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24 Apr 2005: d( 6 Á j¤ : %'=:<> iÁ Á j¤ : %'=: 'Áª B # z zE C @EF Õ$# C! ... E. E. Õ E. Õ E. Õ. Õ. [E. [E. B #. B #. f«z Ñ / ¡ @zRDR@¤H Å zRzHzRp«$ ¢ª@HRHz ¡ ¡¡ zHªD « 4Gg2 S l %' #¤z¿ ( Á @?KJ B|DzR/ #

9 Aug 2005: $%! b%)(P3- % A "& 8$C(/$%B#."&TS %fN % B7 &N # )(.&B#.8 7G+ / %&&)(P%WPb%B&b%B!'&"#G6B&! & "# ... 7= # % &N (< %"! B7 <= # %. #b <b% 8. %BA -b%B!<'&B#.@(<b%B7bB! % 8 )($! %

21 May 2005: BA. B. B. B. A A D. A. DCC. AA A A. ... 3. (b) Now show that γ is an epimorphism (i.e., is surjective).

9 Aug 2005: N$9%9 #%>7#,- %>.- "%-. r29B!7#.- >!$&9B&B/ <B&!-B/a -.I!& 7#. -$& , ,-7#Gp! ... 213%#B>! @}(|B&B&29 s BC!,p & R7#. $.p#{S B&B&! F '- ,$7#B&.-,{S = 4$&5.

24 Apr 2005: YZ;>D?B 86 <? YA m 6 c]dJYJLaYSB 86 <? YA m 6 BED7FA]U aaBED7FA]UbaZ;PRiF 5? ... YZ;>D B B º ¢? X:<;>=J? A A? X:<;>=@? A A m 2mEC 6 c.

26 Sep 2005: Let us compute the commutator of twosuch vector fields F(A,a) and F(B,b). ... Thus. [F(A,a),F(B,b). ]= F(C,c), where (C, c) = (AB BA,Ab Ba).

9 Aug 2005: YL < :k| K2 -J4)? > :B,B"E77)W(& 8-B:I; D 7? ... 7! bP : k2q5),) ) &y ,4/0? x :B,B -,)4sE: )E,B4 j H.

21 May 2005: 8. Part (b) is incorrect: In the figure, the segments labelled A are supposed to beidentified and the segments labelled B are supposed to be identified, but theresulting segments A and ... a},{1, 2, 5}, {1, 2, 6}, {1, 3, 4}, {1, 5, 6},{2, 3, 5}, {2, 4, 6}

9 Aug 2005: B B 9 3 8. 3 @ 6!849 P 6 C48 35476)6 9. ... B B ( 3 8 2 6 @' 2 >=! 6!849 6 ( @ 358 2 6 @' 2 0%&' 6 358!

9 Aug 2005: 6). µc =b P(c|b)b. b P(c|b)(7). σc =b P(c|b)(b µc)(b µc). ... al,. which would produce a P(w|B) as. P(w|B) b. P(w|c)P(c|B). (13).