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\begin{document}

\addtocounter{section}{2}

\begin{center}
{\large Approach to Whitehead's theorem, from Spanier \cite{spanier}} \\
\textbf{Chapter 7, Section 2}
\end{center}

See Hatcher \cite{hatcher}, page 346, for the following three results
and definitions.

\begin{theorem}\label{7-2-1}
Fix $n \geq 1$.  Given a map $\alpha : (E^{n}, S^{n-1}, p_{0})
\rightarrow (X,A,x_{0})$, then $[\alpha]=0$ in $\pi_{n}(X,A,x_{0})$ if
and only if $\alpha$ is homotopic relative to $S^{n-1}$ to some map
$E^{n} \rightarrow A$.
\end{theorem}

\begin{definition}
Fix $n \geq 0$.  A pair $(X,A)$ is said to be \emph{$n$-connected} if
for $0 \leq k \leq n$ every map $\alpha : (E^{k},S^{k-1}) \rightarrow
(X,A)$ is homotopic relative to $S^{k-1}$ to some map $E^{k}
\rightarrow A$.
\end{definition}

\begin{corollary}\label{7-2-2}
Fix $n \geq 0$.  A pair $(X,A)$ is $n$-connected if and only if every
path component of $X$ intersects $A$ and for every point $a \in A$ and
every $1 \leq k \leq n$, $\pi_{k}(X,A,a)=0$.
\end{corollary}



\addtocounter{theorem}{-2}
\addtocounter{section}{4}

\begin{center}
\textbf{Chapter 7, Section 6}
\end{center}

Fix a space $X$ and a subspace $A$.  Fix $n \geq 0$.

\begin{lemma}\label{7-6-1}
If $X$ is obtained from $A$ by adjoining $n$-cells, then $X \times 0
\cup A \times I$ is a strong deformation retract of $X \times I$.
\end{lemma}

%This should seem plausible; for instance, if $A$ is a circle and $X$
%is a disk, then $X \times I$ is a solid cylinder, while $X \times 0
%\cup A \times I$ consists of the bottom and the sides of the
%cylinder.

\begin{proof}
See Hatcher \cite{hatcher} Proposition 0.16, page 15.  (Hatcher's
result actually proves Theorem~\ref{7-6-12}, which generalizes this
lemma.)
\end{proof}

\begin{corollary}\label{7-6-2}
If $X$ is obtained from $A$ by adjoining $n$-cells, then the inclusion
map $A \subset X$ is a cofibration.
\end{corollary}

\begin{lemma}\label{7-6-3}
Let $X$ be obtained from $A$ by adjoining $n$-cells and let $(Y,B)$ be
a pair such that $\pi_{n}(Y,B,b)=0$ for all $b \in B$ if $n \geq 1$
and such that every point of $Y$ can be joined to $B$ by a path if
$n=0$.  Then any map $(X,A) \rightarrow (Y,B)$ is homotopic relative to
$A$ to a map $X \rightarrow B$.
\end{lemma}

\begin{proof}
This result is basically the same as \cite[Lemma 4.6]{hatcher}, so you
can look there for another proof.

Here's the idea: suppose you obtain $X$ from $A$ by attaching a single
$n$-cell, in which case $X=A \cup D^{n}$, with the boundary $S^{n-1}$
of $D^{n}$ contained in $A$.  So a map $(X,A) \rightarrow (Y,B)$
corresponds to maps $A \rightarrow B$ and $D^{n} \rightarrow Y$ which
are compatible on the boundary of $D^{n}$.  In other words, it gives
compatible maps $f: A \rightarrow B$ and $g: (D^{n},S^{n-1}) \rightarrow
(Y,B)$.  Since $\pi_{n}(Y,B,b)=0$ for all $b \in B$, the map $g$ is
homotopic, relative to $S^{n-1}$, to a map $D^{n} \rightarrow B$.  The
map $f$ already has image in $B$, so when you glue $f$ and $g$
together, you get a map which is homotopic, relative to $A$, to a map
$X \rightarrow B$.

If $X$ is obtained from $A$ by attaching more than one $n$-cell, just
do the above procedure on each cell.  Since the interiors of the
cells are disjoint, and since the homotopies are stationary on their
boundaries, the resulting homotopy is continuous.
\end{proof}

\begin{definition}
A \emph{relative $CW$ complex} $(X,A)$ consists of a topological space
$X$, a closed subspace $A$, and a sequence of closed subspaces
$(X,A)^{k}$ for $k \geq 0$ such that 
\begin{enumerate}
\item $(X,A)^{0}$ is obtained from $A$ by adjoining $0$-cells.
\item For $k \geq 1$, $(X,A)^{k}$ is obtained from $(X,A)^{k-1}$ by
adjoining $k$-cells.
\item $X=\bigcup (X,A)^{k}$.
\item $X$ has a topology coherent with $\{(X,A)^{k}\}_{k}$.  (This
means that the topology on $X$ is the finest topology on $X$ with
respect to which the inclusion maps $(X,A)^{k} \rightarrow X$ are
continuous, for all $k \geq 0$.  This is one way of describing the
standard topology on a CW complex.)
\end{enumerate}
\end{definition}

For example, $(X,\emptyset)$ is a relative CW complex if and only if
$X$ is a CW complex.  If $(X,A)$ is a relative CW complex, so is
$(X,(X,A)^{k})$ for any $k$ (because $X$ is obtained from $(X,A)^{k}$
by attaching $(k+1)$-cells, $(k+2)$-cells, etc.).

\addtocounter{theorem}{8}

\begin{theorem}\label{7-6-12}
If $(X,A)$ is a relative CW complex, then the inclusion map $A \subset
X$ is a cofibration.
\end{theorem}

\begin{proof}
Use induction and Corollary~\ref{7-6-2}.
\end{proof}

\begin{theorem}\label{7-6-13}
Let $(X,A)$ be a relative CW complex, with dimension $(X-A) \leq n$,
and let $(Y,B)$ be $n$-connected.  Then any map $(X,A) \rightarrow
(Y,B)$ is homotopic relative to $A$ to a map $X \rightarrow B$. 
\end{theorem}

\begin{proof}
This follows, using induction, from Corollary~\ref{7-2-2},
Lemma~\ref{7-6-3}, and Theorem~\ref{7-6-12}.
\end{proof}

\begin{corollary}\label{7-6-14}
Let $(X,A)$ be a relative CW complex and let $(Y,B)$ be $n$-connected
for all $n$.  Then any map $(X,A) \rightarrow (Y,B)$ is homotopic
relative to $A$ to a map $X \rightarrow B$.
\end{corollary}

\begin{proof}
Let $f: (X,A) \rightarrow (Y,B)$ be a map.  It follows from
Theorems~\ref{7-6-12} and \ref{7-6-13} that there is a sequence of
homotopies 
\[
H_{k} : (X,A) \times I \rightarrow (Y,B), \quad k \geq 0
\]
constructed by induction on $k$ such that
\begin{enumerate}
\item $H_{0}(x,0) = f(x)$ for $x \in X$.
\item $H_{k}(x,1) = H_{k+1}(x,0)$ for $x \in X$.
\item $H_{k}$ is a homotopy relative to $(X,A)^{k-1}$.
\item $H_{k}((X,A)^{k} \times 1) \subseteq B$.
\end{enumerate}
Then a homotopy $H: (X,A) \times I \rightarrow (Y,B)$ with the
required properties is defined by
\begin{gather*}
H(x,t) = H_{k-1}\left( x, \frac{t-(1-1/k)}{(1/k)-1/(k+1)} \right),
\quad 1-\frac{1}{k} \leq t \leq 1-\frac{1}{k+1}, \\
H(x,1) = H_{k}(x,1), \quad x \in (X,A)^{k}.
\end{gather*}
\end{proof}

\begin{lemma}[Main technical lemma]\label{7-6-15}
Fix $n \geq 1$.  If $X$ is obtained from $A$ by adjoining $n$-cells,
then $(X,A)$ is $(n-1)$-connected.
\end{lemma}

(Compare to \cite[Corollary 4.11]{hatcher}: Hatcher's result is
basically the same, but he gets it as a consequence of the cellular
approximation theorem, whereas we use it to prove cellular
approximation.  Hatcher has a different main technical lemma,
\cite[Lemma 4.9]{hatcher}.)

\vspace{0.1in}

\noindent 
Notation. If $e$ is a cell, or more generally a polyhedron
(simplicial complex or $\Delta$-complex or some such thing), then
$\dot{e}$ denotes its boundary.

\begin{proof}
For $k \leq n-1$ let $f: (E^{k},S^{k-1}) \rightarrow (X,A)$ be a map.
Because $f(E^{k})$ is compact, there exist a finite number, say
$e_{1}$, \dots, $e_{m}$, of $n$-cells of $X-A$ such that $f(E^{k})
\subseteq e_{1} \cup \dotsb \cup e_{m} \cup A$.  For $1 \leq i \leq
m$, let $x_{i}$ be a point of $e_{i}-\dot{e}_{i}$.  Each of the sets
$Y = A \cup (e_{1}-x_{1}) \cup \dotsb \cup (e_{m}-x_{m})$ and
$e_{i}-\dot{e}_{i}$ for $1 \leq i \leq m$ intersects $f(E^{k})$ in a
set open in $f(E^{k})$.  There is a triangulation (simplicial complex)
of $E^{k}$, say $K$, such that for every
simplex $s \in K$ either $f(s) \subseteq Y$ or for some $1 \leq i
\leq m$, $f(s) \subseteq e_{i}-\dot{e}_{i}$.  Let $A'$ be the
subpolyhedron of $E^{k}$ consisting of all simplices $s \in K$
such that $f(s) \subseteq Y$, and for $1 \leq i \leq m$ let $B_{i}$
be the subpolyhedron consisting of all simplices $s$ of $K$
such that $f(s) \subseteq e_{i} - \dot{e}_{i}$.  Then $S^{k-1}
\subseteq A'$, $E^{k}=A' \cup B_{1} \cup \dotsb \cup B_{m}$, and if $i
\neq j$, then $B_{i}-A'$ is disjoint from $B_{j}-A'$.  Let $\dot{B}_{i}
= B_{i}\cap A'$ and observe that $(B_{i},\dot{B}_{i})$ is a relative
CW complex, with $\dim (B_{i}- \dot{B}_{i}) \leq k \leq n-1$.

For $1 \leq i \leq m$ the pair $(e_{i}-\dot{e_{i}},
e_{i}-\dot{e_{i}} - x_{i})$ is homeomorphic to $(E^{n}-S^{n-1},
E^{n}-S^{n-1}-0)$ and has the same homotopy groups as
$(E^{n},S^{n-1})$.  By the long exact homotopy sequence of a pair,
$(E^{n},S^{n-1})$ is $(n-1)$-connected.  It follows from
Theorem~\ref{7-6-13} that $f|(B_{i},\dot{B}_{i})$ is homotopic
relative to $\dot{B}_{i}$ to a map from $B_{i}$ to
$e_{i}-\dot{e}_{i}-x_{i}$.  Because $B_{i}-\dot{B}_{i}$ is disjoint
from $B_{j}-\dot{B}_{j}$ for $i\neq j$, these homotopies fit together
to define a homotopy relative to $A'$ of $f$ to some map $f'$ such
that $f'(E^{k}) \subseteq Y$.  Clearly, $A$ is a strong deformation
retract of $Y$.  Therefore $f'$ is homotopic relative to $S^{k-1}$ to
a map $f''$ such that $f''(E^{k}) \subseteq A$.  Then $f \simeq f'
\simeq f''$, all homotopies relative to $S^{k-1}$.  Therefore $(X,A)$
is $(n-1)$-connected.
\end{proof}

\begin{corollary}\label{7-6-16}
If $(X,A)$ is a relative CW complex, then for any $n \geq 0$,
$(X,(X,A)^{n})$ is $n$-connected.
\end{corollary}

\begin{proof}
We prove by induction on $m$ that $((X,A)^{m}, (X,A)^{n})$ is
$n$-connected for $m > n$.  Since $(X,A)^{n+1}$ is obtained from
$(X,A)^{n}$ by adjoining $(n+1)$-cells, it follows from
Lemma~\ref{7-6-15} that $((X,A)^{n+1}, (X,A)^{n})$ is $n$-connected.
Assume $m > n+1$ and that $((X,A)^{m-1},(X,A)^{n})$ is $n$-connected.
By Lemma~\ref{7-6-15}, the pair $((X,A)^{m},(X,A)^{m-1})$ is
$(m-1)$-connected, and since $n<m-1$, it is also $n$-connected.  Then
$\pi_{0}((X,A)^{n}) \rightarrow \pi_{0}((X,A)^{m-1}$ and
$\pi_{0}((X,A)^{m-1}) \rightarrow \pi_{0}((X,A)^{m}$ are both
surjective, when $\pi_{0}((X,A)^{n}) \rightarrow \pi_{0}((X,A)^{m}$ is
also surjective.  Furthermore, for any $x \in (X,A)^{n}$, it follows
from the exactness of the homotopy sequence of the triple $((X,A)^{m},
(X,A)^{m-1}, (X,A)^{n})$, with base point $x$, that $\pi_{k}
((X,A)^{m}, (X,A)^{n}, x) = 0$ for $1 \leq k \leq n$.  By
Corollary~\ref{7-2-2}, $((X,A)^{m}, (X,A)^{n})$ is $n$-connected.

To show that $(X,(X,A)^{n})$ is $n$-connected, if $0 \leq k \leq n$
and $\alpha : (E^{k},S^{k-1}) \rightarrow (X, (X,A)^{n})$, then
because $\alpha (E^{k})$ is compact and $X$ has a topology coherent
with the subspaces $(X,A)^{m}$, there is $m > n$ such that $\alpha
(E^{k}) \subseteq (X,A)^{m}$.  Hence $\alpha$ can be regarded as a map
from $(E^{k},S^{k-1})$ to $((X,A)^{m}, (X,A)^{n})$ for some $m >n$.
Because $((X,A)^{m}, (X,A)^{n})$ is $n$-connected, $\alpha$ is
homotopic relative to $S^{k-1}$ to some map $(E^{k},S^{k-1})
\rightarrow (X,A)^{n}$.
\end{proof}

\begin{theorem}[Cellular approximation]\label{7-6-17}
Given a map $f:(X,A) \rightarrow (X',A')$ between relative CW
complexes which is cellular on a subcomplex $(Y,B)$ of $(X,A)$, there
is a cellular map $g: (X,A) \rightarrow (X',A')$ homotopic to $f$
relative to $Y$.
\end{theorem}

\begin{proof}
It follows from Corollary~\ref{7-6-16}, Theorem~\ref{7-6-13}, and
Theorem~\ref{7-6-12} that there is a sequence of homotopies $H_{k}:
(X,A) \times I \rightarrow (X',A')$ relative to $Y$, for $k \geq 0$,
such that 
\begin{enumerate}
\item $H_{0}(x,0) = f(x)$ for $x \in X$.
\item $H_{k}(x,1) = H_{k+1}(x,0)$ for $x \in X$.
\item $H_{k}$ is a homotopy relative to $(X,A)^{k-1}$.
\item $H_{k}((X,A)^{k} \times 1) \subseteq (X',A')^{k}$.
\end{enumerate}
Then a homotopy $H: (X,A) \times I \rightarrow (X',A')$ with the
desired properties is defined by 
\begin{gather*}
H(x,t) = H_{k-1} \left( x, \frac{t-(1-1/k)}{(1/k) - 1/(k+1)} \right),
\quad 1-\frac{1}{k} \leq t \leq 1-\frac{1}{k+1}, \\
H(x,1) = H_{k}(x,1), \quad x \in (X,A)^{k}.
\end{gather*}
\end{proof}

For example, choosing $Y$ and $B$ to be empty gives the following.

\begin{corollary}\label{7-6-18}
Any map between relative CW complexes is homotopic to a cellular map.
If two cellular maps between relative CW complexes are homotopic,
there is a cellular homotopy between them.
\end{corollary}

If $A$, $A'$, $Y$ and $B$ are all empty, then the theorem says that
every map between CW complexes is homotopic to a cellular map.  If $A
= \{x_{0}\}$ and $A'=\{x_{0}'\}$ so that $f:(X,x_{0}) \rightarrow (X',
x_{0}')$, then let $(Y,B)=(x_{0},x_{0})$; you can conclude that $f$ is
homotopic, via a basepoint-preserving homotopy, to a cellular map.

Hatcher points out the following important consequence \cite[Corollary
4.8]{hatcher}: if $n < k$, then $\pi_{n}(S^{k}) = 0$.  To prove this,
take the standard CW structure on $S^{n}$---that is, a single $0$-cell
and a single $n$-cell.  Do the same for $S^{k}$.  Then any map
$(S^{n},x_{0}) \rightarrow (S^{k},y_{0})$ is homotopic, preserving the
basepoint, to a cellular map.  Since $n<k$, the $n$-skeleton of
$S^{k}$ consists of the single point $y_{0}$; thus every map
$(S^{n},x_{0}) \rightarrow (S^{k},y_{0})$ is homotopic, preserving the
basepoint, to the constant map.

\begin{definition}
Fix $n \geq 1$.  A continuous map $f: X \rightarrow Y$ is called an
\emph{$n$-equivalence} if $f$ induces a one-to-one correspondence
between the path components of $X$ and of $Y$, and if for every $x \in
X$, $f_{*}: \pi_{q} (X,x) \rightarrow \pi_{q}(Y,f(x))$ is an
isomorphism for $q < n$ and an epimorphism for $q=n$.  A map $f: X
\rightarrow Y$ is called a \emph{weak equivalence} or
\emph{$\infty$-equivalence} if $f$ is an $n$-equivalence for all $n
\geq 1$.
\end{definition}

%The following results are immediate:

%\begin{proposition}\label{7-6-19}
%A composite of $n$-equivalences is an $n$-equivalence.
%\end{proposition}

%\begin{proposition}\label{7-6-20}
%Any map homotopic to an $n$-equivalence is an $n$-equivalence.
%\end{proposition}

%\begin{proposition}\label{7-6-21}
%A homotopy equivalence is a weak equivalence.
%\end{proposition}

\addtocounter{theorem}{3}

Let $f: X \rightarrow Y$ be a map and let $Z_{f}$ be the mapping
cylinder of $f$.  Then $f = r \circ i$, where $r : Z_{f} \rightarrow
Y$ is a homotopy equivalence.  Therefore $f$ is an $n$-equivalence if
and only if $i : X \subseteq Z_{f}$ is an $n$-equivalence.  It follows
from the exactness of the homotopy sequence of $(Z_{f},X)$ and from
Corollary~\ref{7-2-2} that $i$ is an $n$-equivalence if and only if
$(Z_{f},X)$ is $n$-connected.

\begin{theorem}\label{7-6-22}
Let $f: X \rightarrow Y$ be an $n$-equivalence \ulp{}$n$ finite or
infinite\urp{} and let $(P,Q)$ be a relative CW complex, with $\dim
(P-Q) \leq n$.  Given maps $g : Q \rightarrow X$ and $h: P \rightarrow
Y$ such that $h|Q = f \circ g$, there exists a map $g': P \rightarrow
X$ such that $g'|Q=g$ and $f\circ g' \simeq h$ relative to $Q$.
\end{theorem}

\begin{proof}
Let $Z_{f}$ be the mapping cylinder of $f$, with inclusion maps $i : X
\subseteq Z_{f}$ and $j:Y \subseteq Z_{f}$, and retraction $r: Z_{f}
\rightarrow Y$ a homotopy inverse of $j$.  Then in the diagram
\[
\begin{diagram}
\node{Q} \arrow{s,l}{g} \arrow{e} 
\node{P} \arrow{s,r}{j \circ h}\\
\node{X} \arrow{e,t}{i} 
\node{Z_{f}}
\end{diagram}
\]
a homotopy $i \circ g \simeq j \circ h|Q$ can be found whose composite
with $r$ is constant.  By Theorem~\ref{7-6-12}, there is a map $h' : P
\rightarrow Z_{f}$ such that $h'|Q = i \circ g$ and such that $r \circ
h' \simeq r\circ j\circ h$ relative to $Q$.  We regard $h'$ as a map
from $(P,Q)$ to $(Z_{f},X)$.  Since $(Z_{f},X)$ is $n$-connected and
$\dim (P-Q) \leq n$, it follows from Theorem~\ref{7-6-13} that $h'$ is
homotopic relative to $Q$ to some map $g' : P \rightarrow X$.  Then
$g'|Q=g$ and 
\[
f\circ g' = r \circ i \circ g' \simeq r \circ h' \simeq r \circ j
\circ h = h,
\]
all the homotopies relative to $Q$.  Hence $g'$ has the desired
properties.
\end{proof}

\begin{corollary}\label{7-6-23}
Let $f: X \rightarrow Y$ be an $n$-equivalence \ulp{}$n$ finite or
infinite\urp{} and consider the map 
\[
f_{*} : [P,X] \rightarrow [P,Y].
\]
If $P$ is a CW complex of dimension $\leq n$, this map is surjective,
and if $\dim P \leq n-1$, it is also injective.
\end{corollary}

\begin{proof}
The first part follows from Theorem~\ref{7-6-22} applied to the
relative CW complex $(P,\emptyset)$.

For the second part, we apply Theorem~\ref{7-6-22} to the relative CW
complex $(P \times I, P \times \dot{I})$.  Given $g_{0}, g_{1}: P
\rightarrow X$ such that $f \circ g_{0} \simeq f \circ g_{1}$, there
is a map $g: P \times \dot{I} \rightarrow X$ such that $g(z,0) =
g_{0}(z)$ and $g(z,1) = g_{1}(z)$ for $z \in P$, and a map $h: P
\times I \rightarrow Y$ such that $h|P \times \dot{I} = f \circ g$.
Since $\dim (P \times I) \leq n$, by Theorem~\ref{7-6-22} there is a
mapping $g': P \times I \rightarrow X$ such that $g'|P \times I = g$.
Then $g'$ is a homotopy from $g_{0}$ to $g_{1}$, showing that $[g_{0}]
= [g_{1}]$.
\end{proof}

\begin{corollary}[Whitehead theorem]\label{7-6-24}
A map between CW complexes is a weak equivalence if and only
if it is a homotopy equivalence.  That is, if $f: X \rightarrow Y$ is
a map between CW complexes, then $f_{*}: \pi_{k} (X,x_{0}) \rightarrow
\pi_{k}(Y,f(x_{0}))$ is an isomorphism for all $x_{0} \in X$ and for
all $k \geq 0$ if and only if $f$ is a homotopy equivalence.
\end{corollary}

\begin{proof}
Certainly every homotopy equivalence is a weak equivalence.
Conversely, if $f: X \rightarrow Y$ is a weak equivalence
between CW complexes, it follows from Corollary~\ref{7-6-23} that $f$
induces bijections
\[
f_{*}: [Y,X] \rightarrow [Y,Y], \quad f_{*}: [X,X] \rightarrow [X,Y].
\]
If $g: Y \rightarrow X$ is any map such that $f_{*}[g] = [1_{Y}]$,
then $f \circ g \simeq 1_{Y}$, and also 
\[
f_{*}[g\circ f] = [f \circ g \circ f] = [1_{Y} \circ f] = [f \circ
1_{X}] = f_{*}[1_{X}].
\]
Therefore $[g\circ f] = [1_{X}]$ or $g \circ f \simeq 1_{X}$, and so
$f$ is a homotopy equivalence.
\end{proof}



\section*{Homework}

\begin{enumerate}
\item Recall that a space $X$ is \emph{simple} (\emph{abelian}, in
Hatcher's terminology) if the action of $\pi_{1}(X,x_{0})$ on
$\pi_{n}(X,x_{0})$ is trivial for all $n \geq 1$.  Prove that
$\mathbb{R}P^{n}$ is simple if and only if $n$ is odd.  (Hatcher's
problem 4 on page 357 might be relevant.)
\item If $m \neq n$, show that $\mathbb{R}P^{2m+1} \times S^{2n+1}$ and 
$\mathbb{R}P^{2n+1} \times S^{2m+1}$ have isomorphic homotopy groups
but non-isomorphic homology groups.
\item Fill in the details in the proofs of Theorems 6.12 and 6.13
above.  (Or fill in the details of any of the proofs above if you
think they need filling in.)
\item Do problems 10 and 12, page 357 of Hatcher.
\end{enumerate}



\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{1}

\bibitem{hatcher}
A. Hatcher, \emph{Algebraic topology}, preprint.

\bibitem{spanier}
E.~H. Spanier, \emph{Algebraic topology}, Springer-Verlag, Berlin, 1981,
  Corrected reprint.


\end{thebibliography}


\end{document}
