Remarks about the mandatory assignment

The 10 mandatory assignments that were handed in by April 28th have now been checked.  All have passed.  I will bring the papers to class on May 4th.

Here are some comments regarding the solutions:

Problem 1(b): If b_1 and b_2 are lifts of c_1 and c_2, when calculating s(c_1+c_2) you can choose the lift of c_1+c_2 to be b_1+b_2.  This simplifies the verification that s(c_1+c_2) = s(c_1) + (c_2).

Problem 2(b): You can construct the deformation retraction of Y onto Z in three steps: First slide (A, B, C) to (0, B-A, C-A) = (0, B', C'), sending A to 0, then scale (0, B', C') to (0, B'/|B'|, C'/|B'|) = (0, B'', C''), so that B'' has length 1, and finally glide C'' linearly to C''', so that (0, B'', C''') is in Z.

It is easier to show that each of these three steps is a deformation retraction, so that the composite is also a deformation retraction, than to check directly that the composite is a deformation retraction.  Why does the explicit formula for the deformation of (A, B, C) to (0, B'', C''') give points that stay within Y?

Problem 2(c): To check that y = (A, B, C) has a neighborhood U in Y such that U, g(U) and g^2(U) are disjoint, let d be the minimum of d(y, g(y)), d(y, g^2(y)) and d(g(y), g^2(y)), and let U be the open ball around y of radius d/2.  Disjointness follows by the triangle inequality.

Warning: The condition that g_1(U) \cap g_2(U) is nonempty means that there exist points u, v \in U with g_1(u) = g_2(v), not that there exists a point u \in U with g_1(u) = g_2(u).

Problem 2(d): By the general theory, p_* : Z = pi_1(Y) --> pi_1(X) is injective, with pi_1(X)/p_* pi_1(Y) = G = Z/3.  Hence there is a short exact sequence 0 --> Z --> pi_1(X) --> Z/3 --> 0.  Let f be a path in Y from the base point y_0 to g(y_0).  Then pf is a loop in X from x_0 = p(y_0), representing a class x in \pi_1(X, x_0) that maps to the generator g of G = Z/3.  Three times pf, i.e., pf*pf*pf, represents 3x and lifts to f*gf*g^2f.  This lift is a loop in Y from y_0.  Under the deformation retraction to Z, the lift f*gf*g^2f corresponds to a generator y of \pi_1(Z) = \pi_1(Y).  Hence p_*(y) = 3x, so x generates \pi_1(X) = Z, and p_* : Z --> Z sends the generator y to 3 times the generator x, i.e., is given by multiplication by 3.

- John

Published May 3, 2016 9:59 AM - Last modified May 3, 2016 11:31 AM