described in terms of the $1$-cells that are traversed by $adec$ and $dfb$.

Figure for Exercise 5.3.6.

Consider a handle decomposition of a surface. The $0$-handles deformation-retract to a collection of points. When we attach a $1$-handle, the composition of the attaching maps with the deformation retraction sends these intervals to one or two points. Use this idea to show that the union of the $0$-handles and $1$-handles is homotopy equivalent to a CW complex with $h_0$ $0$-cells and $h_1$ $1$-cells, where $h_i$ is the number of $i$-handles.

Show that if a surface (possibly with boundary) has a handle decomposition with $h_0$ $0$-handles, $h_1$ $1$-handles, and $h_2$ $2$-handles, then there is a corresponding CW complex with $h_0$ $0$-cells, $h_1$ $1$-cells, and $h_2$ $2$-cells to which it is homotopy equivalent.

Show that a connected surface (possibly with boundary) is homotopy equivalent to a CW complex with a single $0$-cell.

For Figure 5.11(a), give a CW decomposition and then get a homotopy-equivalent CW complex with a single $0$-cell. Use this to compute the fundamental group.

Figure for Exercises 5.3.10 and 5.3.11.

For Figure 5.11(b) (shown together with Figure 5.11(a) above, in Exercise 5.3.10), give a CW decomposition and then get a homotopy-equivalent CW complex with a single $0$-cell. Use this to compute the fundamental group.

Using the notation established in Exercise 5.1.4 (where $X$ is the CW complex $e^0_1 \cup e^0_2 \cup e^1_1 \cup e^1_2 \cup_f e^2_1 \cup_g e^2_2$, with each $2$-cell's boundary divided into an upper half arc $S^1_+$ and a lower half arc $S^1_-$, identified respectively with $e^1_1$ and $e^1_2$), consider the CW complex $Y$ where $Y^1 = X^1$ but $Y = X^1 \cup_f e^2_1$, and $\partial e^2$ is identified via sending $S^1_+$ to $e^1_1$ as before but sending $S^1_-$ to $e^1_1$ by reversing the direction. Geometrically, think of $e^1_1$ as $[-1,1]$ on the $x$-axis, with the map being a vertical projection. Identify geometrically what the quotient space will be. Show that $Y$ is homotopy equivalent, but not homeomorphic, to $S^2 \vee S^1$, with the homotopy equivalence coming from collapsing $e^1_1$ to a point.

Suppose $K$ is a CW complex and $L$ is a subcomplex of $K$; then $K$ is built from $L$ by attaching $k$-cells for various $k$, which allows arguments to reduce to the case of attaching a single cell. The homotopy extension theorem states: suppose $L \subset K$ is a subcomplex, $g : K \to Y$, and there is a homotopy of the restriction $g|_L : L \to Y$, that is, a map $H : L \times I \to Y$ with $H(x,0) = g(x)$. Then there is a homotopy $H' : K \times I \to Y$ with $H'(x,t) = H(x,t)$ for $x \in L$ and $H'(x,0) = g(x)$.

To see why this should be true, assume $K = L \cup_f D^k$, where $f : S^{k-1} \to K$ is continuous. The homotopy extension property asserts that a map defined on $L \times I \cup K \times \left\{ 0\right\}$ extends to a continuous map from $K \times I$; the key is to reduce to the case $L = S^{k-1}$, $K = D^k$.

Lemma. Consider $S^{k-1} \times I \cup D^k \times \left\{ 0\right\} \subset D^k \times I$. There is a deformation retraction of $D^k \times I$ onto $S^{k-1} \times I \cup D^k \times \left\{ 0\right\}$.

(Idea of the lemma's proof, illustrated for $k=2$: let $p = (0,2) \in \mathbb {R}^k \times \mathbb {R} = \mathbb {R}^{k+1}$ and consider rays from $p$. For $b \in B = S^{k-1} \times I \cup D^k \times \left\{ 0\right\}$, the ray from $p$ to $b$ meets $D^k \times I$ in an interval with first point on $D^k \times \left\{ 1\right\}$ and last point $b$; for a point $x \in D^k \times I$, moving $x$ along its own such interval to its endpoint $b \in B$ defines the deformation retraction $H$, and $h(x) = H(x,1)$ gives a continuous map $h : D^k \times I \to B$ with $h|_B = \operatorname {id}|_B$.)

Now suppose $g : K = L \cup_f D^k \to Y$ and there is a homotopy $H : L \times I \to Y$ with $H(x,0) = g(x)$ for $x \in L$. Consider the composition $J : S^{k-1} \times I \cup D^k \times \left\{ 0\right\} \to Y$ given by $J(u,t) = H(f(u), t)$ for $u \in S^{k-1}$, and $J(v, 0) = g(v)$ for $v \in D^k$. The lemma allows extending $J$ to $D^k \times I$ by composing with the deformation retraction.

By applying the lemma and using how $K \times I$ is formed as a quotient space, show that there is an extension $H' : K \times I \to Y$ of $H$.

Use an inductive argument, with induction over the number of cells of $K \setminus L$, to prove the homotopy extension theorem (stated in Exercise 5.3.13): if $L \subset K$ is a subcomplex, $g : K \to Y$, and $H : L \times I \to Y$ is a homotopy of $g|_L$ with $H(x,0) = g(x)$, then there is a homotopy $H' : K \times I \to Y$ extending $H$ with $H'(x,0) = g(x)$.