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