MATH 151A-MIDTERM EXAM- SOLUTION TO PROBLEM 2.
We first show that the action described is a covering space action. The mapping
can be described as a linear transformation
2 0
T =
0 1/2
This is invertible and it defines a Z action on X by nx = T n x. Now to show
that this is a covering space action, we will need to show that for any point (x, y)
there exists an open neighborhood U of (x, y) such that T n (U ) ∩ T m (U ) = ∅ if
n 6= m. By multiplying the inverse of the matrix, this is equivalent to showing
that T k (U ) ∩ U = ∅. So, given a point (x, y) ∈ X we will find a rectangular set V
around (x, y) such that T (V ) ∩ V = ∅. First note that for the intersection to be
trivial we need that if x1 < x < x2 then x2 < 2x1 . So first pick x1 close enough
to x such that x1 < x < 2x1 if and only if 1 < x/x1 < 2. Then choose an x2
such that x1 < x < x2 < 2x1 if and only if 1 < x/x1 < x2 /x1 < 2 (note that
the existence of x2 can be done by density of the rational numbers). In a similar
fashion choose y1 < y < y2 such that such that 1/2y2 < y1 . Then the rectangle
V = (x1 , x2 ) × (y1 , y2 ) has been constructed so that (x, y) ∈ V and T (V ) ∩ V = ∅.
By induction it can be shown that for any positive k, T k (V ) ∩ V = ∅. Finally, since
balls are the basis of the standard topology, we may take a ball containing (x, y)
but contained in V to be the neighborhood U .
Next we show that X/Z is not a Hausdorff space. First define sets A = {(x, y)|xy =
c 6= 0, x > 0, y > 0} ∩ ([1, 2] × R) and B = {(x, y)|xy = c 6= 0, x > 0, y >
0} ∩ (R × [1/2, 1]). Note that these sets map to the same set in the quotient. Take
a sequence of points ak = (1, 2k ) ∈ A and bk = (1/2k , 1) each for k ∈ N. Note that
ak ∈ A and ak → (1, 0). Similarly bk ∈ B and bk → (0, 1). Convergence carries
through a continuous map. Even though both sequences ak and bk are equal in the
quotient, the points they converge to are not. Since convergence is non-unique the
space must not be Hausdorff.
Finally we compute π1 . The main ingredient is Van Kampen’s theorem. Take
0
the sets A = {(x, y)|xy = c 6= 0, x < 0, y > 0} ∩ (R × [1/2, 1]) and let A be the
projection to the quotient space such that A is an infinitely long cylinder. Similarly,
for each of the other positive and negative portions of each axis define sets B, C, D,
each of which are cylinders. Each of these sets is open because the complement is
closed in the quotient space. The pairwise intersections of each will either be the
empty set or a portion or a part of an infinitely long cylinder in R3 . In the latter
case the intersection is path connected because we may take a path in X and project
it through the quotient map to get a path in the quotient space. Finally all triple
intersections are empty and so the triple intersections are trivially path connected.
1
2
MATH 151A-MIDTERM EXAM- SOLUTION TO PROBLEM 2.
So we may apply Van Kampen’s theorem. Since the set A is an infinitely long
cylinder π1 (A) = hai ∼
= Z, where a is a loop that wraps once around the cylinder.
Similarly π1 (B) = hbi, π1 (C) = hci, etc. To compute the normal group N in Van
Kampen’s theorem note that for any intersection of fundamental groups which is
non-trivial, such as π1 (A ∩ B), a loop w ∈ π1 (A ∩ B) is on a part of an open
ended-infinitely long cylinder such that iAB (w) = iBA (w). Thus the group N will
always be trivial and we have that π1 (X/Z) = ha, b, c, di = Z ∗ Z ∗ Z ∗ Z.