Preliminary Examination in Topology: January 2011
Algebraic Topology portion
Instructions: If possible, answer all three questions on both sides of this sheet. If this
is not possible, then two complete solutions is better than three partial solutions.
Time Limit: 90 minutes.
a a
c-1 b d b
d c c-1 c
b
-1 d b
-1 d
a-1 a-1
X1 X2
1. Consider two surfaces X1 and X2 , each obtained by identifying edges of an octagon
as is the figure.
(a) Is either of these surfaces a regular cover of the other? If so, indicate which one,
and list all possibilities for the number of sheets.
(b) Is either of these surfaces an irregular cover of the other? If so, indicate which
one, and list all possibilities for the number of sheets.
Answer to both parts We identify the surfaces by orientation and Euler character-
istic. X1 has two vertices, 4 edges and one face, so χ(X1 ) = −1. (One vertex is the tail
of a, the head of b, and the tail of c, while the other vertex is the head of a, the tail of b,
the head of c, and the head and tail of d.) X2 has only one vertex, so χ(X2 ) = −2. Both
surfaces are non-orientable, thanks to the d edges. So X1 = #3 RP 2 and X2 = #4 RP 2 ,
and X2 is the double cover of X1 . (For instance, you can view X1 as a big RP 2 with
two small RP 2 ’s glued in. Taking the double cover of the big RP 2 gives an S 2 with four
small RP 2 ’s glued in, which is homeomorphic to X2 .) Since all double covers are regular,
X2 is a regular double cover of X1 , with two sheets, but neither surface is an irregular
cover of the other.
1
2
2. Give clear proofs of the following two assertions. We are looking for proofs from first
principles — quoting a theorem will not suffice.
(a) Let f : RP 2 → X × Y be continuous and assume that p1 ◦ f and p2 ◦ f are each
homotopic to constant maps, where p1 : X × Y → X and p2 : X × Y → Y are
the projection maps. Then f is homotopic to a constant map.
Answer Let fi = pi ◦ f , and let F1 : RP 2 × [0, 1] → X,F2 : RP 2 × [0, 1] → Y
be homotopies of f1 and f2 to the a constant map. Then F = F1 × F2 is a null-
homotopy of f . To see that F is continuous, note that (F1 × F2 )−1 (U × V ) =
F1−1 (U ) ∩ F2−1 (V ).
(b) Let p : X e → X be a cover and p(x̃) = x. Let F : D2 → X be a continuous map
of a 2-dimensional disk into X with F (y) = x where y ∈ ∂D2 . Then there is a
lift Fe : D2 → X e of F such that Fe(y) = x̃.
Answer Represent D2 by the square [0, 1] × [0, 1] with y = (0, 0). Let ǫ > 0 be
such that if A ⊂ [0, 1] × [0, 1] has diameter at most ǫ then F (A) lies in an open set
of X that is evenly covered (ǫ is the Lebesgue number of the cover of [0, 1] × [0, 1]
given by {F −1 (U )|U is an evenly covered open subset of X}.) Subdivide [0, 1] ×
[0, 1] into sub-rectangles of dimensions ǫ/2 × ǫ/2. We construct the lift Fe one
subrectangle at a time. Start with [0, ǫ/2] × [0, ǫ/2] = A. Let UA be an open set
of X evenly covered by p and s.t. F (A) ⊂ UA . Let U fA be the slice of this even
cover containing x̃. Lift Fe|A as p̃|U−1
f ◦ F |A . Note that Fe|A (0, 0) = x̃. We follow
A
the same procedure for each subrectangle. Each time we pick a new subrectangle
that intersects the previously defined domain along either its left-hand edge, its
bottom edge, or the union of the two edges. As this overlap is connected, the
overlap will map into a single slice above an evenly covered open set containing
the image of f restricted to this subrectangle. We extend the map over this
rectangle in this slice as above. We continue until Fe has been defined over D2 .
3
3. Let T be a torus and C a homotopically non-trivial, simple closed curve in T . Let X
be the 2-complex obtained by attaching a 1-punctured torus, S, to T along C (that is,
by identifying ∂S with C).
T
X = T UC S
(a) Compute π1 (X).
Answer Use Van Kampen’s theorem, with U being a neighborhood of T and
V being a neighborhood of S. Since π1 (U ) =< a, b|ab = ba >, where a is the
class of C, since π1 (V ) =< e, f >, and since the class of C in π1 (V ) is ef e−1 f −1 ,
π1 (X) =< a, b, e, f |ab = ba, a = ef e−1 f −1 >.
(b) Show that C must lift to a closed loop (as opposed to an open path) on any 2-fold
cover of X.
Answer Any 2-fold cover, p : X̃ → X is regular, hence its fundamental group
is the kernel of a map f : π1 (X) → Z/2Z. By part (a), the class of C in π1 (X) is
a commutator and hence in the kernel of f . Since the class of C in π1 (X) in in
p∗ (π1 (X̃), C lifts to a closed loop.
(c) Give three non-homeomorphic, connected, covering spaces of X and exhibit (ex-
plain or draw) a covering map for each. You needn’t show that the spaces are not
homeomorphic. Are there any others?
Answer Each double cover of X will be gotten by gluing a (possibly discon-
nected) double cover of T to a (possibly disconnected) double cover of S along
two copies of C to obtain a connected space.
There are two double covers of T in which the preimage of C is two curves:
(1) A single torus to which C lifts to two parallel curves.
(2) Two homeomorphic copies of T .
Likewise, there are two possible double covers of S:
(3) A twice-punctured torus to which C lifts to the two punctures.
(4) Two homeomorphic copies of S.
1 and 3, 1 and 4, and 2 and 3 give non-homeomorphic connected covers. 2 and
4 gives a disconnected space, which doesn’t count. These are all of the double
covers.
