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 $j<n$, the projection
  $A\otimes (1-\epsilon)\otimes\cdots\otimes(1-\epsilon):
  A^{\otimes(n+1)}\to A\otimes (IA)^{\otimes n}$
  is a chain map.
  Thus $(IA)_{\bullet}$ is also acyclic, being a retract (as a chain complex)
  of the acyclic chain complex~$A_{\bullet}$.
  Therefore
  $H_*\bigl(R\otimes_A (IA)_{\bullet}\bigr)=\Tor^A_{**}(R,R)$.
  To complete the proof, we show that
  $(E^1,d^1)\cong (IA)_{\bullet}\otimes_A R$.
  
  Given an $A$-module $M$ (with action denoted $\mu_M:A\otimes M\to M$)
  define $R$-modules isomorphism 
  $s_M:A\otimes M\to A\otimes M$ by $s_M:=(A\otimes\mu_M)\circ(\psi\otimes M)$
  with inverse
  $r_M:=(A\otimes\mu_M)\circ(A\otimes c\otimes M)\circ(\psi\otimes M)$.
  They are $A$-module isomorphisms between $A\otimes M$ with action of
  $A$ on the left factor and $A\otimes M$ with diagonal action of~$A$ (which we
  will denote~$A\dtensor M$).
  These correspond to the inverse isomorphisms of $G$-spaces
  $s_X:G\times X\to G\times X$ and 
  $s_X:G\times X\to G\times X$ given by $s_X(g,x)=(g,g\cdot x)$ and
  $r_X(g,x)=(g,g^{-1}\cdot x)$ when a group $G$ acts on a set~$X$.
  Notice that $s_M\big|_{R\otimes M}=r_M\big|_{R\otimes M}=1_{R\otimes M}$
  so passing to the quotient gives induced isomorphisms
  $\bar{s}_M:IA\otimes M\to IA\dtensor M$ and 
  $\bar{r}_M:IA\dtensor M\to IA\otimes M$.
  Write $s_j$ and $r_j$ for
  $A^{\otimes(n+1-j)}\otimes s_{A^{\dtensor(j-1)}}$
  and $A^{\otimes(n+1-j)}\otimes r_{A^{\dtensor(j-1)}}$.
  Set $$s^{(n)}:=s_n\circ s_{n-1}\circ\cdots\circ s_2:
  A_n\cong A^\Delta_n,\qquad{\rm and}\qquad
  \bar{s}^{(n)}:=s_n\circ\bar{s}_{n-1}\circ\cdots\circ\bar{s}_2:
  (IA)_n\cong (IA)^\Delta_n,$$
  where $A^\Delta_n:=A\otimes A^{\dtensor n}$
  and $(IA)^\Delta_n:= A\otimes (IA)^{\dtensor n}$.
  Similarly set
  $$r^{(n)}:=r_2\circ\cdots\circ r_{n-1}\circ r_n:A^\Delta_n\cong A_n
  \qquad{\rm and}\qquad
  \bar{r}^{(n)}:=\bar{r}_2\circ\cdots\bar{r}_{n-1}\circ r_n:
  (IA)^\Delta_n\cong (IA)_n.$$
  Let $\partial:A^\Delta_n\to A^\Delta_{n-1}$ be defined by
  $\partial:=(-1)^{n}s^{(n-1)}\circ d\circ r^{(n)}$
  and
  $\bar\partial:(IA)^\Delta_n\to (IA)^\Delta_{n-1}$
  by~$\bar\partial:=
  (-1)^{n}\bar{s}^{(n-1)}\circ d\circ\bar{r}^{(n)}$.
  Then, by construction, $\partial$ and $\bar{\partial}$ are differentials
  such that $A_{\bullet}\cong A^\Delta_\bullet$ and
  $(IA)_{\bullet}\cong(IA)^\Delta_\bullet$.
  
  For $j<n$, 
  $s^{(n-1)}\circ d_j\circ r^{(n)}=
  A^{\otimes(n-j)}\otimes\bigl(s^{(j-1)}\circ d_j\circ r^{(j)}\bigr)$.
  
  For $j>3$,
  $s_2\circ d_j\circ r_2=A^{\otimes^{n-j}}\otimes\mu_A\otimes
  A^{\otimes j-3}\otimes (s_{A}\circ r_{A})
  =A^{\otimes^{n-j}}\otimes \mu_A\otimes A^{\otimes j-3}
  \otimes A^{\dtensor 2}$.
  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 $j<n$.
  
  When $j=n$ we get
  $s^{(n-1)}\circ d_n\circ r^{(n)}
  =(\mu_A \otimes A^{\dtensor(n-1)})
  \circ(A\otimes r_{A^{\dtensor(n-1)}})$.
  In general, after composing with the isomorphisms
  $A\dtensor M\cong R\otimes_A (A\otimes A\dtensor M)$ and
  $R\otimes_A(A\otimes M)\cong M$, the composite
  $R\otimes_A\bigl((\mu_A\otimes_M)\circ(A\otimes r_M)\bigr):
  A\dtensor M\to M$
  is given by $\mu_M\circ(c\otimes M)$ and so
  $R\otimes_A(s^{(n-1)}\circ d_n\circ r^{(n)})=
  \mu_{A^{\dtensor(n-1)}}
  \circ(c\otimes A^{\dtensor(n-1)})$.
  
  Therefore
  $\bar\partial=
  (-1)^{n}R\otimes_A(s^{(n-1)}\circ d_n\circ r^{(n)})
  =\mu_{A^{\dtensor(n-1)}}\circ(c\otimes A^{\dtensor(n-1)})=d^1$.
  \qed

P146; Lines 8 and 9;
  Replace "(FG)" by "(GF)".

P146; Lines -3
  Replace "M\otimes_S A"" by "M\otimes_S B" and replace "M\otimes_R B"
    by "M\otimes_R A"

P147; Line 12
  "and also the homology" should be "and so the homology"

P147; Display below "In one spectral sequence we have"
  Top line of display: "L_pG" should be "L_qG"
  Left side of bottom line of display: "G(T)_{*q}" should be "G(T)_{0,q}"

P157; L16
  Delete " are referred to"

P158; Theorem13.4.2
  Insert "(on pointed $CW$-complexes)" after cohomology theory

P161; Theorem13.4.3
  Replace "space~$W$" by "$CW$-complex~$W$"

P164; Lemma1.6.4; L-3 and -2
  Replace
    "resolution of~$(\zp)\oplus(\zp)$, the diagonal map $\zp\to(\zp)\oplus(\zp)$
     induces"...
  by
    "resolution of~$(\zp)\otimes(\zp)\cong\zp$, the map $\zp\to(\zp)\otimes(\zp)$
     induces"...