## CW Approximation

A weak homotopy equivalence is a map $f:X\to Y$ that induces isomorphisms $\pi_n(X,x_0)\to\pi_n(Y,f(x_o))$ for all $n\geq 0$ and all choices of basepoint $x_0$.

In other words, Whitehead’s theorem says that a weak homotopy equivalence between CW complexes is a homotopy equivalence. Just to recap, a map $f:X\to Y$ is said to be a homotopy equivalence if there exists a map $g:Y\to X$ such that $fg\cong id_Y$ and $gf\cong id_X$. The spaces $X$ and $Y$ are called homotopy equivalent.

It turns out that for any space $X$ there exists a CW complex $Z$ and a weak homotopy equivalence $f:Z\to X$. This map $f:Z\to X$ is called a CW approximation to $X$.

## Excision for Homotopy Groups

According to Hatcher (Chapter 4.2), the main difficulty of computing homotopy groups (versus homology groups) is the failure of the excision property. However, under certain conditions, excision does hold for homotopy groups:

Theorem (4.23): Let $X$ be a CW complex decomposed as the union of subcomplexes $A$ and $B$ with nonempty connected intersection $C=A\cap B$. If $(A,C)$ is m-connected and $(B,C)$ is n-connected, $m,n\geq 0$, then the map $\pi_i(A,C)\to\pi_i(X,B)$ induced by inclusion is an isomorphism for $i and a surjection for $i=m+n$.

## Miscellaneous Definitions

Suspension: Let $X$ be a space. The suspension $SX$ is the quotient of $X\times I$ obtained by collapsing $X\times\{0\}$ to one point and $X\times\{1\}$ to another point.

The definition of suspension is similar to that of the cone in the following way. The cone $CX$ is the union of all line segments joining points of $X$ to one external vertex. The suspension $SX$ is the union of all line segments joining points of $X$ to two external vertices.

The classical example is $X=S^n$, when $SX=S^{n+1}$ with the two “suspension points” at the north and south poles of $S^{n+1}$, the points $(0,\dots,0,\pm 1)$.

Here are some graphical sketches of the case where $X$ is the 0-sphere and the 1 sphere respectively.

## Cellular Approximation for Pairs

(This is Example 4.11 in Hatcher’s book).

Cellular Approximation for Pairs: Every map $f:(X,A)\to (Y,B)$ of CW pairs can be deformed through maps $(X,A)\to (Y,B)$ to a cellular map $g:(X,A)\to (Y,B)$.

What “map of CW pairs” mean, is that $f$ is a map from $X$ to $Y$, and the image of $A\subseteq X$ under $f$ is contained in $B$CW pair $(X,A)$ means that $X$ is a cell complex, and $A$ is a subcomplex.

First, we use the ordinary Cellular Approximation Theorem to deform the restriction $f:A\to B$ to be cellular. We then use the Homotopy Extension Property to extend this to a homotopy of $f$ on all of $X$. Then, use Cellular Approximation Theorem again to deform the resulting map to be cellular staying stationary on $A$.

We use this to prove a corollary: A CW pair $(X,A)$ is n-connected if all the cells in $X-A$ have dimension greater than $n$. In particular the pair $(X,X^n)$ is n-connected, hence the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i and a surjection on $\pi_n$.

First we note that being n-connected means that the space is non-empty, path-connected, and the first n homotopy groups are trivial, i.e. $\pi_i(X)\cong 0$ for $1\leq i\leq n$.

Proof: First, we apply cellular approximation to maps $(D^i,\partial D^i)\to (X,A)$ with $i\leq n$, thus the map is homotopic to a cellular map of pairs $g$. Since all the cells in $X-A$ have dimension greater than $n$, the n-skeleton of $X$ must be inside $A$. Therefore $g$ is homotopic to a map whose image is in $A$, and thus it is 0 in the relative homotopy group $\pi_i(X,A)$. This proves that the CW pair $(X,A)$ is n-connected. Note that 0-connected means path-connected.

Consider the long exact sequence of the pair $(X,X^n)$:

$\dots\to\pi_n(X^n,x_0)\xrightarrow{i_*}\pi_n(X,x_0)\xrightarrow{j_*}\pi_n(X,X^n,x_0)\xrightarrow{\partial}\pi_{n-1}(X^n,x_0)\to\dots\to\pi_0(X,x_0)$

Since it is an exact sequence, the image of any map equals the kernel of the next. Thus, $\text{Im}(i_*)=\ker j_*=\pi_n(X,x_0)$ (since $\pi_n(X,X^n,x_0)=0$). Thus $i_*$ is surjective. Since $\pi_n(X,X^n,x_0)=0$, the later terms in the long exact sequence are also 0, thus, the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i, since the first n homotopy groups all vanish.

## Cellular Approximation Theorem and Homotopy Groups of Spheres

First we will state another theorem, Whitehead’s Theorem: If a map $f:X\to Y$ between connected CW complexes induces isomorphisms $f_*:\pi_n(X)\to\pi_n(Y)$ for all $n$, then $f$ is a homotopy equivalence. If $f$ is the inclusion of a subcomplex $X\to Y$, we have an even stronger conclusion: $X$ is a deformation retract of $Y$.

The main theorem discussed in this post is the Cellular Approximation Theorem: Every map $f:X\to Y$ of CW complexes is homotopic to a cellular map. If $f$ is already cellular on a subcomplex $A\subset X$, the homotopy may be taken to be stationary on $A$. This theorem can be viewed as the CW complex analogue of the Simplicial Approximation Theorem.

Corollary: If $n, then $\pi_n(S^k)=0$.

Proof: Consider $S^n$ and $S^k$ with their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and one k-cell for $S^k$. Let $[f]\in\pi_n(S^k)$, where $f:S^n\to S^k$ is a base-point preserving map. By the Cellular Approximation Theorem, $f$ is homotopic to a cellular map $g$, where cells map to cells of same or lower dimension.

Since $n, the n-cell $S^n$ can only map to the 0-cell in $S^k$. The 0-cell in $S^n$ (the basepoint) is also mapped to the 0-cell in $S^k$. Thus $g$ is the constant map, hence $\pi_n(S^k)=0$.

## Covering space projection induces isomorphisms

Proposition 4.1 (from Hatcher): A covering space projection $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ induces isomorphisms $p_*:\pi_n(\tilde{X},\tilde{x}_0)\to\pi_n(X,x_0)$ for all $n\geq 2$.

We will elaborate more on this proposition in this blog post. Basically, we will need to show that $p_*$ is a homomorphism and also bijective (surjective and injective).

Homomorphism

$p_*([f]):=[pf]$

$p_*([f]+[g])=[p(f+g)]$

$p(f+g)(s_1,s_2,\dots,s_n)=\begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\ pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1] \end{cases}$

$p_*[f]+p_*[g]=[pf]+[pg]$

$(pf+pg)(s_1,s_2,\dots,s_n)= \begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\ pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1] \end{cases}$, which we can see is the same.

Thus, $p_*$ is a homomorphism.

Surjective

For surjectivity, we need to use a certain Proposition 1.33: Suppose given a covering space $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ and a map $f:(Y,y_0)\to (X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\tilde{f}:(Y,y_0)\to (\tilde{X},\tilde{x}_0)$ of $f$ exists iff $f_*(\pi_1(Y,y_0))\subset p_*(\pi_1(\tilde{X},\tilde{x}_0))$.

Let $[f]\in\pi_n(X,x_0)$, where $f:(S_n,s_0)\to(X,x_0), n\geq 2$. Since $S^n$ is simply connected for $n\geq 2$, $\pi_1(S_n,s_0)=0$. Thus $f_*(\pi_1(S_n,s_0))=0\subset p_*(\pi_1(\tilde{X},\tilde{x}_0))$. By Proposition 1.33, a lift $\tilde{f}:(S_n,s_0)\to (\tilde{X},\tilde{x}_0)$ of $f$ exists, where $p\tilde{f}=f$.

i.e. we have $\boxed{p_*[\tilde{f}]=[p\tilde{f}]=[f]}$. Hence $p_*$ is surjective.

Injective

Let $[\tilde{f}_0]\in\ker p_*$, where $\tilde{f}_0:I^n\to \tilde{X}$ with a homotopy $f_t:I^n\to X$ of $f_0=p\tilde{f}_0$ to the trivial loop $f_1$.

By the covering homotopy property (homotopy lifting property), there exists a unique homotopy $\tilde{f}_t:I^n\to \tilde{X}$ of $\tilde{f}_0$ that lifts $f_t$, i.e. $p\tilde{f}_t=f_t$. There is a lifted homotopy of loops $\tilde{f}_t$ starting with $\tilde{f}_0$ and ending with a constant loop. Hence $[\tilde{f}_0]=0$ in $\pi_n(\tilde{X},\tilde{x}_0)$ and thus $p_*$ is injective.

## The Groupoid Properties of Operation * on Path-homotopy Classes (Proof)

Theorem: The operation * has the following properties:

(1) (Associativity) [f]*([g]*[h])=([f]*[g])*[h], i.e. it doesn’t matter where we place the brackets.

(2) (Right and left identities) Given $x\in X$, let $e_x$ denote the constant path $e_x: I\to X$ mpping all of I to the point x. If f is a path in X from $x_0$ to $x_1$, then $[f]*[e_{x_1}]=[f]$ and $[e_{x_0}]*[f]=[f]$.

(3) (Inverse) Given the path f in X from $x_0$ to $x_1$, let $\bar{f}$ be the path defined by $\bar{f}=f(1-s)$. $\bar{f}$ is called the reverse of f. Then, $[f]*[\bar{f}]=[e_{x_0}]$ and $[\bar{f}]*[f]=[e_{x_1}]$.

We will prove the above statements, of which (1) Associativity is actually the trickiest.

Proof:

We shall prove two elementary lemmas first. (This part is not proved in the book by Munkres).

Lemma 1: If $k: X\to Y$ is a continuous map, and if F is a path homotopy in X between the paths f and f’, then $k\circ F$ is a path homotopy in Y between the paths $k\circ f$ and $k\circ f'$.

Proof of Lemma 1: Since F is a path homotopy in X between paths f and f’, we have by definition that F(s,0)=f(s), F(s,1)=f'(s), F(0,t)=x_0, F(1,t)=x_1.

Then, k F(s,0)=kf(s), kF(s,1)=kf'(s), kF(0,t)=k(x_0), kF(1,t)=k(x_1). Since kF is continuous (composition of two continuous functions), kF is inded a path homotopy in Y between he paths kf and kf’.

Lemma 2: If $k:X\to Y$ is a continuous map and if f and g are paths in X with f(1)=g(0), then

$k\circ (f*g)=(k\circ f)*(k \circ g)$

Proof of Lemma 2:

$k(f*g)(s)=kh(s)$, where h=f*g as defined previously.

$(kf)*(kg)(s)=kh(s)$.

We will first verify property (2) on Right and Left Identities. Let $e_{x_0}$ denote the constant path in I at 0, and we let $i: I\to I$ denote the identity map, which is a path in I from 0 to 1. Then $e_0 * i$ is also a path in I from 0 to 1.

Because I is convex, there is a path homotopy G in I between i and $e_0 *i$ (Straight-line homotopy) Then $f\circ G$ is a path homotopy in X between the paths $f\circ i=f$ and $f\circ (e_0 *i)$ (Lemma 1). Furthermore by Lemma 2, $f\circ (e_0 *i) = (f \circ e_0) * (f \circ i)$ which is equivalent to $e_{x_0} *f$.

A similar argument, using the fact that if $e_1$ denotes the constant path at 1, then $i*e_i$ is path homotopic in I to the path i, shows that $[f]*[e_{x_1}]=[f]$.

To prove (3) (Inverse), we note that the reverse of i is $\bar{i}(s)=1-s$. Then $i*\bar{i}$ is a path in I beginning and ending at 0. The constant path $e_0$ is also beginning and ending at 0. Again, because I is convex, there is a path homotopy H in I between $e_0$ and $i*\bar{i}$ (straight-line homotopy). Then, using lemma 1 and 2, $f\circ H$ is path homotopy between $f\circ e_0=e_{x_0}$ and $f\circ (i*\bar{i})=(f\circ i)*(f\circ\bar{i})=f*\bar{f}$. Very similarly, we can use the fact that $\bar{i}*i$ is path homotopic  in I to $e_1$ to show that $[\bar{f}]*[f]=[e_{x_1}]$.

We will continue the proof of associativity (which is longer) in the next blog post.

## The Groupoid Properties of * on Path-homotopy Classes

This is one of the first instances where algebra starts to appear in Topology. We will continue our discussion of material found in Topology (2nd Economy Edition) by James R. Munkres.

First, we need to define the binary operation *, that will later make * satisfy properties that are very similar to axioms for a group.

Definition: If f is a path in X from $x_0$ to $x_1$, and if g is a path in X from $x_1$ to $x_2$, we define the product f*g of f and g to be he path h given by the equations

$h(s)=\begin{cases}f(2s) &\text{for }s\in [0,\frac{1}{2}], \\ g(2s-1)& \text{for }s\in[\frac{1}{2}, 1]\end{cases}$

Well-defined: The function h is well-defined, at s=1/2, f(1)=x_1, g(0)=x_1.

Continuity: h is also continuous by the pasting lemma.

h is a path in X from x_0 to x_2. We think of h as the path whose first half is the path f and whose second half is the path g.

We will verify that the product operation on paths induces a well-defined operation on path-homotopy classes, defined by the equation $[f]*[g]=[f*g]$

Let F be a path homotopy between f and f’, and let G be a path homotopy between g and g’.

i.e. we have F(s,0)=f(s), F(s,1)=f'(s)
F(0,t)=x_0, F(1,t)=x_1
G(s,0)=g(s), G(s,1)=g'(s)
G(0,t)=x_1, G(1,t)=x_2

We can define:

$H(s,t)=\begin{cases}F(2s,t) &\text{for }s\in[0,\frac{1}{2}],\\ G(2s-1,t)&\text{for }s\in[\frac{1}{2},1]\end{cases}$.

We can check that F(1,t)=x_1=G(0,t) for all t, hence the map H is well-defined. H is continuous by the pasting lemma.

Let’s check that H is the required path homotopy between f*g and f’*g’.

For s in [0,1/2],

H(s,0) = F(2s,0) =f(2s)=h(s)

H(s,1) = F(2s,1) =f'(2s)=h'(s)
h’ := f’ * g’

H(0,t) = F(0,t) = x_0

s in [1/2,1] works fine too:

H(s,0) = G(2s-1,0) = g(2s-1)=h(s)

H(s,1) = G(2s-1,1)= g'(2s-1) = h'(s)

H(1,t) = G(1,t)= x_2

Thus, H is indeed the required path homotopy between f*g and f’*g’. * is almost like a binary operation for a group. The only difference is that [f]*[g] is not defined for every pair of classes, but only for those pairs [f], [g] for which f(1) = g(0), i.e. the end point of f is the starting point of g.

## Homotopy of Paths

For this post we will explain what is a homotopy of paths.

The book above is a nice introductory book on Topology, which includes a section of introductory Algebraic Topology.

Definition: If f and f’ are continuous maps of the space X into the space Y, we say that f is homotopic to f’ if there is a continuous F: X x I -> Y such that

F(x, 0)=f(x) and F(x,1) = f'(x)

for each x. The map F is called a homotopy between f and f’. If f is homotopic to f’, we write $f \simeq f'$.

If f and f’ are two paths in X, there is a stronger relation, called path homotopy, which requires that the end points of the path remain fixed during the deformation. We write $f \simeq_p f'$ if f and f’ are path homotopic.

Next, we will prove that the relations $\simeq$ and $\simeq_p$ are equivalence relations.

If f is a path, we shall denote its path-homotopy equivalence class by [f].

Proof: We shall verify the properties of an equivalence relation, namely reflexivity, symmetry and transitivity.

Reflexivity:

Given f, it is rather easy to see that $f \simeq f$. The map F(x,t) is the required homotopy.

F(x,0)=f(x) and F(x,1)=f(x) is clearly satisfied.

If f is a path, then F is certainly a path homotopy, since f and f itself has the same initial point and final point.

Symmetry:

Next we shall show that given $f \simeq f'$, we have $f' \simeq f$. Let F be a homotopy between f and f’. We can then verify that G(x,t) = F(x, 1-t) is a homotopy between f’ and f.

G(x,0) = F(x, 1)=f’ (x)

G(x,1) = F(x, 0) = f(x)

Furthermore, if F is a path homotopy, so is G.

G(0,t)=F(0, 1-t) = $x_0$

G(1,t)=F(1,1-t) = $x_1$

Transitivity:

Next, suppose that $f \simeq f'$ and $f' \simeq f''$, we show that $f \simeq f''$. Let F be a homotopy between f and f’, and let F’ be a homotopy between f’ and f”. This time, we need to define a slightly more complicated homotopy G: X x I -> Y by the equation

$G(x,t) = \begin{cases} F(x,2t) &\text{for }t\in [0,\frac{1}{2}],\\ F'(x, 2t-1) &\text{for } t\in [\frac{1}{2}, 1].\end{cases}$

First, we need to check if the map G is well defined at t=1/2. When t=1/2, we have F(x,2t) = F(x,1)=f'(x) = F'(x,2t-1).

Because G is continuous on the two closed subsets X x [0, 1/2] and X x [1/2, 1] of XxI, it is continuous on all of X x I, by the pasting lemma.

Thus, we may see that G is the required homotopy between f and f”.

G(x,0)=F(x,0) = f(x)

G(x,1) = F’ (x, 1) = f”(x)

We can also check that if F and F’ are path homotopies, so is G.

G(0,t) = F(0, 2t) = $x_0$

G(1, t)=F'(1, 2t-1) = $x_1$