ERRATA
P14; Thm. 2.7.3
Replace "is a CW-complex" by "has the homotopy type of a CW-complex"
P29: Remark 3
Add "and $B_n(D)$ is projective" to the end of the sentence.
P33 and P37; Statement of excision (P33) and comment below Thm. 5.2.6 (P37)
A6 on P33 should not require that U be open and the comment below Thm. 5.2.6
should be deleted.
P37; Paragraph below Thm. 5.2.7
Throughout the paragraph, A5 should be A6 and A6 should be A5.
P44; Pf. of Thm. 5.6.2
Should not claim that $T\simeq 1$.
Should instead say that by acyclic models $T\circ AW\simeq $AW$.
P44: Remark
Insert "that" after "Note"
P49: Pf. of Thm. 6.3.1
Remove redundant "The proof is a long exercise in cap products".
P54: Prop. 7.1.1
Insert "be" before "a fibration"
P54: 3 lines before Prop. 7.2.2
Insert "$g:$" before "$B\to Z"
P67: diagram near top and 3 lines below diagram
The leftmost vertical map was not intended to have been marked as an isomorphism.
There are two typos in the formula for f three lines below: there is an
asterisk missing; and an extra $\partial$. It should read
$$f=(g_+j++,p)_*\circ((j_+)_*\partial)^{-1}\circ(g_+j_+,p)_*^{-1}$$
P82: L10; definition of $\phi'$
replace f by j(f) to get
$\phi'(\omega, f,t)=(\omega,(\omega,j(f)),t)$
P83: L-7
Replace "$[f,g]$" by"$[f',g']$"
P83: Example 7.8.1
Move Example 7.8.1 to end of Section 7.8 and replace wording by:
Let $\iota_p$ and $\iota_q$ denote the canonical inclusions of $S^p$ and~$S^q$
into $S^p\vee S^q$.
The space formed by attaching a cell to $S^p\vee S^q$ by means of the
Whitehead product $[\iota_p,\iota_q]:S^{p+q-1}\to S^p\vee S^q$
is homotopy equivalent to~$S^p\times S^q$.
To see this observe that by Theorem~7.7.4a' and Ganea's Theorem, the induced
map from the homotopy cofibre of~$[\beta_{S_p},\beta_{S_q}^{-1}]$
to~$S^p\times S^q$ is more than~$p+q$ connected.
Composing with $\susp(S^p\wedge S^q)\to\susp(\Omega S^p\wedge\Omega S^q)$,
which is also connected through this range, shows that the induced map from
the homotpy cofibre of $[\iota_p,\iota_q^{-1}]$ to~$S^p\times S^q$ induces
a homology isomorphism through degree~$p+q$ and is thus a homotopy equivalence.
Since the degree~$-1$ self-map of $S^q$ is a homotopy equivalence, the
homotopy cofibre of~$[\iota_p,\iota_q]$ is also homotopy equivalent
to~$S^p\times S^q$.
P84: Below display beginning "$\Omega X*\Omage Y\approx$"
Insert:
where the first homotopy equivalence is given by
$$(a,b,t)\mapsto\left\{\begin{array}{ll}
\bigl((a,2t),b\bigr)&\mbox{if}\; 0\le t\le1/2;\\
\bigl(a,(b,2t-1)\bigr)&\mbox{if}\; 1/2\le t\le1,
\end{array}\right.
$$
P84; Last display of Section 7.8
Replace "$\psi(\omega,\gamma,t)=$" by "$(\omega,\gamma,t)\mapsto$"
P87; Proof of Lemma 7.9.7
For arbitrary $X$, proof is nonsense since there is no map $-1:X\to X$ to put
in the first diagram. Replace the proof by:
Since the diagram is homotopy commutative, there is an induced map
between the homotopy fibres of the vertical maps.
Applying Ganea's Theorem to the fibration $\Omega X \to PX\to X$ shows that
the homotopy fibre of $\beta_X:\susp\Omega X\to X$ is $\Omega X *\Omega X$,
which by Theorem~7.7.4a' we know is homotopy equivalent to the homotopy fibre
of the second vertical map.
We must show that some induced map of homotopy fibres is a homotopy
equivalence.
Applying the final sentence in the statement of Ganea's Theorem, with the
order of the factors reversed, gives that the induced map
$i:\susp\Omega X\wedge\Omega X\to PX/\Omega X\approx\susp\Omega X$ from the
homotopy fibre of the left vertical map is given by
$(\omega,t,\gamma)\mapsto \bigl(\mu(\omega,\gamma^{-1}),t\bigr)$, where $\mu$
is the multiplication in~$\Omega X$.
Therefore we have an induced map $\sigma$ of homotopy fibres such that
$[\beta_X,\beta_X^{-1}]\circ\sigma \simeq f\circ i$.
Using the map~$\psi$ in the proof of Theorem~7.7.4a' and the definition
of~$f$, we see that $f\circ i=[\beta_X,\beta_X^{-1}]$ and so
$[\beta_X,\beta_X^{-1}]\circ\sigma\simeq[\beta_X,\beta_X^{-1}]$.
However the loop of the right fibration splits, so
$\Omega[\beta_X,\beta_X^{-1}]$ has a left homotopy inverse and thus
$\Omega\sigma$ is homotopic to the identity.
Therefore $\sigma$ induces the identity on homotopy groups and is thus a
homotopy equivalence.
P100; L20
Insert "and the restriction of the trivialization maps to each fibre is a
linear transformation," before "such a bundle is called"
P100; L-4 and L-5
Replace "$p^{-1}(X\times 1)$" and "p^{-1}(X\times 0} by
"$p^{-1}(B'\times 1)$" and "p^{-1}(B'\times 0}" respectively
P106; L-5
Replace "The resulting algebra" by "If $|X|$ is even, the resulting algebra"
P107-108
Insert "associative" before "$R$-algebra" or "Hopf algebra" in the following
places:
P107; L2
P107; L10
P107; L16
P107; L-2
P108; last line of Thm.10.3.1a
P108; first line of Thm.10.3.1c
P107-108
Insert "coassociative" before "$R$-coalgebra" or "Hopf algebra" in the following
places:
P107; L6
P107; L13
P107; L16
P108; L2
P108; last line of Thm.10.3.1b
P108; first line of Thm.10.3.1d
P108; Thm.10.3.2
Insert "associative coassociative" before "Hopf algebra"
P108; Thm.10.3.3
Replace "nonnegatively graded connected" by "connected associative coassociative"
P108; Thm.10.3.3
Change "injective" to "monomorphism" and "surjective" to "an epimorphism".
P132; Exercise11.9.10
Begin first sentence:
"Up to homotopy, the space $CP^\infty$ deloops to give"...
P137; Lemma11.10.3; L2
Insert "surjective" before "morphism"
P139; Theorem11.10.7; part 4
Replace
"$f_*$" and $g_*$" by "$\AH(f)$" and "$\AH(g)$"
P144; Beginning Line -3; Proof of Theorem11.11.3;
Replace from Line -3 on Page 144 to end of proof by:
\def\dtensor{\odot}
By construction, the homotopy cofibre of $Y_{n-1}\to Y_n$ is
$\susp F_{n-1}\approx\susp(\Omega B)^{*n}\approx\susp^n(\Omega B)^{(n)}$
(where $X^{*n}$ denotes the $n$-fold join of~$X$ and $X^{(n)}$ denotes the
$n$-fold smash product).
Therefore
$E^1_{p,q}=\tilde{H}_{p+q}\bigl(\susp^p(\Omega B)^{(p)}\bigr)=
\bigl(\tilde{H}_*(\Omega B)^{\otimes p}\bigr)_q$.
By Ganea's theorem, the composition
$\susp^{n+1}(\Omega B)^{*(n+1)}\approx \susp F_{n}\namedright{\susp j_n}
\susp Y_{n}\namedright{}\susp^{n+1}(\Omega B)^{*n}$
(corresponding to $Y_{n+1}/Y_{n}\to\susp Y_{n}\to\susp Y_{n}/\susp Y_{n-1}$)
is given by the $n$-fold suspension of
$$(\omega_n,\omega_{n-1},\ldots\omega_0,t)\mapsto
\bigl(\omega_n^{-1}\cdot(\omega_{n-1},\ldots,\omega_0),t\bigr)
=(\omega_n^{-1}\cdot\omega_{n-1},\ldots\omega_n^{-1}\cdot\omega_0,t\bigr).$$
Set $A:=H_*(\Omega B)$ and $IA:\Ker\epsilon=\tilde{H}_*(\Omega B)$, where
$\epsilon$ is the augmentation.
Then $d_1$ is given by
$d_1(a_n\otimes\cdots a_0)=
\mu_{\Delta}\bigl(c(a_n),a_{n-1}\otimes\cdots\otimes a_0)$
where $\mu_{\Delta}$ is the diagonal action of the Hopf algebra~$A$
on~$A^{\otimes n}$ (described at the end of Section~10.1) and $c$ is its
canonical conjugation.
The extended module functor $F: R$-modules $\to A$-modules by
$F(M):= A\otimes M$ (with $A$ acting on the first factor) is left adjoint to
the forgetful functor $J: A$-modules $\to R$-modules.
The adjoint pair $F\adj J$ gives rise to an $A$-free acyclic chain
complex~$A_{\bullet}$ in which
$A_n=A^{\otimes(n+1)}$
(see Theorem~12.1.2 and its preceding discussion), and the differential is
given by $d=\sum_{j=0}^n(-1)^{j}d_j$ where
$$d_j(a_n\otimes\cdots\otimes a_{j+1}\otimes a_j\otimes a_{j-1}\otimes\cdots
\otimes a_0)
=a_n\otimes\cdots\otimes a_{j+1}\otimes a_ja_{j-1}\otimes\cdots
\otimes a_0)$$
for $j>0$ and
$$d_0(a_n\otimes\cdots\otimes a_0)
=\epsilon(a_0)(a_n\otimes\cdots\otimes a_1).$$
$IA$ is an ideal in~$A$, so the subspace $(IA)_{\bullet}$ defined by
$(IA)_n:=A\otimes (IA)^{\otimes n}$
forms an $A$-free subcomplex of~$A_{\bullet}$.
Since
$$(d_{j+1}-d_j)(a_n\otimes\cdots\otimes a_{j+1}\otimes 1\otimes a_{j-1}
\otimes\cdots\otimes a_0)=0$$
for $j3$,
$s_2\circ d_j\circ r_2=A^{\otimes^{n-j}}\otimes\mu_A\otimes
A^{\otimes j-3}\otimes (s_{A}\circ r_{A})
=A^{\otimes^{n-j}}\otimes \mu_A\otimes A^{\otimes j-3}
\otimes A^{\dtensor 2}$.
Inductively,
$$\eqalign{
s^{(j-1)}\circ d_j\circ r^{(j)}&=
s_{j-1}\circ\cdots\circ s_2\circ d_j\circ r_2\circ\cdots\circ r_j\cr
&=s_{j-1}\circ\cdots\circ s_3\circ
(A^{\otimes(n-j)}\otimes\mu_A\otimes A^{\dtensor 2})
\circ r_3\circ\cdots\circ r_j\cr
&=\ldots\cr
&=s_{j-1}\circ(A^{\otimes(n-j)}\otimes \mu_A
\otimes A^{\dtensor(j-2)})\circ r_{j-1}\circ r_j.\cr
}$$
In general
$s_M\circ(\mu_A\otimes M)\circ(A\otimes r_M)
=\mu_{(A\dtensor M)}$.
Therefore
$$\eqalign{
s_{j-1}\circ
A^{\otimes(n-j)}\otimes\mu_A\otimes A^{\dtensor(j-2)}\circ
r_{j-1}\circ r_j
&=(A^{\otimes(n-j)}\otimes
\mu_{A^{\dtensor(j-1)}})\circ r_j\cr
&=A^{\otimes(n-j)}\otimes(\mu_{A^{\dtensor(j-1)}}\circ
r_{A^{\dtensor(j-1)}}).\cr
}$$
However, in general $\mu_M\circ r_M=\mu_M\circ(\mu_A\otimes M)
\circ(A\otimes c\otimes M)\circ\psi=\epsilon\otimes M$ by
the definition of the canonical conjugation.
Thus
$s^{(j-1)}\circ d_j\circ r^{(j)}\big|_{A\otimes
(IA)^{\dtensor n}}=0$ for $j